Kristensmith Taylor

CBSE Class 12 Physics Chapter 12  Atom Multiple Choice Questions And Answers

Question 1. The radius of the second orbit in a hydrogen atom is R. What is the radius in the third orbit?

1. 3R
2. 2.25 R
3. 9R
4. R/3

Answer: 2. 2.25 R

Read and Learn More Important Questions for Class 12 Physics with Answers

W.k.t Radius is n^th orbit

rn = n²R’ → (1)

R = 4R’ → (2) [For 2 Orbit, n= 2]

r₃ = 9R’ → (3) [For 2 Orbit, n= 3]

= \(9 \times \frac{R}{4}\)

∴ r₃ = 2.25 R

Question 2. The wavelength of the first line of the Lyman series is X. The wavelength of the first line in the Balmer series is ______.

1. \(\frac{27}{5} \lambda\)
2. \(\frac{5}{27} \lambda\)
3. \(\frac{9}{2} \lambda\)
4. \(\frac{2}{5} \lambda\)

Answer: 1. \(\frac{27}{5} \lambda\)

⇒ \(\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\)

⇒ \(\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{1}-\frac{1}{2^2}\right)=\frac{3}{4} \mathrm{R} \Rightarrow \mathrm{R}=\frac{4}{3} \lambda\)

So, for Balmer series (n₁ = 2, n₂ = 3)

⇒ \(\frac{1}{\lambda^{\prime}}=\mathrm{R}\left(\frac{1}{4}-\frac{1}{9}\right)\)

⇒ \(\frac{1}{\lambda^{\prime}}=\frac{4}{3 \lambda}\left(\frac{5}{36}\right)\)

∴ \(\lambda^{\prime}=\frac{27 \lambda}{5}\)

Question 3. For the first orbit of hydrogen atom the minimum excitation potential is ______ V.

1. 13.6
2. 3.4
3. 10.2
4. 3.6

Answer: 3. 10.2

Question 4. An electron with an energy of 12.09 eV strikes a hydrogen atom In the ground state and gives its all energy to the hydrogen atom. Therefore hydrogen atoms are excited to _______ stale.

1. Fourth
2. third
3. Second
4. First

Answer: 3. Second

⇒ \(E_n=-\frac{13.6}{n^2}\)

Toral energy gained by hydrogen atoms is

-13.6+12.09 =-1.51 eV

So, \(n^2=\frac{-13.6}{-1.51}=9 \Rightarrow n=3\)

So, second excited state.

Question 5. The ratio of energies of electrons in a second excited state to the first excited slate in H-atom:

1. 1:4
2. 9:4
3. 4:9
4. 4:4

Answer: 3. 4:9

⇒ \(E_n=-13.6 \frac{z^2}{n^2}\) (used it)

For the second excited state n = 3 and the first excited state n = 2

∴ \(\frac{E_2}{E_1}=\frac{4}{9}\)

Question 6. For the first orbit of the hydrogen atom, the minimum excitation potential is ______V.

1. 13.6
2. 3.4
3. 10.2
4. 3.6

Answer: 3. 10.2

Question 7. If the potential energy of the electron in the hydrogen atom is \(\frac{-e^2}{4 \pi \varepsilon_0 r}\), then what is its kinetic energy?

1. \(\frac{\mathrm{e}^2}{8 \pi \varepsilon_0 \mathrm{r}}\)
2. \(\frac{-e^2}{4 \pi \varepsilon_0 r}\)
3. \(\frac{-e^2}{8 \pi \varepsilon_0 r}\)
4. \(\frac{e^2}{4 \pi \varepsilon_0 r}\)

Answer: 1. \(\frac{\mathrm{e}^2}{8 \pi \varepsilon_0 \mathrm{r}}\)

⇒ \(k_n=-\frac{U n}{2}\)

∴ \(k_n=+\frac{e^2}{8 \pi \varepsilon_0 r}\)

Question 8. What is the angular momentum of an electron of Li-alom in n = 5 orbit?

1. 6.625 x 10^-34 Js
2. 5.27 x 10^-34 Js
3. 1.325 x 10^-34 Js
4. 16.56 x 10^-34 Js

Answer: 2. 5.27 x 10^-34 Js

⇒ \(\mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi}\)

⇒ \(\frac{5 \times 6.62 \times 10^{34}}{2 \times 3.14}\)

∴ 5.27 x 10^-34 Js

Question 9. A hydrogen atom absorbs 12.1 eV of energy and gets excited to a higher energy level. How many photons are emitted during the downward transition? Assume during each downward transition, one photon is emitted.

1. 2 or 3
2. 1 or 3
3. 1 or 2
4. 5 or more

Answer: 3. 1 or 2

Question 10. 13.6 eV energy is required to separate a hydrogen atom into a proton and an electron. Compute the orbital radius of the corresponding electron.

1. 5.3 x 10^-11 m
2. 10.6 x 10^-11 m
3. 2.65 x 10^-11 m
4. 1.33 x 10^-11 m

Answer: 1. 5.3 x 10^-11 m

⇒ \(\mathrm{E}=-\frac{\mathrm{e}^2}{8 \pi \varepsilon_0 \mathrm{r}}\)

⇒ \(r=-\frac{e^2}{8 \pi \varepsilon_0 E}\)

By solving r = 5.29 x 10^-11 m

or

⇒ \(r_1=0.529 \frac{n^2}{Z} Å\)

∴  0.529 A

Question 11. What is the shortest wavelength present in the Paschcn scries of spectral lines?

1. 821 nm
2. 6563 A
3. 911 nm
4. 656 mm

Answer: 1. 821 nm

⇒ \(\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\)

⇒ \(\frac{1}{\lambda}=1.097 \times 10^7\left(\frac{1}{9}-\frac{1}{\infty}\right)\)

λ = 8.21 x 10^-7 m

λ = 821 x nm

Question 12. The total energy and kinetic energy of an electron in hydrogen atom arc E and K respectively.

1. K = 2 E
2. K = -E
3. \(K=\frac{E}{2}\)
4. K = E

Answer: 2. K = -E

Question 13. The ionization energy of an electron in the third excited state for a hydrogen atom is _______ eV.

1. 13.6
2. 1.51
3. 0.85
4. 3.4

Answer: 3. 0.85

⇒ \(E_n=-\frac{13.6}{n^2}\)

energy of the third excited state

⇒ \(-\frac{13.6}{16}=-0.8 .5 \mathrm{eV}\)

So Ionisation energy

= + 0.85 eV

Question 14. In the Geiger-Marsden scattering experiment, thin gold foil is used to scatter alpha particles because alpha particles will

1. Not suffer more than one scattering and the gold nucleus is 50 times heavier than the alpha particle.
2. Not suffer more than one scattering and the gold nucleus is lighter than the alpha particle.
3. Not suffer more than a few scattering and the gold nucleus is 25 times heavier than the alpha particle.
4. Suffer more than one scattering and the gold nucleus is 25 times heavier than an alpha particle.

Answer: 1. Not suffer more than one scattering; the gold nucleus is 50 times heavier than the alpha particle.

Atoms Assertion And Reason

For question numbers 1 to 5 two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (1), (2), (3), and (4) as given below.

1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is NOT the correct explanation of A
3. A is true but R is false
4. A is false and R is also false

Question 1. Assertion: The mass of the atom is concentrated in the nucleus.

Reason: The mass of a nucleus can be either less than or more than the sum of the masses of nucleons present in it.

Answer: 3. A is true but R is false

Question 2. Assertion: Bohr’s orbits are also called stationary stales.

Reason: In Bohr’s orbits electron revolves in a fixed path.

Answer: 1. Both A and R are true and R is the correct explanation of A

Question 3. Assertion: A hydrogen atom consists of only one electron but its emission spectrum has many lines.

Reason: The Lyman series is found in the emission spectrum.

Answer: 2. Both A and R are true but R is NOT the correct explanation of A

Question 4. Assertion: For the scattering of a-particles at a large angle only the nucleus of the atom is responsible.

Reason: Nucleus is very heavy in comparison to electrons.

Answer: 1. Both A and R are true and R is the correct explanation of A

Question 5. Assertion: Atom arc is not electrically neutral.

Reason: The number of protons and electrons is different

Answer: 4. A is false and R is also false

CBSE Class 12 Physics Chapter 12  Atoms Short Questions And Answers

Question 1. Show that the radius of the orbit in a hydrogen atom varies as n² where n is the principal quantum number of the atom.

Answer:

Coulomb force: The electrostatic attraction force between the electron and nucleus is _______

⇒ \(\mathrm{F}_{\mathrm{e}}=\frac{\mathrm{kZe}^2}{\mathrm{r}_{\mathrm{n}}^2}\) → (1) \(\left\{\mathrm{F}_{\mathrm{e}}=\frac{\mathrm{k} \mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}^2}\right.\)

Centripetal Force = \(\frac{m v_n^2}{r_n}\)

In equilibrium \(\frac{m v_n^2}{r_n}=\frac{k Z e^2}{r_n^2}\)

⇒ \(\mathrm{v}_{\mathrm{n}}^2=\frac{\mathrm{kZe^{2 }}}{\mathrm{mr}_{\mathrm{n}}}\) → (2)

Then form Bohr’s second postulate \(m v_n r_n=\frac{n h}{2 \pi}\)

∴ \(\mathrm{v}_{\mathrm{n}}=\frac{\mathrm{nh}}{2 \pi \mathrm{mr}_{\mathrm{n}}}\) → (3)

From eq (2) and (3)

⇒ \(\frac{n^2 h^2}{4 \pi^2 m^2 r_n^2}=\frac{k Z e^2}{m r_n}\)

⇒ \(r_n=\left(\frac{h^2}{4 \pi^2 e^2 k m}\right) \frac{n^2}{Z}\)

Put the value, π, k, h, m, e

∴ \(r_n=0.529 \times 10^{-8} \times \frac{n^2}{Z} \mathrm{~cm}=0.529 \times \frac{n^2}{Z}Å\)

Question 2. How does one explain, Bohr’s second postulate of quantization of orbital angular momentum using the de Broglie’s hypothesis?

Answer:

The behavior of particle waves can be viewed as analogous to the waves traveling on a siring. Particle waves can lead to standing waves held under resonant conditions. When a stationary string is plucked, several wavelengths are excited but only those wavelengths survive which form a standing wave in the string.

Thus, in a string standing waves arc formed only when the total distance traveled by f a wave is an integral number of wavelengths. Hence, for any electron moving in a circular orbit of radius r_n, the total distance is equal to the circumference of the orbit. 2πr_n .

2πr_n = n → (1)

Here, λ is de Broglie wavelength.

We know

λ = h/p

or λ = h/mv_n → (2)

Where mv_n is the momentum of an electron revolving in the nth orbit,

From equation (2) and equation (1) we get.

2πr_n = nh/mv_n

L = mv_n r_n = nh/2π

Hence, de Broglie’s hypothesis successfully proves Bohr’s second postulate.

Question 3. Given the value of the ground state energy of hydrogen atom as -13.6 eV, find out its kinetic and potential energy in the ground and second excited states.

Answer:

Ground stale energy of hydrogen atom as -13.6 eV,

E_n = Total energy = \(-\frac{13.6 \mathrm{eV}}{n^2}\)

K.e = -T.E and P.e = 2 T.E

For ground stale n = 1, then

⇒ \(\mathrm{T} . \mathrm{E} .=\frac{-13.6 \mathrm{eV}}{(1)^2}=-13.6 \mathrm{eV}\)

K.E. = +13.6 eV

P.E. = -27.2 eV

For second excited state n = 3, then

∴ \(\text { T.E. }=\frac{-13.6}{9}=-1.51 \mathrm{eV}\)

K.E. = +1.51 eV

P.E. = -3.02 eV

Question 4. Given the ground stale energy E₀ = -13.6 cV and Bohr radius r₀ = 0.53 Å. Find out how the dc Broglie wavelength associated with the electron orbiting in the ground stale would change when it jumps into the first excited state.

Answer:

Given ground slate energy E₀ = – 13.6 eV

Energy in the first excited state = \(E_1=\frac{-13.6 \mathrm{eV}}{(2)^2}=-3.4 \mathrm{eV}\)

∴ \(\lambda=\frac{h}{\sqrt{2 \mathrm{mE}_1}}=\frac{6.63 \times 10^{-34}}{\sqrt{2 \times\left(9.1 \times 10^{-31}\right)\left(3.4 \times 1.6 \times 10^{-19}\right)}}\)

∴ \(\lambda=\frac{6.63 \times 10^{-34}}{9.95 \times 10^{25}}=6.66 \times 10^{-10} \mathrm{~m} \simeq 6.7 \times 10^{-10} \mathrm{~m}\)

Question 5. A hydrogen atom initially in the ground slate absorbs a photon with energy 12.5 eV. Calculate the longest wavelength of the radiation emitted and identify the series to which it belongs.

Answer:

We Know, \(\frac{1}{\lambda_{\max }}=R\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)\)

The energy of the incident photon = 12.5 eV

Energy of ground stale = -13.6 eV

∴ The energy of the hydrogen atom after absorption of a photon can be -1.1 eV

This means that electrons can go to the excited state n_i = 3. Now it emits photons of maximum wavelength on going to n_f = 2

⇒ \(\frac{1}{\lambda_{\max }}=\mathrm{R}\left\{\frac{1}{2^2}-\frac{1}{3^2}\right\}\)

⇒ \(\lambda_{\text {mat }}=\frac{36}{5 \mathrm{R}}=\frac{36}{5 \times 1.1 \times 10^7}=6.545 \times 10^{-7} \mathrm{~m}=6.545 \mathrm{~A}^{\circ}\)

It belongs to the Balmer Series.

Question 6. When is the H_α line in the emission spectrum of the hydrogen atom for the Balmer series obtained? Calculate the frequency of the photon emitted during this transition.

Answer: 

H_α is a specific deep-red visible spectral line in the Balmer series with a wavelength of 656.28 nm, it occurs when a hydrogen electron transits from its 3^rd to 2^nd lowest energy level.

This transition produces an H-alpha photon and the 1^st line of the Balmer series.

⇒ \(\frac{1}{\lambda}=R\left[\frac{1}{n_f^2}-\frac{1}{n_i^2}\right]\)

⇒  \(\mathrm{R}\left[\frac{1}{2^2}-\frac{1}{3^2}\right] \Rightarrow \frac{1}{\lambda}=1.097 \times 10^7\left[\frac{5}{36}\right]\)

∴ \(v=\frac{c}{\lambda}=\frac{3 \times 10^8 \times 1.097 \times 10^7 \times 5}{36}\) = 4.57 x 10¹⁴ Hz

Question 7. The energy levels of a hypothetical atom are shown below. Which of the shown transitions will result in the emission of a photon of wavelength 275 nm?

Answer:

Using relation, \(\mathrm{E}=\frac{1242 \mathrm{eV}}{\lambda(\text { in } \mathrm{nm})}\)

Here, = 275nm, \(\mathrm{E}=\frac{1242}{275}=4.5 \mathrm{eV}\)

This energy of photon exists corresponding to ‘B’.

Question 8. State Bohr’s postulate of hydrogen atom which successfully explains the emission lines in the spectrum of hydrogen atom.

Use the Rydberg formula to determine the wavelength of the line.

[Given : Rydberg constant R = 1.03 x 10⁷ m^-1]

Answer:

According to Bohr’s postulate when an e^– jumps from one orbit to another, the energy difference between them is emitted in the form of energy i.e. as a photon which shows the emission lines in the spectrum of hydrogen atoms.

⇒ \(E_i-E_f=h \nu=\frac{h c}{\lambda}\)

Now, the Rydberg formula for the Balmer series is

⇒ \(\frac{1}{\lambda}=R\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)\) where R = Rydberg constant = 1 .09 x 10⁷ m^-1

The H_α– line of the Balmer series is obtained when an e jumps to the second orbit (nf = 2) from
the third orbit (n. = 3).

Further, \(\frac{1}{\lambda}=1.03 \times 10^7\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=6.99 \times 10^{-7}=699 \mathrm{~nm}\)

‘λ’ lies in the visible region.

Question 9. Using Rutherford’s model of the atom, derive the expression for the total energy of a tin electron in the hydrogen atom. What is the significance of total negative energy possessed by the electron?

Answer:

According to Rutherford’s nuclear model of the atom, the electrostatic force of attraction F_e between the revolving electron and the nucleus provides the centripetal force (F_c).

Thus, F_c = F_e

∴ \(\frac{m v^2}{r}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{e^2}{r^2}\) (∵ z = 1)

Thus the relation between the orbit radius and the e- velocity is, \(\mathrm{r}=\frac{\mathrm{e}^2}{4 \pi \varepsilon_0 \mathrm{mv}^2}\)

The K.E. (K) and electrostatic potential energy (U) of electrons in hydrogen atoms are

⇒ \(K=\frac{1}{2} m v^2=\frac{e^2}{8 \pi \varepsilon_0 r}\)

and \(\mathrm{U}=\frac{-\mathrm{e}^2}{4 \pi \varepsilon_0 \mathrm{r}}\) (-ve sign shows, that the nature of force is attractive)

Thus, the total rnech. energy E of an e^– is,

∴ \(\mathrm{E}=\mathrm{K}+\mathrm{U}=\frac{\mathrm{e}^2}{8 \pi \varepsilon_0 \mathrm{r}}-\frac{\mathrm{e}^2}{4 \pi \varepsilon_0 \mathrm{r}}=\frac{-\mathrm{e}^2}{8 \pi \varepsilon_0 \mathrm{r}}\)

Total energy (E) is -ve which shows that e^– is bound to the nucleus, if (E) were +ve then e- would leave the the atom.

Question 10.

1. Stale Bohr’s postulate of the hydrogen atom gives the relationship to the frequency of emitted photons in a transition.
2. An electron jumps from the fourth to the first orbit in an atom. How many maximum number of spectral lines can be emitted by the atom? To which series do these lines correspond?

Answer:

1. According to Bohr’s theory, energy is quantized i.c. for each orbital the corresponding energy is
given as

ΔE = hv

2. There will be 6 spectral lines and they correspond to.

Max. number of spectral lines = \(\frac{n(n-1)}{2}\)

For the fourth orbit n = 4

Then, Max. possible lines = \(\frac{4(4-1)}{2}\)

A-Lyman Series, B-Balmcr series, C-Paschcn series

Question 11. A 12.5 eV electron beam is used to excite a gaseous hydrogen atom at room temperature. Determine the wavelengths and the corresponding series of the lines emitted.

Answer:

When an electron beam of 12.5 eV is used to excite hydrogen, then the probable transition is from n=1 to n=3 (i.e. E₃-E₁ = -1.5l-(-l 3.6) = 12.09e V)

On de-excitation, the electron may jump by following the ways

3 → I, 3 → 2, 2 → 1

Possible wavelength and their corresponding series of lines emitted are

1. From n=3 to n = 1, \(\frac{1}{\lambda}=R\left[\frac{1}{1^2}-\frac{1}{3^2}\right] \Rightarrow \frac{1}{\lambda}=R\left[1-\frac{1}{9}\right]=R\left(\frac{8}{9}\right)\)

or \(\lambda=\frac{9}{8 \mathrm{R}}=1.026 \times 10^{-7} \mathrm{~m}=102.6 \mathrm{~nm}\)

It belongs to the Lyman series.

2. From n = 3 to n = 2,

⇒ \(\frac{1}{\lambda}=R\left[\frac{1}{2^2}-\frac{1}{3^2}\right]=R\left[\frac{1}{4}-\frac{1}{9}\right] \Rightarrow \frac{1}{\lambda}=R\left[\frac{5}{36}\right]\)

∴ \(\lambda=\frac{36}{5 \mathrm{R}}=6.563 \times 10^{-7} \mathrm{~m}=656.3 \mathrm{~nm}\)

It belongs to the Balmer series.

3. From n = 2 to n = 1,

⇒ \(\frac{1}{\lambda}=R\left[\frac{1}{1^2}-\frac{1}{2^2}\right]=R\left[1-\frac{1}{4}\right] \Rightarrow \frac{1}{\lambda}=\frac{3 R}{4}\)

A = 1.215 x 1 0^-7m = 121.5 nm

It belongs to the Lyman series.

Question 12.

1. In A Geiger-Marsden Experiment. Find the distance of the closest approach to the gold nucleus (mass no. = 79)m of a 7.7 Me V α-particle before it comes momentarily to rest and reverses its direction.
2. Plot a graph between several scattered α-particles detected in the gold foil experiment and the angle of scattering. What is the main assumption in plotting this graph?

Answer:

K.E. = P.E

⇒ \(K E_{\alpha}=\frac{k q_1 q_2}{r}\) where q₁ = 79e, q₂ = 2e

So r = \(\mathrm{r}=\frac{\mathrm{kq}_1 \mathrm{q}_2}{\mathrm{KE}_\alpha}=\frac{9 \times 10^9 \times\left(79 \times 1.6 \times 10^{-19}\right)\left(2 \times 1.6 \times 10^{-19}\right)}{7.7 \times 10^6 \times 1.6 \times 10^{-19}}=\frac{18 \times 79 \times 1.6}{7.7} \times 10^{-16} \mathrm{~m}\)

r = 295.4 x 10^-16 = 29.54 m = 30 fm

0 = Scattering angle

N= No. of α-Particles

The plot is Schematic and not according to scale.

Question 13.

1. state Bohr’s quantization condition for defining stationary orbits. How does de Broglie’s hypothesis explain the stationary orbits?
2. Find the relation between the three wavelengths λ_1, λ₂, λ₃ and from the energy level diagram shown below.

Answer:

1. Bohr’s quantization condition: The electron can revolve around the nucleus only in those circular orbits in which the angular momentum of an electron is an integral multiple of \(\frac{h}{2 \pi}\), where h is Planck’s constant. \(\mathrm{mvr}=n \frac{\mathrm{h}}{2 \pi}\) [m = mass, v = velocity of electron]

Their circular orbits are stationary, de Broglie hypothesis: de-Broglie interprets Bohr’s 2^nd postulate in terms of the wave nature of the electron. According to him- The electron can revolve in certain stable orbits for which the angular momentum is some integral multiple of h/2π.

Mathematically, \(2 \pi r_n=\frac{n h}{m v}\)

Therefore, \(\mathrm{mvr}_{\mathrm{n}}=\frac{\mathrm{nh}}{2 \pi}\)

2. \(E_{C B}=\frac{h c}{\lambda_1} \Rightarrow E_{B A}=\frac{h c}{\lambda_2} \Rightarrow E_{C A}=\frac{h c}{\lambda_3}\)

here E_CB = Energy gap between levels B and C

E_CA = Energy gap between levels A and C

E_BA= Energy gap between levels A and B

E_CA = E_CB + E_BA

∴ \(\frac{h c}{\lambda_3}=\frac{h c}{\lambda_1}+\frac{h c}{\lambda_2} \Rightarrow \frac{1}{\lambda_3}=\frac{1}{\lambda_1}+\frac{1}{\lambda_2} \Rightarrow \lambda_3=\frac{\lambda_1 \lambda_2}{\lambda_1+\lambda_2}\)

Question 14. State Bohr’s postulate to explain stable orbits in a hydrogen atom. Prove that the speed with which the electron revolves in nlh orbit is proportional to (1/n).

Answer:

Bohr’s postulate:

Electron revolves around the nucleus in those orbits for which the angular momentum is an
integral multiple of h/2π.

∴ \(\mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi}\)

Speed of e^– in n^th orbit

For hydrogen radius of nth orbit is given by = \(r_n=\frac{\varepsilon_0 n^2 h^2}{\pi m c^2}\)

Form Bohr’s postulate

⇒ \(m v_n r_n=\frac{n h}{2 \pi}\)

⇒ \(m v_n\left(\frac{\varepsilon_0 n^2 h^2}{\pi m e^2}\right)=\frac{n h}{2 \pi}\)

∴ \(\mathrm{V}_{\mathrm{n}}=\frac{\mathrm{c}^2}{2 \varepsilon_0 h n} \Rightarrow \mathrm{V}_{\mathrm{n}} \propto \frac{1}{\mathrm{n}}\)

Question 15. A narrow beam of protons, each having 4.1 MeV energy is approaching a sheet of lead (Z = 82). Calculate :

1. The speed of a proton in the beam, and
2. The distance of its closest approach

Answer:

Energy of proton = 4. 1 MeV, Z = 82

1. Speed of a proton

⇒ \(\frac{1}{2} m_p v_p^2=4.1 \times 10^6 \times 1.6 \times 10^{-19}\)

∴ \(v_p^2=\frac{2 \times 4.1 \times 1.6 \times 10^{-13}}{1.67 \times 10^{-27}}=7.85 \times 10^{14}\)

V_p = 2.8 x 10⁷ m/s

2. Distance of closest approach

⇒ \(K. E=\frac{k e(z e)}{r_0}\)

⇒\(r_0=\frac{K Z e^2}{r_0}=\frac{9 \times 10^{9} \times 82 \times\left(1.6 \times 10^{-19}\right)^2}{4.1 \times 10^6 \times 1.6 \times 10^{-19}}=288 \times 10^{-16}\)

∴r₀ = 2.9 x 10^-14 m^-1

CBSE Class 12 Physics Chapter 12  Long Questions And Answers

Question 1. According to the third postulate of Bohr’s model, when an atom makes a transition from the higher energy stale with quantum number nj to the lower energy stale with quantum number n_f(n_f <n_i), the difference of the energy is carried away by the photon of frequency such that hv = En_i-En_f.

Since both ni and nf are integers, this immediately shows that in transitions between the different atomic levels, light is radiated in various discrete frequencies. For the hydrogen atom spectrum, the Balmcr formula corresponds to n_f = 2 and n_i = 3, 4, 5, etc. This result of Bohr’s model suggested the presence of other scries spectra for hydrogen atoms – those corresponding to the transitions resulting from n_f = 1 and n_i = 2, 3, etc: and n_f = 3 and n_i = 4, 5, etc., and so on. Such series were identified in the course of spectroscopic investigations and are known as the Lyman, Balmer, Paschen, Brackett, and Pfund series. The electronic transitions corresponding to this series are shown

(1). The total energy of an electron in an atom in an orbit is -3.4 eV. Its kinetic and potential energies are respectively

1. 3.4 eV, 3.4 eV
2. -3.4 eV, -3.4 eV
3. -3.4eV,-6.8 eV
4. 3.4 eV, -6.8 eV

Answer: 4. 3.4 eV, -6.8 eV

(2). Given the value of the Rydberg constant is 10⁷ m^-1, the wave number of the last line of the Balmer series in the hydrogen spectrum will be

1. 0.5 x 10⁷ m^-1
2. 0.25 x 10⁷ m^-1
3. 2.5 x 10⁷ m^-1
4. 0.025 x 10⁴ m^-1

Answer: 2. 0.25 x 10⁷ m^-1

(3). The ratio of the wavelength of the last line of Balmcr scries and the last line of the Lyman series

1. 0.5
2. 2
3. 1
4. 4

Answer: 4. 1

(4). The wavelength of Balmer scries lies in

1. Ultraviolet region
2. Infrared region
3. Far infra-red region
4. Visible region

Answer: 4. Visible region

Question 2. Neutrons mid protons tire identical particles in the sense that their masses are nearly the same and the force called nuclear force, does not distinguish between them, Nuclear force is the strongest force. The stability of the nucleus is determined by the neutron Proton ratio or mass defect or packing fraction, The Shape of the nucleus is calculated by quadruple moment and the spin of the nucleus depends on an even or odd mass number. The volume of the nucleus depends on the mass number. The whole mass of the atom (nearly 99%) is centered at the nucleus.

(1). The correct statement about the nuclear force is as follows:

1. Charge independent
2. Short range force
3. Nonconservative force
4. All of these

Answer: 4. All of these

(2). The range of a nuclear force is the order of:

1. 2 x 10^-10 m
2. 1.5 x 10^-20 m
3. 1.2 x 10^-4m
4. 1.4 x 10^-15 m

Answer: 4. 1.4 x 10^-15 m

(4). A force between two protons is the same as the force between a proton and a neutron. The nature of the force is:

1. Electrical force
2. Weak nuclear force
3. Gravitational force
4. Strong nuclear force

Answer: 1. Electrical force

CBSE Class 12 Physics Chapter-3 Current Electricity Multiple Choice Questions And Answers

Question 1. In the circuit given below P≠R and the reading of the galvanometer is the same with switch S open or closed. Then:

1. I_Q=I_R
2. I_R=I_G
3. I_P=I_G
4. I_Q+I_G

Answer: 4. I_Q+I_G

Read and Learn More Important Questions for Class 12 Physics with Answers

Question 2. Two wires A and B of the same material having length in the ratio 1:2 and diameter in the ratio 2:3 are connected in series with a battery. The ratio of the potential differences (V_A/V_B) across the two wires respectively is:

1. 1/3
2. 3/4
3. 4/5
4. 9/8

Answer: 4. 9/8

In Series, I → Same.

V = IR

⇒ \(\frac{V_A}{V_B}=\frac{R_A}{R_B}\)

⇒ \(R=\frac{\rho l}{A}=\frac{4 \rho l}{\pi d^2}\) [4, ρ, l, π – Constant]

∴ \(\frac{R_A}{R_B}=\frac{l_A}{l_B} \times\left(\frac{d B}{d A}\right)^2 = \frac{V_A}{V_{B}}=\frac{1}{2} \times\left(\frac{3}{2}\right)^2 = \frac{V_A}{V_{B}}=\frac{9}{8}\)

CBSE Class 12 Physics Chapter-3 Important Questions

Question 3. Infinity resistance in a resistance box has:

1. A resistance of 105Ω
2. A resistance of 107Ω
3. A resistor of resistance
4. A gap only

Answer: 4. A gap only

Question 4. A battery of 15V and negligible internal resistance is connected across a 50Ω resistor. The amount of energy dissipated as heal to the resistor in one minute is:

1. 122 J
2. 270 J
3. 420 J
4. 720 J

Answer: 2. 270 J

Use \(H=\frac{V^2}{R} t\)

Question 5. In a potentiometer experiment, the balancing length of a cell is 120cm. When the cell is shunted with a 1 Ω resistance, the balancing length becomes 40 cm. The internal resistance of the cell is:

1. 10Ω
2. 7Ω
3. 3Ω
4. 2Ω

Answer: 4. 2Ω

Question 6. Two students A and B calculate the charge flowing through a circuit. A concludes that 300 C of charge flows in 1 minute. B concludes that 3.125 x 10¹⁹ electrons flow in 1 second. If the current measured in the circuit is 5 A, then the correct calculation is done by:

1. A
2. B
3. Both A and Both
4. Neither A nor Both

Answer: 3. Both A and Both

⇒ \(I=\frac{q}{t}\) [for A]

∴ \(I=\frac{ne}{t}\) [for B]

Question 7. The resistances of two wires having the same length and the same area of cross-section arc 2Ω and 8 Ω respectively. If the resistivity of 2Ω wire is 2.65 x 10^-8 m then the resistivity of 8Ω wire is:

1. 10.60 x 10^-4 Ω m
2. 8.32 x 10^-8 Ω m
3. 7.61 x 10^-8 Ωm
4. 5.45 x 10^-8 Ω m

Answer: 1. 10.60 x 10^-4 Ω m

Use \(R=\frac{\rho l}{A}\)

⇒ \(\frac{\rho_2}{\rho_1}=\frac{R_2}{R_1}\)

⇒ \(\rho_2=\frac{8}{2} \times 2.65 \times 10^{-8}\)

∴ \(\rho_2=10.60 \times 10^{-8} \Omega \mathrm{m}\)

CBSE Class 12 Physics Chapter-3 Important Questions

Question 8. The given figure shows an I-V graph of a copper wire of length L and an area of cross-section A. The slope of the curve becomes

1. Less if the length of the wire is increased
2. More if the length of the wire is increased
3. More if a wire of steel of the same dimension is used
4. If the temperature of the wire is increased

Answer: 1. Less if the length of the wire is increased

⇒ \(\frac{I}{V}=\frac{1}{R}=\frac{A}{\rho l}\)

Slope = \(\frac{A}{\rho l}\)

l slope ↓

Question 9. When a potential difference V is applied across a conductor at temperature T. the drift velocity of the electrons is proportional to

1. T
2. √T
3. V
4. √V

Answer: 3. V

∴ \(v_d=\frac{c V}{m l} \bar{\tau}\)

∴ V_d ∝ V

Question 10. A cell supplies a current of 0.9 A through a 20 resistor and a current of 0.3 A through a 70 resistor. What is the internal resistance of cells?

1. 0.5Ω
2. 1.0Ω
3. 1.2Ω
4. 2.0Ω

Answer: 1. 0.5Ω

E = I₁ (R₁ + r) ⇒ E = 0.9(2 + r) →(1)

E = I₂ (R₂ + r) ⇒ E = 0.3 (7 + r) →(2)

From equation (1) and (2)

0.9 (2 + r) = 0.3 (7 + r)

6 + 3r = 7 + r

2r = 1 ⇒ r = 0.5 Ω

Question 11. At what temperature would the resistance of a copper conductor be double its resistance at 0°C? Given temperature coefficient of resistance for copper is 3.9 x 10^-3 °C^-1.

1. 256.4°C
2. 512.8°C
3. 100°C
4. 256.4 K

Answer: 1. 256.4°C

R₁ = R₀ (1 +∝ t)

2R₀ = R₀(1 + 3.9 x 10^-3 x t) ⇒ \(\mathrm{t}=\frac{2-1}{0.0039}\) = 256.4°C

Question 12. A student is asked to connect four cells of emf e each and internal resistance r each in a series of helping conditions. By mistake, he oppositely connects one cell. What will be the effective EMF and effective internal resistance?

1. 4ε, 2r
2. 2ε, 4r
3. 3ε, 2r
4. 4ε, 4r

Answer: 2. 2ε, 4r

Question 13. In a hydrogen atom, the electron is moving in a circular orbit of radius 5.0 x 10^-11 m with a constant speed of 2 x 10⁶ m/s. The electric current formed due to the motion of electrons is _______.

1. 1.12 A
2. 1.02 A
3. 1.02 mA
4. 1.12 mA

Answer: 3. 1.02 mA

∴\(I=\frac{e V}{2 \pi r} ⇒ \) \(I=\frac{1.6 \times 10^{-19} \times 2 \times 10^6}{2 \times 3.14 \times 5 \times 10^{11}} \mathrm{~A}\) ⇒ I = 1.02 mA

CBSE Class 12 Physics Chapter-3 Important Questions

Question 14. A voltmeter of very high resistance is joined in the circuit as shown in the figure. The voltage shown by the voltmeter will be _________.

1. 5 V
2. 2.5 V
3. 10 V
4. 7.5 V

Answer: 1. 5 V

⇒ \(I=\frac{10}{5+5}=1 \mathrm{~A}\)

∴ V = IR = 1 x 5 = 5V

Question 15. The figure shows a part of a closed circuit. If the current flowing through it is 2A. What will be the potential difference between points B and A, V_B– V_A is:

image

1. +2V
2. +1 V
3. -1V
4. -2V

Answer: 4. -2V

Apply KVL

image

⇒ \(2 \times \frac{1}{4}+1+2 \times \frac{1}{4}=V_A-V_B\)

⇒ 2 = V_A-V_B

∴ V_B-V_A = -2V

Question 16. If the current in an electric bulb increases by 2%. what will be the change in the power of a bulb? (Assume that the resistance of the filament of a bulb remains constant).

1. Decreases by 2%
2. Decreases by 4%
3. Increases by 2%
4. Increases by 4%

Answer: 4. Increases by 4%

P = I²R

⇒ \(\frac{\Delta \mathrm{P}}{\mathrm{P}} \times 100=2 \frac{\Delta \mathrm{I}}{\mathrm{I}} \times 100\), R constant

⇒ 2 x 2

⇒ \(\frac{\Delta P}{P} \times 100=4\)

So power will increase by 4%.

Important Questions in Physics Class 12 – Current Electricity

Question 17. Two bulbs of 220 V and 100 W arc first connected in parallel and then in series with a supply of 220 V. Total power in both cases will be ______ respectively.

1. 100 W. 50 W
2. 50 W. 100 W
3. 200 W. 50 W
4. 50 W. 200 W

Answer: 3. 200 W. 50 W

In parallel → P = P₁ + P₂

P= 200 W

∴ Series \(\frac{1}{P}=\frac{1}{P_1}+\frac{1}{P_2}\)

So, P = 50 W

Question 18. Kirchhoff’s junction rule represents ________.

1. Conservation of energy
2. Conservation of linear momentum
3. Conservation of angular momentum
4. Conservation of charge

Answer: 4. Conservation of charge

Question 19. The device has powder P and voltage ‘V’. The connecting wires from the power station to the device have a finite resistance RC. The power dissipated in the connecting wires PC.

1. \(\frac{V^2 R_C}{P}\)
2. \(\frac{\mathrm{PR}_{\mathrm{C}}^2}{\mathrm{~V}}\)
3. \(\frac{P^2 R_C}{V^2}\)
4. \(\frac{V R_C}{P^2}\)

Answer: 3. \(\frac{P^2 R_C}{V^2}\)

Current following

∴ \(I=\frac{P}{V}\)

So power loss across the wire, Pe = I²RC

∴ \(P_e=\frac{P^2 R_e}{V^2}\)

Question 20. Dimension of mobility (μ) is _______.

1. M^-1T²A^-1
2. M¹L³T^-3A^-2
3. M¹L³T^-4A^-1
4. M¹L⁴T^-3A^-1

Answer: 1. M^-1T²A^-1

Important Questions in Physics Class 12 – Current Electricity Assertion And Reason

For question numbers 1 to 4 two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (1). (2). (3) and (4) as given below.

1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is NOT the correct explanation of A
3. A is true but R is false
4. A is false and R is also false

Question 1. Assertion: The total current entering a circuit is equal to that leaving it by Kirchhoff’s law.

Reason: It is based on the conservation of charge.

Answer: 1. Both A and R are true and R is the correct explanation of A

Question 2. Assertion: The terminal potential of a cell is always less than its emf.

Reason: Potential drop due to internal resistance of cell increases the terminal potential difference.

Answer: 4. A is false and R is also false

Question 3. Assertion: The connecting wires are made of copper.

Reason: The electrical conductivity of copper is high.

Answer: 1. Both A and R are true and R is the correct explanation of A

Question 4. Assertion: There is no current in the metal in the absence of an electric field.

Reason: The motion of free electrons occurs randomly.

Answer: 1. Both A and R are true and R is the correct explanation of A

Class 12 Physics Current Electricity Short Answer Questions

Question 1. The variation of the drift velocity (v_d) of electrons in two copper wires A and B of different lengths versus the potential difference (V) applied across their ends

1. What does the slope of the line represent?
2. Which one of the two wires is longer?

Answer:

V_d = μE, V = El

⇒ \(\mathrm{v}_{\mathrm{d}}=\frac{\mu \mathrm{V}}{l}\)

⇒ \(v_{\mathrm{d}}=\left(\frac{\mu}{l}\right) \mathrm{V}\)

compare with y = mx

1. Slope = \(\) = mobility of charge carriers
2. For constant potential difference (V_d)_A > (V_d)_B

That’s why B has a longer wire.

Question 2. Two wires X and Y of the same material and equal lengths having areas of cross-section A and 2A respectively, are connected in parallel across an ideal battery of emf E. What is the ratio of current density (J_X/J_Y) in them?

Answer:

⇒ \(J=\frac{I}{A}=\frac{E}{R A}\)

⇒ \(\mathrm{J}=\frac{\mathrm{E}}{\rho \ell} \times \frac{\mathrm{A}}{\mathrm{A}}\) (\(\mathrm{R}=\frac{\rho \mathrm{l}}{\mathrm{A}}\))

∴ \(J=\frac{E}{\rho l}\)

So, for both wires, E, ρ, and l are the same

∴ \(\frac{\mathrm{J}_x}{\mathrm{~J}_y}=1: 1\)

Question 3. When 5V potential difference is applied across a wire of length 0. 1 m. the drift speed of electrons is 2.5 x 10^-4 m/s. If the electron density in the wire is 8 x 10²⁸m^-3 calculate the resistivity of the material of the wire.

Answer:

Given: V = 5V

Length of wire l = 0.1m

v_d = 2.5 x 10^-4 m/s

Electron density n = 8 x 10²⁸ m^-3

We know the drift velocity and current are related by the formula

i = ne A v_d → (1)

Where n is the electron density, and e is the charge on an electron. A is an area of the cross-section and v_d is drift velocity.

Also, i = V/R and R = ρl/A where p is resistivity. l is the length of the conductor

so, i = VA/ρl → (2)

comparing equation (1) and (2) we get.

VA / ρl = ne A V_d

or ρ = V / ne l v_d

Put values of all.

ρ = 5/0.1 x 8 x 10²⁸ x 1.6 x 10^-19 x 2.5 x 10^-4

ρ = 1.56 x l0^-5 ohm-meter

Question 4. A battery of emf 12 V and internal resistance 2 Ω is connected to a 4 Ω resistor as shown in the figure.

1. Show that a voltmeter when placed across the cell and the resistor, in turn, gives the same reading.
2. To record the circuit’s voltage and current, why is the voltmeter placed in parallel and the ammeter in series in the circuit?

Answer:

1. When the voltmeter is connected across the cell

I = E/(R + r ) = 12 /(2 + 4) = 12/6 = 2 A

V₁= E-Ir = 12- (2 x 2 ) = 8 V

When the voltmeter is connected across the resistor

V₂ = ER /(r + R) = ( 12 x 4)/(4 + 2) = 12 x 4/6 = 8 V

So, v₁ = v₂

2. A Voltmeter is used to measure the potential difference across two points in a circuit since the voltage in the branches remains the same in a parallel connection. also the resistance of the voltmeter is very high due to which a very small current flows through the voltmeter so it is connected in parallel to measure the voltage.

An ammeter is used to measure the current flowing through a component/circuit. Since the current remains the same in series connections and also the resistance of an ammeter is very small it doesn’t affect the current to be measured. So an ammeter is connected in series to measure current.

Question 5. Two cells of EMFs 1.5 V and 2.0 V having internal resistance 0.2 Ω. and 0.3 Ω respectively are connected in parallel. Calculate the cnif and internal resistance of the equivalent cell.

Answer:

Given E₁ = 1.5 V and r₁ = 0.2 Ω

E₂ = 2 V and r₂ = 0.3 Ω

Equivalent emf = \(\frac{E_1 r_2+E_2 r_1}{r_1+r_2}=\frac{1.5 \times 0.3+2 \times 0.2}{0.5}=\frac{0.85}{0.5}=1.7 \mathrm{~V}\)

∴ Equivalent resistance = \(\frac{r_1 r_2}{r_1+r_2}=\frac{0.2 \times 0.3}{0.2+0.3}=\frac{0.06}{0.5}=0.12 \Omega\)

r_eq, = 0.12 Ω

Question 6.

1. Define the term ‘relaxation time’ in a conductor.
2. Define the mobility of a charge carrier. What is its relation with relaxation lime?

Answer:

1. Relaxation time is the average time interval between two successive collisions of an electron in a conductor when current flows.
2. Mobility of a charge carrier is defined as the drift velocity of the charge carrier per unit electric field i.e μ = v_d/E = eτ/m

Question 7. A metal rod of square cross-sectional area A having length ( has current 1 flowing through it and a potential difference of V volt is applied across its ends. Now the rod is cut parallel to its length into two identical pieces and joined as shown in Figure 2, What potential difference must be maintained across the length 2l so that the current in the rod is still 1?

Answer:

As, \(V=I R=\frac{I_\rho l}{A}\)

Now, \(\frac{V_{\text {new }}}{V}=\frac{l_{\text {new }}}{A_{\text {new }}} \times \frac{A}{l}=\frac{2}{1/2}\) OR V_new = 4V

Now 4V potential difference must be maintained across the length 2l to maintain the same current in the rod.

Question 8. Two wires one of copper and the other of manganin have the same resistance and equal length. Which wire is thicker and why?

Answer:

Manganin is an alloy of Cu with manganese and nickel. Since manganese and nickel have a resistivity greater than copper, pure copper has lower resistivity as compared to alloy manganin. For the same resistance and equal length, manganin wire is thicker than copper.

∴ \(R=\rho \frac{l}{A} \Rightarrow \rho=\frac{R A}{l}\) (∵ \(\rho \propto A\))

Question 9. The plot of the variation of potential difference across a combination of three identical cells in series, versus current is shown below. What is the emf and internal resistance of each cell?

Answer:

V = Terminal voltage across cell combination

The terminal voltage across a cell can be obtained by subtracting the potential drop across the internal resistance of the cell from the emf of the cell.

V = E – IR

When I = 0, = V = E

From graph it is found that when I = 0, V = 6V ⇒ E = 6V

As I = 1A, V = 0 from graph

As V = E-Ir ⇒ 0 = 6-1.r => 6 = 1.r

r = 6 Ω

As 3 identical cells are connected in series so EMF and internal resistance of each cell are 2V and 2 Ω respectively.

Question 10.

1. Derive an expression for the drift velocity of free electrons.
2. How does the drift velocity of electrons in a metallic conductor vary with an increase in temperature? Explain.

Answer:

1. Drift velocity is defined as the average velocity with which free electrons in a conductor drift in a direction opposite to the direction of the applied electric field. When a conductor is subjected to an electric field E. each electron experiences a force.

⇒ \(\vec{F}=-\mathrm{e} \vec{E}\)

and acquires an acceleration

⇒ \(\mathrm{a}=\frac{\overrightarrow{\mathrm{F}}}{\mathrm{m}}=\frac{-\mathrm{e} \overrightarrow{\mathrm{E}}}{\mathrm{m}}\) → (1)

Here m = mass of the electron, e = charge. E = electric field. The average time difference between two consecutive collisions is known as the relaxation time of an electron.

⇒ \(\tau=\frac{\tau_1+\tau_2+\ldots \ldots+\tau_n}{n}\) → (2)

As v = u + at (from equations of motion)

The drift velocity v_d is defined as

⇒ \(\vec{v}_{\mathrm{d}}=\frac{\vec{v}_1+\vec{v}_2+\ldots+\vec{v}_n}{n}\)

⇒ \(\overrightarrow{\mathrm{v}}_{\mathrm{d}}=\frac{\left(\overrightarrow{\mathrm{u}}_1+\overrightarrow{\mathrm{u}}_2+\ldots .+\overrightarrow{\mathrm{u}}_{\mathrm{n}}\right)+\mathrm{a}\left(\tau_1+\tau_2+\ldots .+\tau_{\mathrm{n}}\right)}{\mathrm{n}}\)

⇒ \(\vec{v}_d=0+\frac{a\left(\tau_1+\tau_2+\ldots .+\tau_n\right)}{n}\)

(∵ average thermal velocity = 0)

∴ \(\overrightarrow{\mathrm{v}}_{\mathrm{d}}=0+\mathrm{at}\)

⇒ \(\overrightarrow{\mathrm{v}}_{\mathrm{d}}=-\left(\frac{\mathrm{e} \overrightarrow{\mathrm{E}}}{\mathrm{m}}\right) \tau\) ⇒ (\(\left|\vec{v}_{\mathrm{d}}\right|=\left(\frac{e \tau}{m}\right) \vec{E}\))

2. According to the drift velocity expression, relaxation time is the time interval between successive collisions of an electron. On increasing temperature, the electrons move faster and more collisions occur more quickly. Hence, relaxation lime decreases with an increase in temperature which implies that drift velocity also decreases with temperature.

Question 11. Two identical cells of cmf 1.5 V each joined in parallel to supply energy to an external circuit consisting of two resistances of 7 Ω each joined in parallel. A very high resistance voltmeter reads the terminal voltage of cells to be 1.4 V. Calculate the internal resistance of each cell.

Answer:

A high resistance voltmeter means that no current flows through the voltmeter (practically very little current). When two batteries are connected in parallel, then

⇒ \(E_{eq}=\frac{E_1 r_2+E_2 r_1}{r_1+r_2}\)

Here r₁ = r₂ = r

E₁ = E₂ = I.5V (given)

⇒ \(\mathrm{E}_{\mathrm{eq}}=\frac{1.5 \times \mathrm{r}+1.5 \times \mathrm{r}}{2 \mathrm{r}}\)

E_eq = 1.5 V

Now \(\left.\begin{array}{l}
\mathrm{R}_1=7 \Omega \mathrm{R}_2=7 \Omega
\end{array}\right] \text { given }\)

So \(\frac{1}{R_{\mathrm{eq}}}=\left(\frac{1}{7}+\frac{1}{7}\right) \Omega\)

⇒ \(\mathrm{R}_{\mathrm{eq}}=\frac{7}{2}=3.5 \Omega\)

∵ \(I=\frac{\text { terminal voltage }}{\text { equivalent resistance }}\)

V = terminal voltage = 1.4 (given) = voltmeter reading

So, \(I=\frac{1.4}{3.5}=0.4 \mathrm{~A}\)

Now V = E_eq– 1 x r_eq ⇒ 1.4= 1.5- 0.4 x r_eq

0.4 x r_eq = 0.1

As r_eq = r/2 (∵ \(\frac{1}{r_{e q}}=\frac{1}{r}+\frac{1}{r}\))

So r of each cell = 0.5Ω

Question 12.

1. Define the term ‘conductivity’ of a metallic wire. Write it’s ST unit.
2. Using the concept of free electrons in a conductor, derive the expression for the conductivity of a wire in terms of number density and relaxation lime. Hence obtain the relation between current density and the applied electric field \(\vec{E}\).

Answer:

1. The conductivity of a material equals the reciprocal of the resistance of its wire of unit length and unit area of cross-section.

Alternatively:

The conductivity (σ) of a material is the reciprocal of its resistivity (ρ)

(Also acept \(\sigma=\frac{1}{\rho}\))

Its SI Units (\(\frac{1}{\text { olnm-metrc }}\)) or hom^-1m^-1 or (mho m^-1) or sciemen m^-1.

2. The Acceleration, \(\vec{a}=-\frac{e}{m} \vec{E}\)

The average drift velocity, V_d, is given by \(\vec{v}_{\mathrm{d}}=-\frac{\mathrm{e} \vec{\mathrm{E}}}{\mathrm{m}} \tau\)

(τ = average time between collisions or average relaxation lime)

If n is the number of free electrons per unit volume, then-current I is given by

⇒ \(I=n e A\left|v_d\right|\)

⇒ \(I=n e A\left(\frac{e \tau}{m}\right) E = I=\left(\frac{n e^2 \tau}{m}\right) E A\) (\(v_{\mathrm{d}}=\left(\frac{\mathrm{e} \tau}{\mathrm{m}}\right) \mathrm{E}\))

But I = J A (J = current density)

We, therefore, get

⇒ \(\mathrm{J}=\frac{\mathrm{ne}^2}{\mathrm{~m}} \tau \mathrm{E}\), The term \(\frac{n e^2}{m} \tau\) is conductivity.

∴ \(\sigma=\frac{\mathrm{ne}^2 \tau}{\mathrm{m}}\) ⇒ J = σE

Question 13.

Use Kirchhoff’s rules, to calculate the current in the arm AC of the given circuit.

Answer:

Applying Kirchhoff’s junction rule at node A

I₃ = I₁+I₂ → (1)

Applying Kirchhoff’s KVL in loop EFCAE

-30 I₁+40-40 I₃ = 0

3 I₁+4 I₃ = 4 → (2)

Applying KVL in loop EFDBE

-30 I₁+20 I₂ – 80 = 0

-3 I₁ + 2 I₂ = 8 → (3)

from eq (1) we put the value of I₃ in eq (2)

3 I₁ +4(I₁ + I₂) = 4

7 I₁ +4 I₂ = 4 → (4)

from eq (3) and (4) we get

I₁ =-12/13 A

Putting I₁ in eq (4) we get

⇒ \(\mathrm{I}_2=\frac{34}{13} \mathrm{~A}\)

Then form eq(1) we get \(\mathrm{I}_3=\frac{22}{13} \mathrm{~A}\)

Question 14.

1. The potential difference applied across a given resistor is altered so that the heat produced per second increases by a factor of 9. By what factor does the applied potential difference change?
2. In the figure shown, an ammeter A and a resistor of 4 Ω are connected to the terminals of the source. The emf of the source is 12 V having an internal resistance of 2 Ω. Calculate the voltmeter and ammeter readings.

Answer:

1. \(\mathrm{H}=\frac{\mathrm{V}^2}{\mathrm{R}} \mathrm{t}\) (initially) →(1)

After altering potential differences, we can write,

⇒ \(H^{\prime}=\frac{V^{\prime 2}}{R} t\)

∵ H’ = 9H we get,

⇒ \(9 \mathrm{H}=\frac{\mathrm{V}^{\prime 2}}{\mathrm{R}} \mathrm{t}\) → (2)

Solving (1) and (2) we get, V’ = 3V

2.

Total current ‘I’ in the circuit, \(I=\frac{E}{R+r}=\frac{R}{4+2}=2 A\)

Now potential differences across batteries is

V + Ir = E

V + 2 x 2 = 12

∴ V = 8 Volt

Class 12 Physics Current Electricity Long Answer Questions

Question 1. Wheatstone bridge is an arrangement of four resistances P, Q, R, and S connected as shown in the figure. Their values are so adjusted that the galvanometer G shows no deflection. The bridge is then said to be balanced when this condition is achieved. In the setup shown here, points B and D are at the same potential and it can be shown that P/Q = R/S This is called the balancing condition. If any three resistances are known, the fourth can be found. The practical form of the Wheatstone Bridge is a slide wire bridge or Meter Bridge. Using this the unknown resistance can be determined as \(S=\left(\frac{100-\ell}{\ell}\right) \times R\), where l is the balancing length of the meter bridge.

(1). In a Wheatstone bridge circuit, P = 5 Ohm, Q = 6 Ohm, R = 10 Ohm and S = 5 Ohm. What is the value of additional resistance to be used in series with S, so that the bridge is balanced?

1. 9 Ohm
2. 7 Ohm
3. 10 Ohm
4. 5 Ohm

Answer: 2. 7 Ohm

(2). A Wheatstone bridge consisting of four arms of resistance P, Q, R, S is most sensitive when

1. All the resistance is equal
2. All the resistance is unequal
3. The resistance P and Q arc equal but R>>P and S>>Q
4. The resistance P and Q are equal but R<<P and S<<Q

Answer: 1. All the resistance is equal

(3). The percentage error in measuring resistance with a meter bridge can be minimized by adjusting the balancing point close to

1. 0
2. 20 cm
3. 50 cm
4. 80 cm

Answer: 3. 50 cm

(4). In the meter bridge experiment, the ratio of the resistance of the left and light gap is 2 : 3. The balance point from the left is

1. 20 cm
2. 50 cm
3. 40 cm
4. 60 cm

Answer: 3. 40 cm

2. Relation between V, E, and r of a cell

Emf of a cell is the potential difference between two electrodes of the cell when no current is drawn from the cell. Internal resistance is the resistance offered by the electrolyte of a cell when the electric current flows through it. The internal resistance of a cell depends upon the following factors:

1. Distance between the electrodes
2. Nature and temperature of the electrolyte
3. Nature of electrodes
4. Area of electrodes.

For a freshly prepared cell, the value of internal resistance is generally low and goes on increasing as the cell is put to more use. The potential difference between the two electrodes of a cell in a closed circuit is called terminal potential difference and its value is less than the emf of the cell during discharging and more than the emf of the cell during charging of the cell in a closed circuit. It can be written as V = E – Ir or V = E + Ir

(1). The terminal potential difference of two electrodes of a cell is equal to cmf  the cell when

1. 1 ≠ 0
2. 1 = 0
3. Both (1) and (2)
4. Neither (1) and (2)

Answer: 2. Both (1) and (2)

(2). A cell of cmf E and internal resistance r gives a current of 0.5 A with an external resistance of 12 Ohm and a current of 0.25 A with an external resistance of 25 Ohm. What is the value of the internal resistance of the cell?

1. 5 Ohm
2. 1 Ohm
3. 7 Ohm
4. 3 Ohm

Answer: 2. 1 Ohm

(3). If external resistance connected to a cell has been, increased to 5 times, the potential difference across the terminals of the cell increases from 10 V to 30 V. Then the cmf of the cell is

1. 30 V
2. 60 V
3. 50 V
4. 40 V

Answer: 60 V

(4). During the charging of the cell the correct relation is

1. E = V + Ir
2. E = V – Ir
3. V = E + Ir
4. V = E – Ir

Answer: 3. V = E + Ir

3. A battery is a combination of two or more cells. In the following figure, a single battery is represented in which two cells of emf ε₁, and ε₂, and internal resistance r₁ and r₂ respectively are connected.

Answer the following Questions:

(1). The equivalent emf of this combination is:

1. \(\frac{\varepsilon_1 r_1+\varepsilon_2 r_2}{r_1+r_2}\)
2. \(\frac{\varepsilon_1 r_1-\varepsilon_2 r_2}{r_1+r_2}\)
3. \(\frac{\varepsilon_1 r_2-\varepsilon_2 r_1}{r_1+r_2}\)
4. \(\varepsilon_1-\varepsilon_2\)

Answer: 3. \(\frac{\varepsilon_1 r_2-\varepsilon_2 r_1}{r_1+r_2}\)

(2). For terminalB to be negative:

1. \(\varepsilon_1 r_2>\varepsilon_2 r_1\)
2. \(\varepsilon_1 r_2<\varepsilon_2 r_1\)
3. \(\varepsilon_1 r_1>\varepsilon_2 r_2\)
4. \(\varepsilon_2 r_2=\varepsilon_1 r_1\)

Answer: 2. \(\varepsilon_1 r_2<\varepsilon_2 r_1\)

(3). The current in the internal circuit is

1. \(\frac{\varepsilon_1+\varepsilon_2}{r_1+r_2}\)
2. \(\frac{\varepsilon_1-\varepsilon_2}{r_1+r_2}\)
3. \(\frac{\varepsilon_1}{r_1}-\frac{\varepsilon_2}{r_2}\)
4. \(\frac{\varepsilon_1}{r_2}-\frac{\varepsilon_2}{r_1}\)

Answer: 1. \(\frac{\varepsilon_1+\varepsilon_2}{r_1+r_2}\)

(4). The equivalent internal resistance of the combination is:

1. \(\frac{r_1+r_2}{r_1 r_2}\)
2. \(r_1+r_2\)
3. \(\frac{r_1 r_2}{r_1+r_2}\)
4. \(r_1-r_2\)

Answer: 3. \(\frac{r_1 r_2}{r_1+r_2}\)

Question 4.

1. Define the term drift velocity.
2. Based on electron drift, derive an expression for the resistivity of a conductor in terms of the number density of free electrons and relaxation lime. On what factors do the resistivity of a conductor depend?
3. Why are alloys like Constantan and manganin arc used for making standard resistors?

Answer:

1. Drift velocity is defined as the average velocity with which the electrons drift towards the positive terminal under the effect of the applied electric field.

2. We know that the current flowing through the conductor is:

I = n A e v_d

∴ \(\mathrm{l}=\mathrm{neA}\left(\frac{\mathrm{eE \tau}}{\mathrm{m}}\right)\)

Using \(E=\frac{V}{l}\)

⇒ \(I=n e A\left(\frac{e V}{m l}\right) \tau=\left(\frac{n e^2 A \tau}{m /}\right) V=\frac{1}{R} V\)

I ∝ V → which is Ohm’s law

Where \(R=\frac{m l}{n A e^2 \tau}\) is constant for a particular temperature and is called the resistance of the conductor.

∴ \(R=\left(\frac{m}{n e^2 \tau}\right) \frac{l}{A}=\frac{\rho l}{A} \Rightarrow \rho=\left(\frac{m}{n e^2 \tau}\right)\)

Where ρ is the specific resistance or resistivity of the material of the wire. It depends on the number of free electrons per unit volume and temperature

3. Alloys like constantan and manganin are used for making standard resistors, because:

• They have a high value of resistivity and
• The temperature coefficient of resistance is negligible

Question 5.

1. Plot a graph showing the variation of voltage v/s the current drawn from the cell. How can one get information from this plot about the cmf of the cell and its internal resistance?
2. Two cells of cmf’s E₁, and E₂, and internal resistance r₁, and r₂ arc connected in parallel. Obtain the expression for the emf and internal resistance of a single equivalent cell that can replace this combination.

Answer:

1. The terminal potential difference across the cell,

V = E – Ir or V = – rl + E

Comparing the above relation with the equation of a straight line i.e. y = mx + c, it follows that the graph between (along the x-axis) and V (along the y-axis) will be a straight line having a slope equal to -r and making intercept equal to E on the y-axis. Thus we get information about the EMF of the cell and its internal resistance from this plot.

2.

I = I₁ + I₂ → (1)

v = V_B1-V_B2

V = Polcnlial difference across terminal B₁ and B₂

V = E₁ – I₁r₁ – for first cell

V = E₂ – I₂r₂ – for second cell

and I = I₁ + I₂ as per above eq(1)

So \(I=\frac{E_1-V}{r_1}+\frac{E_2-V}{r_2}=\left(\frac{E_1}{r_1}+\frac{E_2}{r_2}\right)-V\left(\frac{1}{r_1}+\frac{1}{r_2}\right)\) → (2)

\(I=\frac{E_{eq .}-V}{r_{eq .}}\) → (3)

By equation (2) and (3)

⇒ \(\frac{E_{eq .}-V}{r_{eq .}}=\left(\frac{E_1}{r_1}+\frac{E_2}{r_2}\right)-V\left(\frac{1}{r_1}+\frac{1}{r_2}\right)\)

Now, \(\frac{1}{r_{eq}}=\frac{1}{r_1}+\frac{1}{r_2}\) and \(\frac{E_{eq}}{r_{eq}}=\frac{E_1}{r_1}+\frac{E_2}{r_2}\)

∴ \(E_{eq}=\frac{E_1 r_2+E_2 r_1}{r_1+r_2}\) and \(r_{eq}=\frac{r_1 r_2}{r_1+r_2}\)

CBSE Class 12 Physics Chapter 4 Moving Charges And Magnetism Multiple Choice Question And Answers

Question 1. The moving coil galvanometer G₁ and G₂ have the following particulars respectively:

N₁ = 30 , A₁ = 3.6 x 10^-3m² , B₁ = 0.25 T

N₂ = 42 , A₂ = 1.8 x 10^-3m², B₂ = 0.50 T

The spring constant is the same for both the galvanometers, The ratio of current sensitivities of G₁ and G₂ is:

1. 5:7
2. 7:5
3. 1:4
4. 1:1

Answer: 1. 5:7

Read and Learn More Important Questions for Class 12 Physics with Answers

⇒ \(\frac{\mathrm{I}_{\mathrm{S}_1}}{\mathrm{I}_{\mathrm{S}_2}}=\frac{\mathrm{N}_1 \mathrm{~A}_1 \mathrm{~B}_1}{\mathrm{~K}_1} \times \frac{\mathrm{K}_2}{\mathrm{~N}_2 \mathrm{~A}_2 \mathrm{~B}_2}\)

K₁ = K₂

So \(\left(\frac{N_1}{N_2}\right)\left(\frac{A_1}{A_2}\right)\left(\frac{B_1}{B_2}\right)\)

= \(\frac{30}{42} \times \frac{3.6 \times 10^{-3}}{1.8 \times 10^{-3}} \times \frac{0.25}{0.50}\)

= \(\frac{5}{7} \times 2 \times \frac{1}{2}\)

= 5:7

CBSE Class 12 Physics Chapter 4 Question 2. A current 1 is flowing through the loop as shown in the figure ( MA = R, MB = 2R ). The magnetic field at the center of the loop is \(\left(\frac{\mu_0 I}{R}\right)\) times:

1. \(\frac{5}{16}\) in to the plane of paper
2. \(\frac{5}{16}\) Out to the plane of paper
3. \(\frac{7}{16}\) in to the plane of paper
4. \(\frac{7}{16}\) out to the plane of paper

Answer: 4. \(\frac{7}{16}\) out to the plane of paper

B = B_DA(x)+B_AC(x)

= \(\frac{3}{4} \times \frac{\mu_0 \mathrm{I}}{2 \mathrm{R}}+\frac{1}{4} \times \frac{\mu_0 \mathrm{I}}{2(2 \mathrm{R})}\)

= \(\frac{3}{8} \frac{\mu_0 I}{R}+\frac{1}{16} \frac{\mu_0 I}{R}\)

= \(\frac{\mu_0 \mathrm{I}}{\mathrm{R}}\left(\frac{3}{8}+\frac{1}{16}\right)\)

= \(\frac{\mu_0 I}{R}\left(\frac{6+1}{16}\right)\)

B = \(\frac{7}{16} \frac{\mu_0 I}{R}\)

Question 3. A long straight wire in the horizontal plane carries a current of 15A in north to south direction. The magnitude and direction of the magnetic field at a point 2.5 m east of the wire respectively are:

1. 1.2 μT, vertically upward
2. 1.2 μT, vertically downward
3. 0.6 μT, vertically upward
4. 0.6 μT vertically downward

Answer: 1. 1.2 μT, vertically upward

B = \(\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{r}}=\frac{4 \pi \times 10^{-7} \times 15}{2 \pi \times 2.5}=1.2 \mu \mathrm{T}\) (Vertically upwards)

CBSE Class 12 Physics Chapter 4 Question 4. An electron is projected with velocity \(\vec{V}\) along the axis of a current carrying a long solenoid. Which of the following statements is true?

1. The path of the electron will be circular about the axis
2. The electron will be accelerated along the axis
3. The path of the electron will be helical
4. The electron will continue to move at the same velocity v along the axis of the solenoid.

Answer: 4. The electron will continue to move at the same velocity v along the axis of the solenoid.

Question 5. If the speed v of a charged particle moving in a magnetic field \(\vec{B}\) (\(\vec{v}\) is perpendicular to \(\vec{B}\)) is halved, then the radius of its path will:

1. Not change
2. Become two times
3. Become one-fourth
4. Become half

Answer: 4. Become half

∴ \(r=\frac{m v}{q B}\)

Question 6. Which of the following is not affected by the presence of a magnetic field?

1. A current-carrying conductor
2. A moving charge
3. A stationary charge
4. A rectangular current loop with its plane parallel to the field

Answer: 3. A stationary charge

Question 7. Identical thick wires and two identical thin wires, all of the same material and the same length form a square in three different ways P, Q, and R as shown. Due to the current in these loops, the magnetic field at the center of the loop will be zero with ease of

1. P and R only
2. Q and R only
3. P and Q only
4. P, Q, and R

Answer: 1. P and R only

Question 8. A circular coil carrying a certain current produces a magnetic field B₀ at its center. The coil is now rewound to have three turns and the same current is passed through it. The magnetic field at the center is

1. 3B₀
2. \(\frac{B_0}{3}\)
3. \(\frac{B_0}{9}\)
4. 9B₀

Answer: 4. 9B₀

Lel coil is of N turns and radius R, \(B_0=\frac{\mu_0 I}{2 R}\)

2πR = length of wire

Now, 2R = 3 x 2πR’

So, \(R^{\prime}=\frac{R}{3}\)

∴ \(B^{\prime}=\frac{\mu_0 I \times 3}{2\left(\frac{R}{3}\right)}=9\left(\frac{\mu_0 I}{2 R}\right)=9 B_0\)

CBSE Class 12 Physics Chapter 4Question 9. A long solenoid carrying current produces a magnetic field B along its axis, if the number of turns in the solenoid is halved and the current in it is doubled, the new magnetic field will be:

1. B/2
2. B/2
3. 2B
4. 4B

Answer: 2. B/2

⇒ \(\overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})\)

⇒ \(\overrightarrow{\mathrm{F}}=1.6 \times 10^{-19}[(4 \hat{\mathrm{i}}+3 \hat{\mathrm{k}}) \times(3 \hat{\mathrm{k}}+4 \hat{\mathrm{i}})]\)

⇒ \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
4 & 0 & 3 \\
4 & 0 & 3
\end{array}\right|\)

⇒ \(\hat{\mathrm{i}}(0)-\hat{\mathrm{j}}(12-12)+\hat{\mathrm{k}}(0)\)

= 0

So, \(\overrightarrow{\mathrm{F}}=0\)

Question 10. A current-carrying square loop is suspended in a uniform magnetic field acting in the plane of the loop. If the force on one arm of the loop is F, the net force on the remaining three arms of the loop will be:-

1. \(3{\vec{F}}\)
2. \(-3{\vec{F}}\)
3. \({\vec{F}}\)
4. \(-\vec{F}\)

Answer: 4. \(-\vec{F}\)

Question 11. The electron performs circular motion in a circle of radius r, perpendicular to a uniform magnetic field B. The kinetic energy gained by this electron in half the revolution is __________.

1. \(\frac{1}{2} m v^2\)
2. \(\frac{1}{4} m v^2\)
3. Zero
4. πrBeV

Answer: 3. Zero

Work done is zone

Because v ⊥ B

So ΔK = W = 0.

Question 12. At a place, an electric field and a magnetic field are in a downward direction. There an electron moves in a downward direction. Hence this electron.

1. Will bend towards left
2. Will bend towards the right
3. Will gain velocity
4. Will lose velocity

Answer: 4. Will lose velocity

Question 13. When a charged particle moves in a magnetic field its kinetic energy.

1. Remains Constant
2. Increases
3. Can decrease
4. Become zero

Answer: 1. Remains Constant

Question 14. A charged particle moves with velocity v in a uniform magnetic field B. The magnetic force acting on it will be maximum when.

1. \(\vec{V}\) and \(\vec{B}\) arc in same direction.
2. \(\vec{V}\) and \(\vec{B}\) are in opposite direction.
3. \(\vec{V}\) and \(\vec{B}\) are mutually perpendicular.
4. \(\vec{V}\) and \(\vec{B}\) make an angle of 45° with each other.

Answer: 3. \(\vec{V}\) and \(\vec{B}\) are mutually perpendicular.

Question 15. There are 100 turns per cm length in a very long solenoid. It carries a current of 2.5 A. The magnetic field at its center on the axis is __________ T.

1. 3.14 x 10^-2
2. 9.42 x 10^-2
3. 6.28 x 10^-2
4. 12.56 x 10^-2

Answer: 1. 3.14 x 10^-2

B = μ₀ nI

⇒ \(4 \times 3.14 \times \frac{100}{10^{-2}} \times 2.5 \times 10^{-7}\)

⇒ 3.14 x 10^-2T

Question 16. A toroid wound with 100 turns per m of wire carries a current of 3 A. The core of the toroid is made of iron having relative magnetic permeability μ_r = 5000. The magnetic field inside the iron is __________. (μ₀ = 4π x 10^-7 T mA^-1)

1. 0.15 T
2. 1.5 x 10^-2T
3. 0.47 T
4. 1.88 T

Answer: 4. 1.88 T

B = μ₀ nI

= 5000 x 4 x 3.14 x 10^-7 x 100 x 3

B = 1.88 T

Question 17. Two parallel long thin wires, each carrying current I are kept at a separation r from each other. Hence the magnitude of force per unit length of one wire due to the other wire is _________.

1. \(\frac{\mu_0 I^2}{2 \pi r}\)
2. \(\frac{\mu_0 I^2}{r^2}\)
3. \(\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{r}}\)
4. \(\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{r}^2}\)

Answer: 1. \(\frac{\mu_0 I^2}{2 \pi r}\)

Question 18. Two concentric rings are kept in the same plane. The number of turns in both rings is 25. Their radii are 50 cm and 200 cm and they carry electric currents of 0.1 A and 0.2 A respectively, in mutually opposite directions. The magnitude of the magnetic field produced at their center is __________T.

1. 2μ₀
2. 4μ₀
3. \(\frac{10}{4} \mu_0\)
4. \(\frac{5}{4} \mu_0\)

Answer: 4. \(\frac{5}{4} \mu_0\)

⇒ \(\overrightarrow{\mathrm{B}}=\overrightarrow{\mathrm{B}}_1+\overrightarrow{\mathrm{B}}_2\)

⇒ \(\frac{\mu_0 N_1 I_1}{2 R_1}-\frac{\mu_0 N_2 I_2}{2 R_2}=\frac{\mu_0 \times 25}{2}\left(\frac{0.1}{0.5}-\frac{0.2}{2}\right)=\frac{5}{4} \mu_0\)

Question 19. The current sensitivity of the galvanometer is inversely proportional to _________.

1. Torsional constant
2. Number of turns
3. Area
4. Magnetic field

Answer: 1. Torsional constant

Question 20. Parallel currents ______ and antiparallel currents _______.

1. Attract, Attract
2. Repel, attract
3. Attract, repel
4. Repel, repel

Answer: 3. Attract, repel

Question 21. A solenoid of length 0.5 m has a radius of 1 cm and it is made up of 500 turns. If the magnitude of the magnetic field inside the solenoid is 6.28 x 10^-3 T then it carries a current of _________ A.

1. 4
2. 5
3. 2
4. 10

Answer: 2. 5

B = μ₀nI

Solve I

I = \(\frac{\mathrm{B}}{\mu_{\mathrm{o}} \mathrm{n}}\left[\mathrm{n}=\frac{\mathrm{N}}{\ell}=\frac{500}{0.5}=1000 \text { turns } / \mathrm{m}\right]\)

∴ \(\frac{6.28 \times 10^{-3}}{4 \pi \times 10^{-7} \times 1000}=5 \mathrm{~A}\)

Moving Charges And Magnetism Assertion And Reason

For question numbers 1 to 7 two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (1). (2). (3) and (4) as given below.

1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is NOT the correct explanation of A
3. A is true but R is false
4. A is false and R is also false

Question 1. Assertion: When a charged particle moves with velocity v in a magnetic field B (v⊥B). the force on the particle does not work.

Reason: The magnetic force is perpendicular to the velocity of the particle.

Answer: 1. Both A and R are true and R is the correct explanation of A

Question 2. Assertion: The magnetic moment of the toroid is zero.

Reason: The magnetic field outside the volume of the carrying toroid is zero.

Answer: 2. Both A and R are true but R is NOT the correct explanation of A

Question 3. Assertion: Two charge particles at rest experience only electrostatic force.

Reason: Charges at rest can only produce an electrostatic field.

Answer: 1. Both A and R are true and R is the correct explanation of A

Question 4. Assertion: A moving charged particle gets energy from a magnetic field.

Reason: Magnetic force works on moving charged particles.

Answer: 4. A is false and R is also false

Question 5. Assertion: The coil is wound over the metallic frame in a moving coil galvanometer.

Reason: The metallic frame helps in making steady deflection without any oscillation.

Answer: 1. Both A and R are true and R is the correct explanation of A

Question 6. Assertion: When the magnet is brought near iron nails, only translatory force acts on it.

Reason: The field due to a magnet is generally uniform.

Answer: 4. A is false and R is also false

Question 7. Assertion: The higher the range of an ammeter, the smaller its resistance.

Reason: To increase the range of the ammeter, the additional shunt needed to be connected across it.

Answer: 1. Both A and R are true and R is the correct explanation of A.

Moving Charges And Magnetism Short Question And Answers

Question 1.

1. Write the relation for the force acting on a charged particle q moving with velocity \(\vec{v}\) in the presence of a magnetic field \(\vec{B}\).
2. A proton is accelerated through a potential difference V, subjected to a uniform magnetic Held acting normal to the velocity of the proton. If the potential difference is doubled, how will the radius of the circular path described by the proton in the magnetic field change?

Answer:

1. \(\vec{F}=q(\vec{v} \times \vec{B})\)

2. \(r=\frac{m v}{q B}=\frac{\sqrt{2 m(K E)}}{q B}\) [mv = P = \(\sqrt{2 \mathrm{mK}}\)] (K → K.E)

= \(\frac{\sqrt{2 m(q V)}}{q B} r \propto \sqrt{V}\)

Thus if V₂ = 2V₁ = r₂ = √2r₁.

Question 2. In the figure given below, wire PQ is fixed while the square loop ABCD is free to move under the influence of currents flowing in them. State with reason, in which direction does the loop begin to move or rotate?

Answer:

Forces on side AB and CD cancel each other, Now force on AD is attractive due to the similar direction of current, and on side BC force is repulsive due to the opposite direction of current.

As side AD is nearer to wire PQ as compared to BC so force on side AD (attractive) is more than the force on side BC ( repulsive), so square loop ABCD does move towards wire PQ.

Question 3.

1. A proton and an electron traveling along parallel paths enter a region of uniform magnetic field, acting perpendicular to their paths. Which of them will move in a circular path with higher frequency?
2. What can be the cause of the helical motion of a charged particle?

Answer:

1. We know that, frequency

‍ \(v=\frac{\mathrm{q} B}{2 \pi \mathrm{m}} \text { and } \mathrm{m}_{\mathrm{e}}<<\mathrm{m}_{\mathrm{p}}\)

So electrons will move with a higher frequency

2. When a charged particle moves in a uniform external magnetic field, with velocity not perpendicular or parallel to the magnetic field, (Means : 0° < θ < 90°) then the charged particle experiences a force also along with torque and performs the helical motion.

Question 4. Draw the magnetic field lines due to a current passing through a long solenoid.

Answer:

Question 5. Find the condition under which the charged particles moving with different speeds in the presence of electric and magnetic field vectors can be used to select charged particles of a particular speed.

Answer:

Consider a charge ’q’ moving with velocity \(\vec{v}\) in the presence of both electric field \(\vec{E}\) and magnetic field \(\vec{B}\) experience a force given as

⇒\(\overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{E}}+\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})=\overrightarrow{\mathrm{F}}_{E}+\overrightarrow{\mathrm{F}}_{\mathrm{B}}\)

Assume, that \(\vec{E}\) and \(\vec{B}\) are ⊥ to each other and also to the velocity of the particle.

Directions of electric force \(\left(\vec{F}_{E}\right)\) and magnetic force \(\left(\vec{F}_{B}\right)\) are just opposite.

∴ \(\vec{F}=q(E-v B) \hat{j}\)

if magnitudes of electric and magnetic force are equal then, the net force on the particle is zero and it will move undeflected in the fields.

qE = qvB or V = E/B

The above condition is used to select charged particles of a particular velocity.

Question 6.

1. Write the expression for the magnitude of the magnetic force acting on a charged particle moving with velocity v in the presence of magnetic field B.
2. A neutron, an electron, and an alpha particle moving with equal velocities enter a uniform magnetic field going into the plane of the paper as shown. Trace their paths in the field and justify your answer.

Answer:

1. Force acting on a charged particle q which is moving with velocity v in magnetic field B is given by \(\vec{F}=q(\vec{v} \times \vec{B})\)

Fleming’s left-hand rule gives the direction of the force. The direction of the force is perpendicular to the plane containing velocity \(\vec{v}\) and magnetic field \(\vec{B}\).

2. A charged particle experiences a force when it enters the magnetic field. Due to the presence of a magnetic field, the charged particle will move in a circular path because the force is perpendicular to the velocity of the charged particle, required centripetal force will be provided by magnetic force.

The radius of the circular path in which the charged particle is moving is “given by r = mv/qB since v and B are constant so the radius of the path of the particle is proportional to their mass to charge ratio.

Alpha-particle will trace a circular path in an anti-clockwise sense and its deviation will be in the direction of \((\vec{v} \times \vec{B})\)

Nculion will pass without any deviation as the magnetic field does not exert force on the neutral particle.

Electron will trace a circular path in a clockwise sense as its deviation is in the direction opposite to \((\vec{v} \times \vec{B})\) with a smaller radius due to mass or charge.

Question 7. Two long straight parallel conductors are carrying steady current I₁ and I₂ separated by a distance d. if the currents are flowing in the same direction, show how the magnetic field set up in one produces an attractive force on the other. Obtain the expression for this force. Hence define one ampere.

Answer:

We consider two long conductors X₁ Y₁ and X₂Y₂ placed parallel to each other at a distance of d apart. The current I₁ and I₂ are flowing as shown in the diagram. The magnetic field at point P (on the conductor X₁Y₁ due to current I₂ flowing through the long conductor X₂Y₂) is g given by

⇒ \(\mathrm{B}_2=\frac{\mu_0 \mathrm{I}_2}{2 \pi \mathrm{d}}\) →(1)

According to the right-hand rule, the direction of the magnetic field B₂ at point P is perpendicular to the plane of the paper and in an inward direction. Now, the conductor X₁ Y₁ carrying current I₁ lies in the magnetic field B, produced by the conductor X₂Y₂.

Since F=BIl, the force experienced by the unit length of the conductor X₁Y₁ due to magnetic field B₂ is given by

⇒ \(\mathrm{F}=\mathrm{B}_2 \cdot\left(\mathrm{I}_1 \cdot \ell\right)=\frac{\mu_0 \mathrm{I}_1 \mathrm{I}_2}{2 \pi \mathrm{d}}\) → (2)

or

⇒ \(\frac{F}{l}=\frac{\mu_0 I_1 I_2}{2 \pi d}\)

Applying Fleming’s left-hand rule, it follows that the force F on the conductor X₁Y₁ acts in the plane of the paper and towards the left. If we proceed similarly then it can be proved that the conductor X₂Y₂ experiences an equal force in the plane of the paper but towards the right.

Therefore, the two parallel conductors carrying current in the same direction attract each other.

Definition of Ampere: Let I₁ = I₂ = I A and d = 1 m

Then from eq (1), we have that

F = 2x 10^-7 N/mK

This means that one ampere of current is that much current which when flown through two infinitely long parallel conductors separated by one meter in free space, causes a force of 2 x 10^-7 N per meter on each conductor.

Question 8. A rectangular coil of sides ‘l’ and ‘b’ carrying a current I is subjected to a uniform magnetic field \(\vec{B}\) acting perpendicular to its plane. Obtain the expression for the torque acting on it.

Answer:

Consider a rectangular conducting loop (PQRS) of length l and b breadth b placed in a uniform magnetic field. Let I be the current flowing in the loop in a clockwise direction. Let at any instant the angle between the magnetic field and normal to the rectangular coil is θ.

Force acting on the ann PQ in the loop,

⇒ \(\overrightarrow{\mathrm{F}}_1=\mathrm{I}(\vec{\ell} \times \overrightarrow{\mathrm{B}}) \Rightarrow \mathrm{F}_1=\mathrm{IB} \ell\) (Directed inside the sheet of paper)

Similarly, the force acting on the arm RS of the loop,

⇒ \(\overrightarrow{\mathrm{F}}_2=\mathrm{I}(\vec{\ell} \times \overrightarrow{\mathrm{B}}) \Rightarrow \mathrm{F}_2=\mathrm{IB} \ell\) (Directed outside the sheet of paper)

Force \(\overrightarrow{\mathrm{F}}_3\) acting on the arm QR and force \(\overrightarrow{\mathrm{F}}_4\) acting on the arm SP of the loop are equal, opposite and act along the same line, hence they cancel each other. Therefore only two forces \(\overrightarrow{\mathrm{F}}_1\) and \(\overrightarrow{\mathrm{F}}_2\) act on the loop. \(\) and form a couple and try to rotate the loop clockwise. The magnitude of the torque(τ) due to forces \(\overrightarrow{\mathrm{F}}_1\) and \(\overrightarrow{\mathrm{F}}_2\) is given by

τ = Magnitude ol the either force x Perpendicular distance between forces

τ = IlB x b sin θ { l x b = A, area of the loop}

r = I A B sinθ

In vector form \(\vec{\tau}=I(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}})\)

If the loop has N turns, then the net torque acting on the loop is

τ = BINA sinθ

τ = M B sinθ

∴ \(\vec{\tau}=\overrightarrow{\mathrm{M}} \times \overrightarrow{\mathrm{B}}\)

CBSE Class 12 Physics Chapter 4 Question 9. How is a galvanometer converted into a voltmeter and an ammeter? Draw the relevant diagrams and find the resistance of the arrangement in each case. Take the resistance of the galvanometer as G.

Answer:

Conversion of Galvanometer into Voltmeter:

Voltmeter is used to measure p.d. so it is connected in parallel, in a circuit and its resistance should be infinite in the ideal case, thus to maximize the resistance of the galvanometer according to the required range we connect a suitable resistance in series as shown:

If the potential difference between the points to be measured = V and if

the galvanometer gives full-scale deflection when current “Ig” passes through it. Then.

⇒ \(V=I_g\left(R_g+R_X\right) \Rightarrow V=I_g R_g+I_g R_X \Rightarrow V-I_g R_g=I_g R_X\)

⇒ \(R_X=\left(V-I_g R_g\right) / I_g\)

⇒ \(R_x=\frac{V}{I_g}-R_g\)

Also, the equivalent resistance of the voltmeter: R_v = R_x + R_g

Conversion of Galvanometer into Anrmeter:

An ammeter is a current measuring device so its resistance should be zero in the ideal case, thus to minimize its resistance according to the required range, we connect a suitable resistance in parallel with it, which is called a shunt: \(I_S=\left(I-I_g\right)\)

The potential difference across the shunt: \(V_g=I_g R_g\)

But \(V_S=\left(I-I_g\right) R_S\)

V_s = V_g [in parallel combination]

⇒  \(\mathrm{R}_{\mathrm{S}}\left(\mathrm{I}-\mathrm{I}_{\mathrm{g}}\right)=\mathrm{I}_{\mathrm{g}} \mathrm{R}_{\mathrm{g}}\)

⇒ \(R_S=\frac{I_g}{I-I_g} R_g\)

Thus equivalent resistance \(G^{\prime}=\frac{R_g \cdot R_S}{R_g+R_S}\)

Question 10.

1. State Biot – Savart law and express this law in the vector form.
2. Two identical circular coils, P and Q each of radius R. carrying currents I A and √3A respectively, are placed concentric and perpendicular to each other lying in the XY and YZ planes. Find the magnitude and direction of the net magnetic field at the center of the coils.

Answer:

1. Biot-Savart Law (BSL)

According to the Biot Savart Law, the magnitude of a magnetic field \(\mathrm{d} \overrightarrow{\mathrm{B}}\) is proportional to the current I. the element length \(\overrightarrow{\mathrm{dl}}\) and inversely proportional to the square of the distance r. Its direction is perpendicular to the plane containing \(\overrightarrow{\mathrm{dl}}\) and \(\overrightarrow{\mathrm{r}}\).

The field at point P due to the current element

⇒ \(\mathrm{dB}=\frac{\mu_0}{4 \pi} \frac{\mathrm{Id} \ell \sin \theta}{\mathrm{r}^2}\)

⇒ \(\frac{\mu_0}{4 \pi}=10^{-7} \mathrm{Tm} / \mathrm{A}\)

μ₀= permeability of free space (air/vacuum)

Vector form:

2.

Given

I₁ = I A (current in coil P) : I₂ = √3 A (current in coil Q)

The magnetic field at the center of the circular current-carrying coil is given by

B = μ₀ I/2R

So, \(B_p=\frac{\mu_0 I}{2 R}=\frac{\mu_0}{2 R}\)

∴ \(\mathrm{B}_{\mathrm{Q}}=\frac{\sqrt{3} \mu_0}{2 \mathrm{R}}\)

net field at centre(B_R)

⇒ \(B_R=\sqrt{B_p^2+B_Q^2}=\sqrt{\left(\frac{\mu_0}{2 R}\right)^2+\left(\frac{\sqrt{3} \mu_0}{2 R}\right)^2}\)

So, \(B_R=\frac{\mu_0}{R}\), Direction \(\tan \theta=\frac{\mathrm{B}_{\mathrm{p}}}{\mathrm{B}_{\mathrm{Q}}}=\frac{1}{\sqrt{3}}\)

θ = 30°

Question 11.

1. State Ampere’s circuital law.
2. Use this law to find the magnetic field due to a straight infinite current-carrying wire.
3. How are the magnetic field lines different from the electrostatic field lines?

Answer:

1. Ampere’s circuital law: The line integral of the magnetic field over a closed loop is p0 times the total current threading through that loop

⇒ \(\oint \overrightarrow{\mathrm{B}} \cdot \mathrm{d} \vec{l}=\mu_0(\Sigma \mathrm{I})\)

2. Magnetic field due to infinitely long straight current carrying conductor:

Apply Ampere’s law to find out the magnetic field at point ’p’

⇒ \(\oint \overrightarrow{\mathrm{B}} \cdot \mathrm{d} \vec{l}=\mu_0 \Sigma \mathrm{I}\)

⇒ \(\oint \mathrm{Bd} l \cos \theta=\mu_0 \mathrm{I} \quad\left\{\begin{array}{c}
\theta=0^{\circ} \\
\cos \theta=1 \\
\Sigma \mathrm{I}=\mathrm{I}
\end{array}\right.\)

⇒ \(\mathrm{B} \oint \mathrm{d} l=\mu_0 \mathrm{I} \Rightarrow \mathrm{B} \times 2 \pi \mathrm{r}=\mu_0 \mathrm{I}\)

⇒ \(B=\frac{\mu_0 I}{2 \pi r}\)

3. The magnetic field lines form continuous closed loops, whereas electrostatic field lines never form closed loops.

Question 12.

1. A point charge q moving with speed v enters a uniform magnetic field B that is acting into the plane of the paper as shown. What is the path followed by the charge q and in which plane does it move?
2. How does the path followed by the charge get affected if its velocity has a component parallel to \(\vec{B}\)?
3. If an electric field \(\vec{E}\) is also applied such that the particle continues moving along the original straight-line path, what should be the magnitude and direction of the electric field \(\vec{E}\)?

Answer:

1. Charge q moves in a circular path. It does move in X-Y plane.

2. If velocity has a component parallel to B then charge q moves in helical path.

3. \(\overrightarrow{\mathrm{V}}=-\mathrm{V} \hat{\mathrm{i}}\)

[∵ The particle is moving along negative x-direction]

⇒ \(\vec{B}=-B \hat{k}\)

∵ The magnetic field is perpendicular to the plane of the paper directed inwards i.e. negative z-direction.

∴ Force acting due to magnetic field \(F_m=q(\vec{v} \times \vec{B})=q[-v \hat{i} \times(-B \hat{k})]\)

∴ \(\overrightarrow{\mathrm{F}}_{\mathrm{m}}=-q v B \hat{\mathrm{j}}\) (∵ \(\hat{i} \times \hat{k}=-\hat{j}\))

The magnitude of F_m = qvB, in -Y direction

For the undeflected motion of particles.

Fe = qE should be applied in \(y(+\hat{j})\) direction.

CBSE Class 12 Physics Chapter 4 Moving Charges And Magnetism Long Questions And Answers

Question 1. Amperes law gives a method to calculate the magnetic field due to a given current distribution. According to it, the circulation \(\oint \vec{B} \cdot d \vec{l}\) of the resultant magnetic field along a closed boundary is equal to p0 times the total current crossing the area bounded by the closed curve provided the electric field inside the loop remains constant. Ampere’s law is more useful under certain symmetrical conditions. Consider one such case of a long straight wire with a circular cross-section (radius R) carrying current 1 uniformly distributed across this cross-section.

(1). The magnetic field at a radial distance r from the center of the wire in the region r > R, is

1. \(\frac{\mu_0 I}{2 \pi r}\)
2. \(\frac{\mu_0 I}{2 \pi R}\)
3. \(\frac{\mu_0 \mathrm{IR}^2}{2 \pi \mathrm{r}}\)
4. \(\frac{\mu_0 \mathrm{Ir}^2}{2 \pi \mathrm{R}}\)

Answer: 1. \(\frac{\mu_0 I}{2 \pi r}\)

(2). A long straight wire of a circular cross-section (radius a) carries a steady current I and the current I is uniformly distributed across this cross-section. Which of the following plots represents the variation of the magnitude of magnetic field B with distance r from the center of the wire?

Answer: 1.

(3). A long straight wire of radius R carries a steady current I. The current is uniformly distributed across its cross-section. The ratio of magnetic field al R/2 and 2R is

1. 1/2
2. 2
3. 1/4
4. 1

Answer: 4. 1

(4). A long straight wire of a very very thin radius carries a steady current I. How magnetic field at a distance from this wire change if the value of the current in the wire is doubled?

1. Doubled
2. Halved
3. Three times
4. None of the above

Answer: 1. Boubled

Question 2. Read the case study carefully and answer the questions that follow:

A galvanometer is a device used to detect current in an electric circuit. It cannot as such be used as an ammeter to measure current in a given circuit. This is because a galvanometer is a very sensitive device, It gives a full-scale deflection for a current of the order of μA.

Moreover, for measuring currents, the galvanometer has to be connected in series, and it has a large resistance, this will change the value of the current in the circuit. To overcome these difficulties, we connect a small resistance Rs, called shunt resistance, in parallel with the galvanometer coil, so that most of the current passes through the shunt.

Now to use a galvanometer as a voltmeter, it has to be connected in parallel with the circuit element across which we need to measure p.d. Moreover, it must draw a very small current, otherwise, it will appreciably change the voltage that we are measuring. To ensure this a large resistance R is connected in series with the galvanometer.

Based on the information given above, answer the following questions:

(1). A sensitive galvanometer like a moving coil galvanometer can be converted into an ammeter or a voltmeter by connecting a proper resistance to it. Which of the following statements is true:-

1. A voltmeter is connected in parallel and current through it is negligible
2. An ammeter is connected in parallel and the potential difference across it is small
3. A voltmeter is connected in series and the potential difference across it is small
4. An ammeter is connected in series in a circuit and the current through it is negligible

Answer: 1. A voltmeter is connected in parallel and current through it is negligible

(2). By mistake a voltmeter is connected in series and an ammeter is connected in parallel with a resistance in an electrical circuit. What will happen to the instruments?

1. Voltmeter is damaged
2. Ammeter is damaged
3. Both are damaged
4. None is damaged.

Answer: 4. None is damaged.

(3). A galvanometer coil has a resistance of 15 Ω and gives full-scale deflection for a current of 4 mA. To convert it to an ammeter of range 0 to 6 A.

1. 10 m Ω resistance is connected in parallel to the galvanometer.
2. 10 m Ω resistance is to be connected in series with the galvanometer.
3. 0.1 Ω resistance is to be connected in parallel to the galvanometer.
4. 0.1 Ω resistance is to be connected in series with the galvanometer.

Answer: 1. 10 m Ω resistance is connected in parallel to the galvanometer.

(4). A long straight wire of a very very thin radius carries a steady current I. How magnetic field at a distance from this wire change if the value of the current in the wire is doubled?

1. More
2. Equal
3. Less
4. Zero

Answer: 1. More

Question 3.

1. An α-particle, a deuteron, and a proton enter into a uniform magnetic field normally with the same kinetic energy and describe circular paths. Find the ratio of radii of their paths.

2. Give the direction of the magnetic field acting on the current-carrying coil ACDE shown in the figure so that the coil is in unstable equilibrium.

3. Why do we use a low resistance ammeter in a circuit to measure current?

Answer:

⇒ \(K=\frac{q^2 B^2 r^2}{2 m}\)

⇒ \(r=\sqrt{\frac{2 m K}{q^2 B^2}}\)

⇒ \(r=\frac{\sqrt{2 m K}}{q B}\)

K.B.2 → constant

⇒ \(r \propto \frac{\sqrt{m}}{q}\)

⇒ \(\begin{array}{c|c|c}
\alpha \rightarrow{ }_2 \mathrm{He}^{4} & \mathrm{d} \rightarrow{ }_1 \mathrm{H}^2 & \mathrm{P} \rightarrow{ }_1 \mathrm{H}^{1} \\
\mathrm{m}^{\prime}=4 \mathrm{~m} & \mathrm{~m}^{\prime \prime}=2 \mathrm{~m} & \mathrm{~m}=\mathrm{m} \\
\mathrm{q}^{\prime}=2 \mathrm{q} & \mathrm{q}^{\prime \prime}=\mathrm{q} & \mathrm{q}=\mathrm{q}
\end{array}\)

So, \(r_\alpha: r_d: r_p=\frac{\sqrt{4 m}}{2 q}: \frac{\sqrt{2 m}}{q}: \frac{\sqrt{m}}{q}\)

1:2:1

2. For an unstable equilibrium direction of the area and magnetic field must be opposite so the magnetic field must be in the Z direction.

3. So that maximum current flows through the ammeter.

Question 4.

1. Draw the magnetic field lines due to a circular wire carrying current I.
2. A square loop of sides 5 cm carrying a current of 0.2 A in the clockwise direction is placed at a distance of 10 cm from an infinitely long wire carrying a current of I A as shown. Calculate
   1. The resultant magnetic force, and
   2. The torque, if any, acting on the loop

Answer:

1. Magnetic field due to a circular wire carrying current I.

2. (1). Force on the side be an ad are same in magnitude and opposite in direction. So they cancel each other.

⇒ \(\overrightarrow{\mathrm{F}}_2=-\overrightarrow{\mathrm{F}}_4\)

⇒ \({\vec{F}}_2+\overrightarrow{\mathrm{F}}_{+}=0\)

Force on side (ab)

⇒ \(F_1=\left(\frac{\mu_0 I_1 I_2}{2 \pi d_1}\right) \times l\)

⇒ \(F_1=\left(\frac{4 \pi \times 10^{-7} \times 1 \times 0.2}{2 \pi \times 10}\right) \times 5\)

⇒ \(\mathrm{F}_1=2 \times 10^{-8} \mathrm{~N}\) (towards left)

Force on side (cd)

⇒ \(\mathrm{F}_3=\left(\frac{\mu_0 \mathrm{I}_1 \mathrm{I}_2}{2 \pi \mathrm{d}_2}\right) \times l\)

⇒ \(F_3=\left(\frac{4 \pi \times 10^{-7} \times 1 \times 0.2}{2 \pi \times 15}\right) \times 5 \Rightarrow F_3=\frac{4}{3} \times 10^{-8} \mathrm{~N}\)

The resultant force on the loop,

F = F₁-F₂ (∵ F₁ and F₃ are oop.)

\(\)\left(2-\frac{4}{3}\right) \times 10^{-8} \Rightarrow \mathrm{F}=\frac{2}{3} \times 10^{-8} \mathrm{~N} (towards left)

(2). Torque on square loop ‘Abcd’

τ = MB sinθ (∵ θ = 0)

τ = 0

Question 5.

1. With the help of a neat and labeled diagram, explain the principle and working of a moving coil galvanometer.
2. What is the function of a uniform radial field and how it is produced?
3. Define the current sensitivity of a galvanometer. How is current sensitivity increased?

Answer:

1. Principle: When the current-carrying coil is placed in a magnetic field it experiences a torque.

τ = NI AB → (1)

It consists of a plane coil of many turns suspended in a radial magnetic field. When a current is passed in the coil it experiences a torque which produces a twist in the
suspension.

This deflection is directly proportional to the torque

τ= KΦ → (2)

By equation (1) and (2)

∵ N1AB = KΦ;

⇒ \(I=\left(\frac{K}{N A B}\right) \phi\);

K = elastic torsional constant of the suspension

⇒ \(\mathrm{I}=\mathrm{C} \phi ; \mathrm{C}=\frac{\mathrm{K}}{\mathrm{NAB}}\)

C ⇒ Galvanometer constant

So I ∝ Φ

τ ∝ I

Galvanometer:

An instrument used to measure the strength of current by measuring the deflection of the coil due to torque produced by a magnetic field.

τ ∝ i∝θ

Current Sensitivity (CS)

It is defined as the deflection per unit current.

⇒ \(C S=0 / I=\frac{N A B}{K}\)

Voltage Sensitivity (VS)

It is defined as deflection per unit voltage.

⇒ \(V S=\phi / V=Φ / I R=\left(\frac{N A B}{K R}\right)\)

2. The function of radial field, is.

1. To maximize the deflecting torque acting on the current carrying coil.
2. To increase the strength of the magnetic field.

The radial magnetic field is produced by using a concave magnetic pole. Also cylindrical soft iron core helps in the production of the radial magnetic field.

3. Current sensitivity is the deflection shown by the galvanometer for a unit current flow.

⇒ \(I_S=\frac{\phi}{I} \text { or } \frac{N A B}{K}\)

Where Φ is the deflection in the coil.

The current sensitivity of the galvanometer can be increased by

1. Increasing the number of turns (N )
2. Increasing magnetic induction (B)
3. Increasing area of coil (A)
4. Decreasing the couple per unit twist of the spiral springs.

Question 6.

1. Use it to obtain the expression for the magnetic field at an axial point situated at distance d from the center of a circular coil of radius R carrying current I.
2. Also, find the ratio of the magnitudes of the magnetic field of this coil at the center and at an axial point for which x = R√3.

Answer:

1. Vector form of Biot-Savart’s law

⇒ \(d \vec{B}=\frac{\mu_0 I}{4 \pi r^2}(d \vec{l} \times \hat{r})\left\{\hat{r}=\frac{\vec{r}}{r}\right.\)

The magnetic field at an axial point of a current-carrying circular loop:

⇒ \(d\vec{B}=\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{r}^3}(\mathrm{~d} \vec{l} \times \overrightarrow{\mathrm{r}})\)

Magnetic field due to the small element at point ‘P’

⇒ \(\mathrm{dB}=\frac{\mu_0 \mathrm{Id} l \sin \theta}{4 \pi r^2}\)

Angle between \(d\vec{l}\) and \(\hat{r}\) is always 90 ( = 90°). The direction of the magnetic field is perpendicular to the plane of \(d\vec{l}\) and \(\hat{r}\).

⇒ \(\mathrm{dB}=\frac{\mu_0 \mathrm{Id} l}{4 \pi \mathrm{r}^2}\left\{\sin 90^{\circ}=1\right.\)

The total magnetic field at point P

⇒ \(\mathrm{B}_{\text {axis }}=\int_0^{2 \pi \mathrm{R}} \mathrm{dB} \sin \phi\) ⇒ \(\mathrm{B}_{\mathrm{axis}}=\int_0^{2 \pi \mathrm{R}} \frac{\mu_0 \mathrm{Id} l}{4 \pi \mathrm{r}^2} \times \frac{\mathrm{R}}{\mathrm{r}}\left\{\sin \phi=\frac{\mathrm{R}}{\mathrm{r}}\right.\)

⇒ \(\mathrm{B}_{\mathrm{axis}}=\frac{\mu_0 \mathrm{IR}}{4 \pi \mathrm{r}^3} \int_0^{2 \pi \mathrm{R}} \mathrm{d} l\) ⇒ \(B_{\text {axis }}=\frac{\mu_0 I R}{4 \pi r^3} \times 2 \pi R\) ⇒ \(\mathrm{B}_{\mathrm{axis}}=\frac{\mu_0 \mathrm{IR}^2}{2 \mathrm{r}^3}\)

⇒ \(B_{\mathrm{axis}}=\frac{\mu_0 I R^2}{2\left(R^2+x^2\right)^{3 / 2}}\left\{r=\left(R^2+x^2\right)^{1 / 2}\right.\)

⇒ \(B_{\text {axis }}=\frac{\mu_0 \text { NIR }^2}{2\left(R^2+x^2\right)^{3 / 2}}\)

2. At the center of the current loop

⇒ \(B_0=\frac{\mu_0 N I}{2 R}\)

At the axial point x = R√3

⇒₁\(B_1=\frac{\mu_0 NI R^2}{2\left(R^2+3 R^2\right)^{3 / 2}}=\frac{\mu_0 NI R^2}{2 \times 8 R^3}\)

⇒ \(\frac{B_0}{B_1}=\frac{\frac{\mu_0 N I}{2 R}}{\frac{\mu_0 N I}{2 \times 8 R}}=8: 1\)

CBSE Class 12 Physics Chapter 5  Magnetism And Matter Multiple Choice Questions And Answers

Question 1. The vertical component of the earth’s magnetic field at a place is \(\frac{1}{\sqrt{3}}\) times the horizontal component. The angle trip of dip at that place is:

1. 0°
2. 30°
3. 45°
4. 60°

Answer: 2. 30°

Read and Learn More Important Questions for Class 12 Physics with Answers

Question 2. Inside a bar magnet, the magnetic field lines

1. Are not present
2. Are parallel to the cross-sectional area
3. Are in the direction from N-poIcs to S-pole
4. Are in the direction from S-pole to N-pole

Answer: 4. Are in the direction from S-pole to N-pole

Question 3. A magnetic needle is kept nonparallel to the magnetic field in non-uniform magnetic field experiences.

1. A force but not a torque
2. Torque but not a force
3. Both a force and a torque
4. Neither a force nor a torque

Answer: 3. Both a force and a torque

Question 4. When a paramagnetic substance is brought near a north pole or a south pole of a bar magnet, it __________.

1. Experience repulsion
2. Docs do not experience attraction or repulsion
3. Experience attraction
4. Experience attraction or repulsion depending upon which pole is brought near to it.

Answer: 3. Experience attraction

Initially m’ = q_m x l

after bending

m’ = q_m x 2r

⇒ \(m^{\prime}=\frac{m}{l} \times 2 \times \frac{l}{\pi}\)  [ πr = l ]

∴ \(m^{\prime}=\frac{2 m}{\pi}\)

question 5. A straight steel wire of length l has magnetic moment m. If the wire is bent in the form of a semicircle, the new value of the magnetic dipole moment is.

1. m
2. \(\frac{\mathrm{m}}{\pi}\)
3. \(\frac{m}{2}\)
4. \(\frac{2 \mathrm{~m}}{\pi}\)

Answer: 4. \(\frac{2 \mathrm{~m}}{\pi}\)

Initially

m = q_m x l

⇒ \(\mathrm{m}^{\prime}=\mathrm{q}_{\mathrm{m}} \times \frac{l}{2}\)

(q_m will not change)

⇒ \(\mathrm{m}^{\prime}=\frac{\mathrm{m}}{2}\)

∴ \(\vec{P}\) remains constant.

Question 6. A bar magnet of length l, pole strength ‘P’, and magnetic moment ‘\(\vec{m}\)‘ is split into two equal pieces each of length l/2. The magnetic moment and pole strength of each piece respectively ________ and ________.

1. \(\vec{m}, \frac{p}{2}\)
2. \(\frac{\vec{m}}{2}, p\)
3. \(\frac{\vec{m}}{2}, \frac{p}{2}\)
4. \(\vec{m}, p\)

Answer: 2. \(\frac{\vec{m}}{2}, p\)

Question 7. For superconductors, μ_r = _________.

1. Zero
2. Infinite
3. Positive
4. Negative

Answer: 1. Zero

Question 8. Which one of the following represents Curie’s law?

1. \(M=\frac{C B_0}{T}\)
2. \(M=\frac{C \chi}{T}\)
3. \(M=\frac{C \chi}{T-T_C}\)
4. \(M=\frac{C T}{B_0}\)

Answer: 1. \(M=\frac{C B_0}{T}\)

Question 9. Meissner effect is observed in _________ substance.

1. Paramagnetic
2. Ferromagnetic
3. Superconducting
4. Permanent magnetic

Answer: 3. Superconducting

Question 10. According to Gauss’s law for magnetism, the net magnetic flux through any closed surface is _________.

1. Equal to ε₀
2. Infinite
3. Zero
4. Equal to μ₀

Answer: 3. Zero

CBSE Class 12 Physics Chapter 5  Magnetism And Matter Assertion And Reason

For question numbers 1 to 4 two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (1), and (2). (3) and (4) as given below.

1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is NOT the correct explanation of A
3. A is true but R is false
4. A is false and R is also false

Question 1. Assertion: A bar magnet experiences a torque when placed in a magnetic field.

Reason: A bar magnet exerts a torque on itself due to its magnetic field.

Answer: 3. A is true but R is false

Question 2. Assertion: The magnetic moment of the helium atom is zero.

Reason: All the electrons are paired in helium atom orbitals.

Answer: 1. Both A and R are true and R is the correct explanation of A

Question 3. Assertion: An atom behaves as a magnetic dipole.

Reason: It is because an atom contains equal positive and negative charges.

Answer: 3. A is true but R is false

Question 4. Assertion: When the radius of a circular loop carrying current is doubled, its magnetic moment becomes four times.

Reason: Magnetic moment depends on the area of the loop.

Answer: 2. Both A and R are true but R is NOT the correct explanation of A

CBSE Class 12 Physics Chapter 5  Magnetism And Matter Short Questions And Answers

Question 1. The magnetic moment of a circular coil carrying current I, having N turns, each of radius r, is M. Find the magnetic moment of the same coil if it is unwound and rewound into a coil having 2N turns for the same current.

Answer:

M = NIA = NIπ²

According to question N(2πr) = 2N(2πr’)

⇒ \(r^{\prime}=\frac{r}{2}\)

So, New magnetic movement M’ = \((2 \mathrm{~N}) \mathrm{I} \pi\left(\frac{\mathrm{r}}{2}\right)^2=\frac{\mathrm{NI} \pi \mathrm{r}^2}{2}=\frac{\mathrm{M}}{2}\)

Question 2. The bar magnet of magnetic moment 6 J/T is aligned at 60° with a uniform external magnetic field of 0.44 T. Calculate

1. The work done in turning the magnet to align its magnetic moment
   1. Normal to the magnetic field,
   2. Opposite to the magnetic field, and
2. The torque on the magnet in the final orientation in case (2)

Answer:

Magnetic moment M = 6J/T

Magnetic field B = 0.44 T

(1). Work done in rotating the magnet normal to the magnetic field

W = MB(cosθ₁ – cosθ₂)

where θ₁ =60° and θ₂ = 90°

W = 6 x 0.44 (cos60°- cos90°) = 6 x 0.44 (0.5- 0) = 1 .32 J

(2). Work done in rotating the magnet opposite to the magnetic field

W = MB(cosθ₁ – cosθ₂)

Where θ₁ =60°and θ₂ = 180°

W = 6 x 0.44 (cos60°- cos 180°) = 6 x 0.44 (0.5 + 1 ) = 3.96 J

2. Torque in case (2) τ = MB sin θ₂ ⇒ τ = 6 x 0.44 x sin 180° = 0

Question 3.

1. An iron ring of relative permeability |.ir has windings of insulated copper wire of n turns per meter. When the current in the windings is I, find the expression for the magnetic field in the ring.
2. The susceptibility of a magnetic material is 0.9853. Identify the type of magnetic material. Draw the modification of the field pattern by keeping a piece of this material in a uniform magnetic field.

Answer:

1. Consider a ring of radius r having n turns per meter, If n is the number of turns per meter, then the total number of turns in the ring = perimeter times the number of turns = 2πrn = N

Current enclosed = NI = 2πrn I.

By Ampere’s circuital law, \(\oint \vec{B} \cdot d \vec{l}=\mu_0 I\)

B x 2πr = μ₀ 2π rnI or B= μ₀nl

2. Given, susceptibility, χ = 0.9853.

The material is paramagnetic.

If a piece of this material is kept in a uniform magnetic field, then the field pattern gets modified as follows:

The lines of force tend to pass through the material rather than the surrounding air.

Question 4. Different diamagnetic, paramagnetic, and ferromagnetic substances based on their properties.

Answer:

CBSE Class 12 Physics Chapter 5 Magnetism And Matter Long Questions And Answers

Question 1. When the atomic dipoles are aligned partially or fully, there is a net magnetic moment in the direction of the field in any small volume of the material. The actual magnetic field inside the material placed in a magnetic field is the sum of the applied magnetic field and the magnetic field due to magnetization. This field is called magnetic intensity (H).

H = B/μ₀-M

where M is the magnetization of the material p0, is the permittivity of vacuum and B is the total magnetic field. The measure that tells us how a magnetic material responds to an external field is given by a dimensionless quantity is appropriately called the magnetic susceptibility; for a certain class of magnetic materials, the intensity of magnetization is directly proportional to the magnetic intensity.

(1). Identify the wrongly matched quantity and unit pair.

1. Pole strength – A-m
2. Magnetic susceptibility – dimensionless number
3. Intensity of magnetisation – A/m
4. Magnetic permeability- henry-m

Answer: 4. Magnetic permeability- henry-m

(2). A bar magnet has a length of 3 cm, a cross-sectional area of 2 cm², and a magnetic moment of 3 A-m². The intensity of magnetization of a bar magnet is

1. 2 x 10⁵ A/m
2. 3 x 10⁵ A/m
3. 4 x 10⁵ A/m
4. 5 x 10⁵ A/m

Answer: 4. 5 x 10⁵ A/m

(3). A solenoid has a core of material with a relative permeability of 500 and its windings carry a current of A. The number of turns in the solenoid is 500 per meter. The magnetization of the material is nearly:-

1. 2.5 x 10³ A/m
2. 2.5 x 10⁵ A/m
3. 2 x 10³ A/m
4. 2 x 10⁵ A/m

Answer: 2. 2.5 x 10⁵ A/m

(4). The relative permeability of iron is 6000. Its magnetic susceptibility is

1. 5999
2. 6001
3. 6000 x 10^-7
4. 6000 x 10^-7

Answer: 1. 5999

Electromagnetic Induction Class 12 Pdf

CBSE Class 12 Physics Chapter 6 Electromagnetic Induction Multiple Choice Questions And Answers

Question 1. Lenz’s law is the consequence of the law of conservation of:

1. Energy
2. Charge
3. Mass
4. Momentum

Answer: 1. Energy

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Question 2. The emf induced in a 10H inductor in which current changes from 11 A to 2A in 9 x 10^-1 s is :

1. 10⁴ V
2. 10³ V
3. 10² V
4. 10 V

Answer: 3. 10² V

⇒ \(\mathrm{e}=-\mathrm{L} \frac{\Delta \mathrm{I}}{\Delta \mathrm{t}}\)

∴ \(e=\frac{-10 \times(2-11)}{9 \times 10^{-1}}=10^2 \mathrm{~V}\)

Question 3. A metal plate is getting heated. Which one of the following statements is incorrect?

1. It is placed in a space-varying magnetic field that does not vary with time.
2. A direct current is passing through the plate.
3. An alternating current is passing through the plate.
4. It is placed in a time-varying magnetic field.

Answer: 1. It is placed in a space-varying magnetic field that does not vary with time.

Question 4. The magnetic flux linked with a coil is given by Φ = 5t²+ 3t + 10, where Φ is in Weber and l is in second. The induced emf in the coil at t = 5 sec will be

1. 53 V
2. 43 V
3. 10 V
4. 6 V

Answer: 1. 53 V

⇒ \(\mathrm{e}=-\frac{\mathrm{d} \phi}{\mathrm{dt}}\)

e = 10t + 3

e = 10(5) + 3 = 53V

Question 5. The magnetic flux linked with a coil changes with time t (second) according to Φ = 6t²– 5t + 1, where Φ is in Wb. At l = 0.5 S, the induced current in the coil is ______ The resistance of the circuit is 10 Ω.

1. 1 A
2. 0.1 A
3. 0.1 mA
4. 10 A

Answer: 2. 0.1 A

⇒ \(\mathrm{e}=-\frac{\mathrm{d} \phi}{\mathrm{dt}}\)

e = -(12t-5)

e = -(12 x 0.5-5)

e = 1

⇒ \(|\mathrm{e}|=1 \mathrm{~V}\)

∴ \(I=\frac{e}{R}=\frac{1}{10}=0.1 \mathrm{~A}\)

Question 6. A square conducting coil of area 100 cm2 is placed normally inside a uniform magnetic field of 103 Wbm-2. The magnetic flux linked with the coil is ______ wb.

1. 10
2. 10^-5
3. 10-g
4. 0

Answer: 1. 10

= BA

= 10³ x 100 x 10^{– 4}

= 100 weber

Electromagnetic Induction Class 12 PDF

Question 7. The self-inductance of two solenoids A and B having equal lengths are the same. If the number of turns in two solenoids A and B arc 100 and 200 respectively. The ratio of radii of their cross-section will be

1. 2:1
2. 1:2
3. 1:4
4. 4:1

Answer: 1. 2:1

⇒ \(L=\frac{\mu_r \mu_0 N^2 \pi r^2}{l}\)

⇒ \(r^2=\frac{L \times l}{\mu_r \mu_0 N^2 \pi}\)

l, L, r, 0, – constant

Question 8. A magnet is moving towards a coil along its axis and the emf induced in the coil is s. If the coil also starts moving towards the magnet with the same speed, the induced emf will be

2

4

Answer: 2. 2

Question 9. The dimensional formula of self-inductance is __________.

M1L2T-2A-2

M1L1T2A-2

M1L1T-2A-2

M1L1T-1A-1

Answer: 1.

Question 10. The mutual inductance of the system of two coils is 5 mH. The current in the first coil varies according to the equation T = I0 sin cot, where I0 = 10A and co = 100 TT rad/s. The value of maximum induced emf in the second coil is __________.

2V

5V

V

4V

Answer: 2. 5V

= -M(I0 cos t) [for max. value cos t = 1]

= -5 x 10-3 x 10 x 100 x (1)

e = -5V

|e| = 5V

Question 11. Current of 2A passing through a coil of 100 turns gives rise to a magnetic flux of 5 x 10-3 Wb per turn. The magnetic energy associated with coil is __________.

5 x 10-3 J

0.5 x 10-3

5 J

0.5 J

Answer: 4. 0.5 J

= \(\frac{1}{2} \mathrm{\phi}{I}=\frac{1}{2} \times 5 \times 10^{-3} \times 100 \times 2=0.5 \mathrm{~J}\)

Question 12. The flux linked per turn of a coil of N turns changes from (|)1 and <f)2- If die total resistance ol the circuit including the coil is R, the induced charge in the coil.

Answer: 2. \(-N \frac{\left(\phi_2-\phi_1\right)}{R}\)

Question 13. The magnitude of the induced cmf is equal to the lime rate of change of ____.

Electric flux

Magnetic force

Magnetic flux

Electric force

Answer: 3. Magnetic flux

Question 14. Unit of induced emf is _____.

Tesla

Volt/second

weber/second

volt

Answer: 4. Volt

Electromagnetic Induction Class 12 PDF

Question 15. One conducting wire of length 50 cm is moving perpendicular to uniform magnetic field of 0.2 T. with constant velocity of 10 ms 1 cmf induced between two ends of a wire is _________ V.

1.0

0.1

0.01

10

Answer: 1. 1.0

e = -1 Volt

|e| = 1 V

CBSE Class 12 Physics Chapter 6 Electromagnetic Induction Class 12 Electromagnetic Induction Assertion And Reason

For question numbers 1 to 5 two statements arc given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (1), (2), (3) and (4) as given below.

1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is NOT the correct explanation of A
3. A is true but R is false
4. A is false and R is also false

Question 1. Assertion: Induced cmf in two coils made of wire of the same length and thickness, one of copper and another of aluminium is the same. The current in copper coil is more than the aluminium coil.

Reason: Resistance of aluminium coil is more than that of copper coil.

Answer: 1. Both A and R are true and R is the correct explanation of A

Question 2. Assertion: If the current in an inductor is zero at any instant, then the induced emf may not be zero.

Reason: An inductor keeps the flux (i.e. current) constant.

Answer: 1. Both A and R are true and R is the correct explanation of A

Question 3. Assertion: Self-induction is called the inertia of electricity.

Reason: The inductor opposes the change in current.

Answer: 1. Both A and R are true and R is the correct explanation of A

Question 4. Assertion: A small magnet takes a longer time to fall into a hollow metallic tube without touching the wall.

Reason: There is an opposition of motion due to the production of eddy currents in metallic tubes.

Answer: 1. Both A and R are true and R is the correct explanation of A

Question 5. Assertion: Faraday’s laws are consequences of the conservation of energy.

Reason: In a purely resistive AC circuit, the current lags behind the EMF in phase.

Answer: 4. A is false and R is also false

CBSE Class 12 Physics Chapter 6 Class 12 Electromagnetic Induction Short Questions And Answers

Question 1. Two identical circular discs, one of copper and another of aluminium, are rotated about their geometrical axes with the same angular speed in the same magnetic field acting perpendicular to their planes. Compare the (1) induced cmf, and (2) induced current produced in discs between their centre and edge. Justify your answers,

Answer:

Emf will be same for both, because it depends on N, A, B, ro and that arc same for both disc But induced current \(I=\frac{\mathrm{e}}{\mathrm{R}}=\frac{\mathrm{NAB}\omega}{\mathrm{R}}\)

Copper has less resistance so induced current will be higher in copper.

Question 2. A rectangular loop PQMN with movable arm PQ of length 10 cm and resistance 2£2 is placed in a uniform magnetic field of 0.1 T acting perpendicular to the plane of the loop as is shown in the figure. The resistances of the arms MN, NP and MQ are negligible. Calculate the

1. emf induced in the arm PQ and
2. Current induced in the loop when arm PQ is moved with velocity 20 m/s

Answer:

1. Induced emf in arm PQ

e = -B/v

c = -0.1 x 10 x 1 0-2 x 20

e = -0.2 V

2. Current induced in loop

Question 3. Predict the polarity of the capacitor in the situation described below:

image

Answer:

Plate A (+ve) and plate B(-ve)

Question 3. An aeroplane is flying horizontally from west with a velocity of 900 km/hour. Calculate the potential difference developed between the ends of its wings having a span of 20m. The horizontal component of the Earth’s magnetic field is 5 x Hr4 T and the angle of dip is 30°.

Answer:

Potential difference developed between the ends of the wings ‘e’ = Blv

Given Velocity (v) = 900 km/hour

= 250 m/s

Wing span (l) = 20 m

Vertical component of Earth’s magnetic field

Bv = BH tan (5

= 5 x 10-4 (tan 30°) tesla

∴ Potential Difference

e = BVl

e = 5 x 10-4 (tan 30°) x 20 x 250

e = 1 .44 volt

1. Question 4.
2. Define self-inductance. Write its SI unit.
3. A long solenoid with 15 turns per cm has a small loop of area 2.0 cm2 placed inside the solenoid normal to its axis. If the current canned by the solenoid changes steadily from 2.0 A to 4.0 A in 0. 1 s. what is the induced cmf in the loop while the current is changing?

Answer:

1. Self inductance of a coil is numerically equal to cmf induced in the coil, when rate of
change of current is unity.

Or

L = N<(>/I

It is equal to the magnetic flux linked with coil when a unit current flow through it.

The S.I unit of L is henry (H).

2. Here, number of turns per unit length.

n= N/l = 15 turns/cm = 1500 turns/m

A = 2 cm2 = 2 x 10-4 m2

dl/dt = (4- 2)/0. 1 or dl/dt = 20 As-1

⇒ \(|\mathrm{e}|=\frac{\mathrm{d} \phi}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{BA})\) (∵ \(B=\frac{\mu_0 N I}{l}\) )

⇒ \(|\mathrm{e}|=\frac{\mathrm{Ad}}{\mathrm{dt}}\left(\mu_0 \frac{\mathrm{NI}}{l}\right)=\mathrm{A} \mu_0\left(\frac{\mathrm{N}}{l}\right) \frac{\mathrm{dI}}{\mathrm{dt}}\)

|e| = (2 x 10-4) x 4 x 10-7 x 1500 x 20V

|e| = 7.5 x 10-6 V

Question 5.

1. Define mutual inductance.
2. A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20A in 0.5 s. what is the change of flux linked with the other coil?

Answer:

1. Mutual Inductance : It is numerically equal to the magnetic flux linked with one coil (secondary coil) when a unit current flows through the other coil (primary coil).

2. Magnetic flux

initial = MI initial = 0 as I initial = 0

and final = Mfinal = 1.5 x 20 = 30Wb

Change in flux d= 1-2 = 30-0 = 30 weber.

Question 6. Define the self-inductance of a coil. Obtain the expression for the energy stored in an inductor L connected across a source of cmf.

Answer:

It is numerically equal to the induced emf per unit rate of change of current through a given coil.

i.e. \(\mathrm{L}=\frac{\mathrm{e}}{-\frac{\mathrm{dI}}{\mathrm{dt}}}\)

Energy stored in an inductor: Work has to be done by battery against the opposing induced cmf in establishing a current in an inductor. The energy supplied by the battery is stored in the inductor.

Let current flowing through the circuit at any instant (t) be i.

Rate of change of current at that time = \(\frac{\mathrm{di}}{\mathrm{dt}}\)

image

magnitude of induced emf in the inductor,

Work done by the battery in time dt,

dW = Vdq = edq

dW = eidt

dW = Lidi

Total work done by the battery in increasing current from 0 to I

This work done is stored in an inductor in the form of magnetic energy.

Question 7. Define mutual inductance between a pair of coils. Derive an expression for the mutual inductance of two long coaxial solenoids of the same length wound one over the other.

[OR. prove that M12 = M21]

Answer:

Mutual Inductance: It is numerically equal to the magnetic flux linked with one coil (secondary coil) when unit current flows through the other coil.

Derivation:

image

For I coil, <|)1 = M12, I2 (1)

where <j)1 is the flux through I coil.

Similarly,

1 = (B2.A1)N1

= \(\)

= \(\) (2)

From (1) and (2)

Now, 2 = M21I1 (3)

where ())2 is the flux through 2nd coil due to Ist coil.

Also. <|)-, = B1 A1 N2 cos0

= \(\) (4)

(0 = 0°)

From (3)and (4)

∴ M12 = M21

CBSE Class 12 Physics Chapter 6 Class 12 Electromagnetic Induction Long Questions And Answers

Question 1. The emf induced across the ends of a conductor due lo its motion in a magnetic field is called motional emf. It is produced due to the magnetic Lorcnlz force acting on the free electrons of the conductor. For a circuit shown in figure, if a conductor of length f moves with velocity v in a magnetic field B perpendicular to both its length and the direction of the magnetic field, then all the induced parameters are possible in the circuit.

image

(1). Which rule is used to find direction of induced current?

Answer:

Fleming’s right hand rule

(2). Bicycle generator creates 1 .5 V at 15 km/hr. Find the value of EMF generated at 10 km/hr ?

Answer:

e = NABo

e

x = 1 volt.

(3). A O.i m long conductor carrying a current of 50 A is held perpendicular to a magnetic field of 1.25 mT. Find the mechanical power required to move the conductor with a speed of 1 m s-1 ?

Answer:

P = F x v = (IlB)v = 50 x 0.1 x 1.25 x 10-3 x 1

= 6.25 x 10-3 W = 6.25 mW

(4). A conducting rod of length f is moving in a transverse magnetic field of strength B with velocity v. The resistance of the rod is R. Find the expression of induced current in the rod?

Answer:

Question 2. Mutual inductance is the phenomenon of inducing cmf in a coil, due to a change of current in the neighboring coil. The amount of mutual inductance that Jinks one coil to another depends very much on the relative positioning of the two coils, their geometry and relative separation between them. Mutual inductance between the two coils increases pr, limes if the coils arc wound over an iron core of relative permeability pr.

image

(1). A short solenoid of radius a, number of turns per unit length n, and length L is kept coaxially inside a very long solenoid of radius b, number of turns per unit length n2. Write the expression for mutual inductance of the system?

Answer:

p07ta2n,n2L

(2). If a change in current of 0.01 A in one coil produces a change in magnetic flux of 2 x 10-2 Weber in another coil. Find the mutual inductance between coils?

Answer:

= MI

M = 2H

(3). How can mutual inductance between two coils can be increased?

Answer:

By increasing the number of turns in the coils

M N2

(4). When a sheet of iron is placed in between the two co-axial coils. How the mutual inductance between the coils will change?

Answer:

Increases

M r

Question 3. Explain the meaning of the term mutual inductance. Consider two concentric circular coils. one of radius R and the other of radius r (R > r) placed coaxially with centres coinciding with each other. Obtain the expression for the mutual inductance of the arrangement.

A rectangular coil of area A, having number of turns N is rotated at T revolutions per second in a uniform magnetic field B, the field being perpendicular to the coil. Prove that the maximum emf induced in the coil is 2 7tf NBA.

Answer:

1. Let current (I1) is flowing in first outer coil. The magnetic field at centre of coil

Flux passing through second inner coil

2 = B1 A2

From definition of mutual induction

2 = M21 I1

Similarly \(M_{12}=\frac{\mu_0 \pi r^2}{2 R}\)

M12 = M21 = M

2. Magnetic flux linked with coil = <j) = NBAcosG

The induced cmf.e = -d(|)/dt (∵ \(\omega =\frac{\mathrm{d} \theta}{\mathrm{dt}}\\) )

For maximum induced emf, sin0 = 1

∴ e = NBA(2f) => e = 27ilNBA

Question 4.

Draw a labelled diagram of an ac generator. Obtain the expression for the cmf induced in the rotating coil of N turns each of cross-sectional area A, in the presence of a magnetic field B.

A horizontal conducting rod 10 m long extending from cast to west is falling with a speed
5.0 ms-1 at right angles to the horizontal component of the Earth’s magnetic field,
0.3 x 10-4 Wb m-2. Find the instantaneous value of the emf induced in the rod.

Answer:

1. image

B1 and B2 are carbon brushes. S1 and S2 are slip rings.

Let at any instant (t), the angle between \(\vec{B}\) and \(\vec{A}\) is 0, then the magnetic flux linked with coil at that instant.

= NBA cos

N = Number of turns in the coil

A = Area of the coil

0) = Angular velocity of coil

[ ∵ 0 = rot ]

cj)B = NBA cos rot

Thus induced emf, \(\mathrm{e}=-\frac{\mathrm{d} \phi_{B}}{\mathrm{dt}}\)

or e = NBA sint [ ∵ = t]

2. Given: L = 10 m, v = 5 m/s, BH = 0.3 x 10-4 Weber

Induced emf (e) = BH vL

= 0.3 x 10-4 x 5 x 10

= 15 x 10-4 volt

= 1.5 mV

Electromagnetic Induction Class 12 PDF

Question 5.

When a bar magnet is pushed towards (or away) from the coil connected to a galvanometer. the pointer in the galvanometer deflects. Identify the phenomenon causing this deflection and write the factors on which the amount and direction of the deflection depends. Slate the laws describing this phenomenon.

Sketch the change in flux and emf when a conducting rod PQ of resistance R and length l moves freely to and fro between A and C with speed v on a rectangular conductor placed in uniform magnetic field as shown in the figure.

image

Answer:

1. When a bar magnet pushed towards or away from coil, magnetic flux passing through coil
change with lime and causes an induced emf. This phenomenon is called EMI.

Induced emf in the coil is given as

Direction and amount of deflection depend on the motion of magnet whether it is moving
towards the coil or away from the coil and the speed with which it moves.

Farady’s law of Electromagnetic Induction:

Whenever there is a change in magnetic flux linked with a coil, an emf is induced in the coil. The induced emf is proportional to the rate of change of magnetic flux linked with the coil.

Where K is constant and negative sign represents oppositions to change in flux. In SI system is in weber, t in seconds, in volt and K = 1

∴ \(\varepsilon=-\frac{\mathrm{d} \phi}{\mathrm{dt}}\)

If the coil has N turns, then

2. Case 1 : When PQ moves forward

(1). For 0 x < b

Magnetic field, B exists in the region

∴ Area of loop PQRS = l x

∴ Magnetic flux linked with loop PQRS.

<]) = BA = Blx

(|) = Blx (1) [b > x 0]

(2). For 2b > x > b

B =0

∴ Flux linked with loop PQRS is uniform and given by

‘ = Blb (x = b)

Forward jounery

Thus, for 2b x bar

Flux, = Bbl [constant]

Return journey

For b x 2b

Flux, = constant = Bbl [Decreasing]

Graphical representation

case – 2 For b> x 0

As = Blx = \(\frac{\mathrm{d} \phi}{\mathrm{dt}}=\frac{\mathrm{Bl}\mathrm{dx}}{\mathrm{dt}}=\mathrm{Bv} l\) ( ∵ \mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}\(\))

Induced emf, e = \(\mathrm{c}=-\mathrm{d} \phi / \mathrm{d} \mathrm{l}=-\mathrm{vB}\left(\Rightarrow \frac{\mathrm{d} \phi^{\prime}}{\mathrm{dt}}=0\right)\)

For 2b x b

As = bb => c = 0

Forward journey,

For b > x 0

c = -vbl,

For 2b x b.e= 0

Return journet,

For b > x 0

e= vBl

For 2b x n2e=0

image

CBSE Class 12 Physics Chapter 14  Semiconductor Electronics Multiple Choice Questions And Answers

Question 1. Vm is the maximum voltage between the ends of the secondary terminal of a transformer used in a half-wave rectifier. When the PN junction diode is reverse-biased, what will be the potential difference between the two ends of the diode?

1. Zero
2. \(\frac{V_m}{\sqrt{2}}\)
3. V_m
4. 2 V_m

Answer: 2. \(\frac{V_m}{\sqrt{2}}\)

Read and Learn More Important Questions for Class 12 Physics with Answers

Question 2. The band gaps of an insulator, semiconductor, and conductor arc respectively Eg₁, Eg_2, and Eg₃ The relationship between them can be given as _______.

1. Eg₁ = Eg₂ = Eg₃
2. Eg₁ < Eg₂ < Eg₃
3. Eg₁ > Eg₂ > Eg₃
4. Eg₁ < Eg₂ > Eg₃

Answer: 3. Eg₁ > Eg₂ > Eg₃

Question 3. A potential barrier of 0.2V exists across a pn junction. If the depletion region is 5.0 x 10^-7 m wide, then the intensity of the electric field in this region is ___ V/m

1. 1 x 10⁵
2. 4 x 10⁵
3. 1 x 10⁶
4. 2 x 10⁵

Answer: 2. 4 x 10⁵

Question 4. When will the conductivity of a Ge semiconductor decrease?

1. On adding acceptor impurity
2. On adding donor impurity
3. On making UV light incident
4. On decreasing the temperature

Answer: 4. On decreasing the temperature

Question 5. In an n-type semiconductor, which of the following statements is true:

1. Electrons are the majority carriers and tri-valent atoms are the dopants.
2. Electrons are the majority carriers and pentavalent atoms are the dopants.
3. Holes are minority carriers and pentavalent atoms are the dopants.
4. Holes are majority carriers and tri-valent atoms arc the dopants.

Answer: 2. Electrons arc majority carriers and pentavalent atoms are the dopants.

Question 6. When a forward bias is applied to a p-n junction, it

1. Raises the potential barrier
2. Reduces the majority carrier current to zero
3. Lowers the potential barrier
4. None of the above

Answer: 3. Lowers the potential barrier

Question 7. In half-wave rectification, what is the output frequency if the input frequency is 50 Hz?

1. 0
2. 100 Hz
3. 50 Hz
4. 25 Hz

Answer: 3. 50 Hz

Question 8. _______ as an impurity, when added in Si or Gc P-type semiconductor is obtained.

1. Antimony
2. Arsenic
3. Phosphorus
4. Boron

Answer: 4. Boron

Question 9. In an intrinsic semiconductor, the density of free electrons is nc and the number density of holes is nh then ________.

1. n_c >> n_h
2. n_c = 2n_h
3. n_c = n_h
4. n_c << n_h

Answer: 3. n_c = n_h

Question 10. In an unbiased p-n junction, holes diffuse from the p-region to the n-region because,

1. Free electrons in the n-region attract them
2. They move across the junction by the potential difference
3. Hole concentration in the p-region is higher as compared to the n-region
4. All the above

Answer: 3. Hole concentration in the p-region is higher as compared to the n-region

CBSE Class 12 Physics Chapter 14  Semiconductor Electronics Short Questions And Answers

Question 1.

1. How does the energy gap of an intrinsic semiconductor change when doped with a trivalent impurity?
2. In a p-n junction under equilibrium, there is no net current. Why?

Answer:

1. Then some allowed energy levels are produced, situated in the energy gap slightly above the valence band, this arc is called acceptor energy levels. Due to this energy band gap decreases.
2. Net current (diffusion and drift current) is zero for both electrons and holes because the diffusion current is equal and opposite to the drift current for both carriers.

Question 2. Explain the formation of the barrier potential in a p-n junction.

Answer:

The loss of electrons from the n-region and the gain of electrons by the p-region causes a difference in potential across the junction of the two regions.

• The polarity of this potential is such as to oppose the further flow of earners so that a condition of equilibrium exists.
• The p-n junction at equilibrium and the potential across the junction. The n-material has lost electrons, and the p-material has acquired electrons. The n-material is thus positive relative to the p-material.

Since this potential tends to prevent the movement of electrons from the n region into the p region, it is often called a barrier potential.

Question 3.

1. Draw the circuit diagram of an illuminated photodiode and its I-V characteristics.
2. How can a photodiode be used to measure the light intensity?

Answer:

2. The photocuring is directly proportional to the intensity of light this can be used for measuring the intensity of incident light.

Question 4.

1. In the following diagram, is the junction diode forward-biased or reverse-biased?

2. Name two important processes that occur during the formation of a pn junction.

Answer:

1. Reverse biased as the p-side is connected to the lower potential and the n-side to the higher potential.

Two important phenomena occur during the formation of the pn junction:

2. (1) Diffusion: Due to the concentration gradient the majority of charge carriers from the p side (holes) and the majority of charge carriers (electrons) on the n side start diffusing towards their opposite side.

(2) Drift: As electron-hole pairs neutralize near the junction they leave behind immobile ions, these ions create an electric field from the n side to the p side due to which the minority charge carrier starts moving (drifting).

Question 5. Draw the circuit diagram of a half-wave rectifier and explain its working.

Answer:

Working of half wave rectifier: The secondary coil of a transformer supplies the desired alternating voltage across terminals A and B. When the voltage at A is in a positive cycle, the diode is in forward bias and it conducts current.

When A is at a negative cycle, the diode is in reverse bias and it does not conduct current only in the positive half-cycle of alternative current (ac), there is a current through the load resistor R_L, and we gel output voltage.

Thus, We get a half-wave rectified output. So this circuit is called a half-wave rectifier.

Circuit Diagram:

Graphy:

Question 6. Draw the circuit diagram of a full wave rectifier along with the input and output waveforms. Briefly explain how the output voltage/current is unidirectional.

Answer:

Working: When the positive half cycle of input a.c. signal flows through the primary coil, induced emf is set up in the secondary coil due to mutual induction. The direction of induced emf is such that the upper end of the secondary coil becomes positive while the lower end becomes negative.

• Thus, diode D₁ is forward biased and diode D₂ is reverse biased, so the current due to diode D₁ flows through the circuit. The output voltage which varies per the input half cycle is obtained across the load resistance (R_L).
• During the negative half cycle of input a.c. signal, diode D₁ is reverse biased and diode D₂ is forward biased. The current due to diode D₂ flows through the circuit. The output voltage is obtained across the load resistance (RL ). The input and corresponding output voltage are predicted.

Since both the halves of input a.c. (wave) arc rectified, so the junction diode is called a full wave rectifier.

Circuit Diagram:

Graph:

Question 7. Distinguish between a conductor and a semi-conductor based on the energy band diagram.

[E_g = Forbidden Energy gap]

Question 8. Show on a plot, the variation of resistivity of

1. A conductor, and
2. A typical semiconductor as a function of temperature.

Using the expression for the resistivity in terms of number density and relaxation time between the collisions, explain how resistivity in the case of a conductor increases while it decreases in a semiconductor, with the rise of temperature.

Answer:

Variation of resistivity With Temperature

The resistivity of a material is given by

∴ \(\rho=\frac{m}{n e^2 \tau}\)

n ⇒ Number density

τ ⇒ Relaxation time

On increasing the temperature of the conductor relaxation time decreases while the number density remains constant. Due to this resistivity of the conductor increases.

In the case of semiconductors, on increasing the temperature relaxation time decreases but the number density increases. An increase in the number density is more effective than a decrease in the relaxation time. That’s why the resistivity of semiconductors decreases.

Question 9. Explain with the help of a diagram the formation of depletion region and barrier potential in a p-n junction.

Answer:

It is clear that the N-type semiconductor has a P-TYPC excess of free electrons and the P-type has an excess of holes therefore joined through a special process, electrons migrate towards the P-side and holes migrate towards the N-side due to concentration gradient.

The departure of an electron from the N-side to the P-side leaves a positive donor ion on the N-side and likewise hole leaves a negative acceptor ion on the P-side resulting in the formation of a depletion layer having widths 10^-7m.

Depletion layer: It is the layer near the junction in which electrons arc absent on the n side and holes are absent on the p side.

Potential barrier: Due to the accumulation of immobile ions near the junction an electric potential difference (V_b) develops between the n side and the p side which acts as a barrier for further diffusion of electrons and holes.

V_b = E_i x d (volt)

Question 10. Draw the energy band diagrams for conductors, semiconductors, and insulators. Which band determines the electrical conductivity of a solid? How is the electrical conductivity of a semiconductor affected by the rise in its temperature? Explain.

Answer:

Energy Band Diagram:

The conduction band determines the electrical conductivity of the solid.

On increasing the temperature of semiconductor electrical conductivity increases. When we increase the temperature, some of the covalent bonds get broken and e-hole pairs are generated.

CBSE Class 12 Physics Chapter 14  Semiconductor Electronics Long Questions And Answers

Question 1. Consider a thin p-type silicon (p-Si) semiconductor wafer. By adding precisely a small quantity of pentavalent impurity, part of the p-Si wafer can be converted into n-Si. There are several processes by which a semiconductor can be formed.

The wafer now contains p-region and nregion and a metallurgical junction between p-, and n- region. Two important processes occur during the formation of a p-n junction: diffusion and drift.

We know that in an n-type semiconductor, the concentration of electrons (number of electrons per unit volume) is more compared to the concentration of holes. Similarly, in a p-type semiconductor, the concentration of holes is more than the concentration of electrons.

During the formation of the p-n junction, and due to the concentration gradient across p-, and n- sides, holes diffuse from p- side to n-sidc (p → n) and electrons diffuse from n-sidc to p-sidc (n → p). This motion of charge carriers gives rise to diffusion current across the junction.

(1). How can a p-type semiconductor be converted into an n-type semiconductor?

1. Adding pentavalent impurity
2. Adding trivalent impurity
3. Not possible
4. Heavy doping

Answer: 1. Adding pentavalent impurity

(2). Which of the following is true about n-type semiconductors?

1. The concentration of electrons is less than that of holes.
2. The concentration of electrons is more than that of holes.
3. The concentration of electrons is equal to that of holes.
4. None of the above.

Answer: 2. Concentration of electrons is more than that of holes.

(3). Which of the following is the reason for diffusion current?

1. Diffusion of holes from p to n
2. Diffusion of electrons from n to p
3. Both (1) and (2)
4. None of these

Answer: 3. Both (1) and (2)

(4). What are the processes that occur during the formation of a p-n junction?

1. Drift
2. Diffusion
3. Both (1) and (2)
4. None of these

Answer: 3. Both (1) and (2)

CBSE Class 12 Physics Chapter 13 Nuclei Multiple Choice Questions And Answers

Question 1. If the radii \({ }_{13}^{27} \mathrm{Zn}\) abd \({ }_{30}^{64} \mathrm{Zn}\) nucleus are R1 and R2 respectively, Then \(\frac{R_1}{R_2}\) = ________.

1. \(\frac{27}{64}\)
2. \(\frac{3}{4}\)
3. \(\frac{9}{16}\)
4. \(\frac{13}{30}\)

Answer: 2. \(\frac{3}{4}\)

Read and Learn More Important Questions for Class 12 Physics with Answers

⇒ \(R=R_0 A^{\frac{1}{3}}\)

∴ \(\frac{R_1}{R_2}=\left(\frac{A_1}{A_2}\right)^{1 / 3}=\left(\frac{27}{64}\right)^{1 / 3}=\frac{3}{4}\)

Question 2. In the radioactive transformation \({ }_z^A X \rightarrow{ }_{Z-1}^A X_1 \rightarrow{ }_{Z-1}^{A-4} X_2 \rightarrow_Z^{A-4} X_3\) which are the successively emitted radioactive radiations?

1. β^–, α, β^–
2. α, β^–, β^–
3. β^–, β^–, α
4. α, α, β^–

Answer: 1. β^–, α, β^–

Question 3. Which are the isotone, isobar and isotope nuclei respectively of \({ }_6^{12} \mathrm{C}\) from among \({ }_6^{14} \mathrm{C},{ }_5^{12} \mathrm{~B},{ }_7^{13} \mathrm{~N}\)?

1. \({ }_5^{12} \text { B, }{ }_6^{14} \mathrm{C,} { }_7^{13} \mathrm{~N}\)
2. \({ }_6^{14} \mathrm{C},{ }_7^{13} \mathrm{~N},{ }_5^{12} \mathrm{~B}\)
3. \({ }_7^{13} \text { N, }{ }_5^{12} \mathrm{B,} { }_6^{14} \mathrm{~C}\)
4. \({ }_6^{14} \mathrm{C},{ }_5^{12} \mathrm{~B,} { }_7^{13} \mathrm{~N}\)

Answer: 3. \({ }_7^{13} \text { N, }{ }_5^{12} \mathrm{B,} { }_6^{14} \mathrm{~C}\)

Question 4. The binding energy per nucleon is almost constant for the nuclei having atomic mass number ________.

1. 30 < A < 240
2. 30 < A < 170
3. 170 < A < 230
4. 156 < A < 192

Answer: 2. 30 < A < 170

Question 5. The energy equivalent to 1 gram (g) substance is _________ J.

1. 9 x 10⁷
2. 9 x 10¹⁰
3. 9 x 10¹³
4. 9 x 10⁸

Answer: 3. 9 x 10¹³

∴ \(E=m c^2=\frac{1}{1000} \times 9 \times 10^{16}=9 \times 10^{13} \mathrm{~J}\)

Question 6. \({ }_{80}^{198} \mathrm{Hg} \text { and }{ }_{79}^{197} \mathrm{Au}\) are example of __________.

1. Isotopes
2. Isomers
3. Isobars
4. Isotones

Answer: 4. Isotones

Question 7. In a curve of binding energy per nucleon versus mass number (A), the maximum value of Ebn is 8.75 MeV/nucleon the value of a corresponding atomic mass number is _______.

1. 56
2. 235
3. 238
4. 171

Answer: 1. 56

Question 8. The saturation property of the nuclear forces is because they are:

1. Charge independent forces
2. Non-central forces
3. Spin-dependent forces
4. Short-range forces

Answer: 4. Short-range forces

CBSE Class 12 Physics Chapter 13 Nuclei Assertion And Reason

For question numbers 1 to 5 two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (1), (2), (3), and (4) as given below.

1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is NOT the collected explanation of A
3. A is true but R is false
4. A is false and R is also false

Question 1. Assertion: Energy is released in nuclear fission.

Reason: The total binding energy of the fission fragments is larger than the total binding energy of the parent.

Answer: 1. Both A and R are true and R is the correct explanation of A

Question 2. Assertion: The nucleus may emit negatively charged particles.

Reason: The nucleus contains a negative charge also.

Answer: 3. A is true but R is false

Question 3. Assertion: Nuclear binding energy per nucleon is in the order \({ }_4^9 \mathrm{Be}>{ }_3^7 \mathrm{Li}>{ }_2^4 \mathrm{He}\)

1. Reason: Binding energy per nucleon increases linearly with differences in several neutrons and protons.
2. Answer: 4. A is false and R is also false

Question 4. Assertion: Isobars arc the clement having the same mass number but a different atomic number.

Reason: Neutrons and protons are present inside the nucleus.

Answer: 2. Both A and R are true but R is NOT the collected explanation of A

Question 5. Assertion: The density of all the nuclei is the same.

Reason: The radius of the nucleus is directly proportional to the cube roots of the mass number

Answer: 1. Both A and R are true and R is the correct explanation of A

CBSE Class 12 Physics Chapter 13 Nuclei Short Questions And Answers

Question 1. Obtain approximately the ratio of the nuclear radii of the gold isotope \({ }_{79} \mathrm{Au}^{197}\) and silver isotopes \({ }_{47} \mathrm{Ag}^{107}\)

Answer:

A₁ = 197 and A₂ = 107 R = R₀A^1/3

⇒ \(\frac{R_1}{R_2}=\left[\frac{A_1}{A_2}\right]^{1 / 3}=\left[\frac{197}{107}\right]^{1 / 3}=1.225\)

Question 2. The relation R = R₀ A^1/3. where R₀ is a constant and A is the mass number of A nucleus, showing that the nuclear matter density is nearly constant. (i.e independent of A)

Answer:

Density of nucleus matter = \(\frac{\text { Mass of nucleus }}{\text { Volume of nucleus }}\)

⇒ \(\rho=\frac{\mathrm{mA}}{\frac{4}{3} \pi \mathrm{R}^3}\) [R = R A1/3]

⇒  \(\rho=\frac{3 \mathrm{~m}}{4 \pi \mathrm{R}_0^3}\) = \(\rho=\frac{\mathrm{A} \times 1.66 \times 10^{27} \mathrm{~kg}}{\frac{4}{3} \times 3.14 \times\left(1.2 \times 10^{-15}\right)^3 \mathrm{~A}} \approx 10^{17} \mathrm{~kg} / \mathrm{m}^3\)

The density of the nucleus is constant.

Question 3. Draw a graph showing the variation of the potential energy of a pair of nucleons as a function of their separation to indicate the region in which the nuclear force is (1) Attractive, (2)Repulsive

Answer:

The potential energy is minimum at a distance r() of about 0.8 fm. This means that the force is attractive for distances larger than 0.8 fm and repulsive if they are separated by distances less than 0.8 fm.

Question 4. What is nuclear force? Write its specific properties.

Answer:

Nuclear Force: The nuclear force is the strongest force of nature that holds nucleons in the nucleus of an atom.

Properties: Nuclear forces act between a pair of neutrons, a pair of protons, and a neutron-proton pair with the same strength, i.e. nuclear force is charge-independent.

1. The nuclear forces are very short-range forces, hence show saturation property.
2. The nuclear forces are dependent on the spin angular momentum of nucleons.
3. Nuclear forces arc non-central forces.

Question 5. Differentiate between nuclear fission and fusion. Give an example of each. Which of the above reactions take place in a nuclear reactor?

Answer:

1. Nuclear Fission: Nuclear fission is the phenomenon of splitting a heavy nucleus (usually A > 230) into two or more lighter nuclei.

Example: \({ }_{92} \mathrm{U}^{235}+{ }_0 \mathrm{n}^1 \rightarrow_{92} \mathrm{U}^{236} \rightarrow_{56} \mathrm{Ba}^{141}+{ }_{36} \mathrm{Kr}^{92}+3 { }_0\mathrm{n}^1+\mathrm{Q}\)

2. Nuclear Fission: The process of combining of two lighter nuclei to form one heavy
nucleus is called nuclear fusion.

Example: \(4_1 \mathrm{H}^1 \rightarrow{ }_2 \mathrm{He}^4+2_1 \mathrm{e}^0+2 v+26.7 \mathrm{MeV}\)

A nuclear reactor is based on a controlled chain reaction.

Question 5.

1. Explain the processes of nuclear fission and fusion using the plot of binding energy per nucleon (BE/A) versus the mass number A.
2. Why are neutrons preferred as better projectiles in causing nuclear reactions?

Answer:

1. The value of binding energy per nucleon gives a measure of the stability of that nucleus greater the binding energy per nucleon of a nucleus, the more stable the nucleus.

The above graph shows, the binding energy per nucleon drawn against mass number A. The saturation effect of nuclear forces is properly responsible for the approximate constancy of binding energy in the range 30 < A < 170.

It is clear from the curve that the binding energy per nucleon of the fused nuclei is more than that of the lighter nuclei taking part in nuclear fusion. Hence energy is released in this process.

In nuclear fission, the sum of the masses of the final products is less than the sum of the masses of the reactant components. Hence energy is released in this process.

2. These are uncharged particle

Question 6.

1. Write the basic nuclear process involved in the emission of β⁺ in a symbolic form, by a radioactive nucleus.
2. In the reactions given below:
   1. \({ }_6^{11} \mathrm{C} \rightarrow{ }_y^{z} \mathrm{B}+x+v\)
   2. \({ }_6^{12} \mathrm{C}+{ }_6^{12} \mathrm{C} \rightarrow{ }_{\mathrm{a}}^{20} \mathrm{Ne}+{ }_1^{\mathrm{c}} \mathrm{He}\)

Answer:

1. \({ }_z X^{A} \rightarrow{\beta^+}{ }_{z-1} Y^{A}+{ }_{+1} e^{0}+v+\text { energy }\)

In the emission of β⁺, mass no. (A) remains the same but atomic no. (Z) decreases by 1 unit.

2. (1). \(x \longrightarrow{ }_{+1} e^0, y \longrightarrow 5, z \longrightarrow 11\)

(2). \(\mathrm{a} \longrightarrow 10, \mathrm{~b} \longrightarrow 2, \mathrm{c} \longrightarrow 4\)

CBSE Class 12 Physics Chapter 13 Nuclei Long Questions And Answers

Question 1. When the nucleus of an atom splits into lighter nuclei through a nuclear reaction the process is termed as nuclear fission. This decay can be natural spontaneous splitting by radioactive decay or can be simulated in a lab by achieving the necessary conditions (bombarding with neutrons). The resulting fragments tend to have a combined mass that is less than the original.

(1). This splitting of a nucleus into smaller nuclei is:

1. Fusion
2. Fission
3. Half-life
4. Gamma-radiation

Answer: 2. Fission

(2). Name the moderator used in the nuclear reactor.

1. Plutonium
2. Thorium
3. Graphite
4. Berilium

Answer: 3. Graphite

(3). Which isotope of Uranium can sustain the chain reaction?

1. U-230
2. U-235
3. U-245
4. U-225

Answer: 2.U-235

(4). Which of the following atomic particles has the least mass?

1. Proton
2. Electron
3. Deuleron
4. Neutron

Answer: 2. Electron

Question 2. Uranium-235 (²³⁵U) is an isotope of uranium making up about 0.72 % of natural uranium. It is a fissile material i.c., and it can sustain a fission chain reaction. U-235 has a half-life of 703.8 million years. The main use for uranium today is for fuel in nuclear power plants. Depleted uranium is used in bullets and larger projectiles to make them hard and dense enough to reach through armored targets. Inhaling large concentrations of this radioactive element can cause lung cancer. It is also a toxic chemical, its ingestion can cause damage to kidneys, and its radioactive properties could cause cancers of bones or liver.

(1). A nuclei having the same number of neutrons but a different number of protons/atomic numbers and called:

1. Isobars
2. Isomers
3. Isotones
4. Isotopes

Answer: 3. Isotones

(2). For a nuclear fission process, suitable nuclei are:

1. Any nuclei
2. Heavy nuclei
3. Lighter nuclei
4. Nuclei lying in the middle of the periodic table

Answer: 2. Heavy nuclei

(3). Mass-energy equation was propounded by:

1. Newton
2. Madam Curie
3. C.V. Raman
4. Einstein

Answer: 4. Einstein

(4). Name the antiparticle of electron:

1. Positro
2. α-Particle
3. proton
4. β- Particle

Answer: 1. Positro

Class 12 Physics Chapter 8 Electromagnetic Waves (Emw) Multiple Choice Questions And Answers

Class 12 Physics Chapter 8 Question 1. In a certain region electric field \(\vec{E}\) and magnetic field \(\vec{B}\) are perpendicular to each other. An electron enters the region perpendicular to the direction of both E and B and moves undeflected. The speed of the electron is _________.

1. \({\vec{E}} \cdot {\vec{B}}\)
2. \(|{\vec{E}} \times {\vec{B}}|\)
3. \(\frac{|{\vec{E}}|}{|{\vec{B}}|}\)
4. \(\frac{|{\vec{B}}|}{|{\vec{E}}|}\)

Answer: 1. \({\vec{E}} \cdot {\vec{B}}\)

Read and Learn More Important Questions for Class 12 Physics with Answers

Question 2. Which wave travels with the speed of light _________

1. Sound wave
2. Heatwave
3. Shock wave
4. Microwave

Answer: 4. Microwave

Question 3. A television network uses _________

1. Microwaves
2. High-Frequency Radio waves
3. Light waves
4. Sound waves

Answer: 2. High-frequency radio waves

Question 4. Light waves are _______

1. Longitudinal
2. Transverse
3. Both Longitudinal and transverse
4. Mechanical

Answer: 4. Mechanical

Question 5. An electromagnetic wave passing through the space is given by equations E = E₀ sin (ωt – kx), B = B₀ sin (ωt- kx) which of the following is true?

1. E₀B₀ = ωk
2. E_0ω = B₀k
3. E₀k = B_0ω
4. E_0ωk = B₀

Answer: 3. E_0ω = B₀k

Question 6. The wavelength range of Heal waves is __________

1. 400 nm to 1 nm
2. 1 mm to 700 nm
3. 0. 1 m to 1 mm
4. 700 nm to 400 nm

Answer: 2. 1 mm to 700 nm

Question 7. The maximum value of B in an electromagnetic wave is equal to 6 x 10^-8T. Thus the maximum value of \(\vec{E}\) is __________.

1. 2 Vm^-1
2. 18 Vm^-1
3. 2.5 Vm^-1
4. 6 Vm^-1

Answer: 2. 18 Vm^-1

E = cB

E = 3 x 10⁸ x 6 x 10^-8

E = 18 Vm^-1

Question 8. Two oppositely charged particles oscillate about their mean equilibrium position in free space, with a frequency of 10⁹ Hz. The wavelength of the corresponding electromagnetic wave produced is _______.

1. 0.3 m
2. 3 x 10¹⁷ m
3. 10⁹ m
4. 3.3 m

Answer: 1. 0.3 m

⇒ \(\lambda=\frac{c}{v}=\frac{3 \times 10^8}{10^9}\)

∴ 0.3 m

Question 9. For a radiation of 6 GHz passing through air, the wave number (number of waves) per 1 m length is _______ (1 GHz = 10⁹ Hz)

1. 5
2. 3
3. 20
4. 30

Answer: 3. 20

Wave No. \(\bar{v}=\frac{1}{\lambda}\) ∵ \(\left[v=\frac{c}{\lambda}\right]\) \(\left[\frac{c}{v}=\frac{1}{\lambda}\right]\)

∴ \(\bar{v}=\frac{v}{c}=\frac{6 \times 10^9}{3 \times 10^8}=20\)

Question 10. Which one of the following is an equation of magnetic energy density?

1. \(\frac{B^2}{2 \mu_0}\)
2. \(\frac{1}{2} \mu_0 B^2\)
3. \(\frac{2 B^2}{\mu_0}\)
4. \(\frac{\mathrm{B}^2}{\mu_0}\)

Answer: 1. \(\frac{B^2}{2 \mu_0}\)

Question 11. Dimension of \(\) is same as dimension of ( where μ = magnetic constant, ε = Dielectric constant)

1. Velocity
2. Square of velocity
3. Acceleration
4. Momentum

Answer: 2. Square of velocity

Class 12 Physics Chapter 8 Electromagnetic Waves Assertion And Reason

For questions numbers 1 to 10 two statements are given labeled Assertion (A) and the other labeled Reason (R). Select the correct answer to these questions from the codes (1), (2), (3), and (4) as given below.

1. Both (A) and (R) are correct, (R) is the correct explanation of (A).
2. Both (A) and (R) are correct, (R) is not the correct explanation of (A).
3. (A) is correct; (R) is incorrect.
4. (A) is incorrect; (R) is incorrect

Question 1. Assertion: Electromagnetic waves do not require a medium for their propagation.

Reason: They cannot travel in a medium.

Answer: 3. (A) is correct; (R) is incorrect.

Question 2. Assertion: A changing electric field produces a magnetic field.

Reason: A changing magnetic field produces an electric field.

Answer: 2. Both (A) and (R) are correct, (R) is not the correct explanation of (A).

Question 3. Assertion: X-rays travel with the speed of light.

Reason: X-rays are electromagnetic rays.

Answer: 1. Both (A) and (R) are correct, (R) is the correct explanation of (A).

Question 4. Assertion: Environmental damage has increased the amount of ozone in the Atmosphere.

Reason: The increase of ozone increases the amount of ultraviolet radiation on the earth.

Answer: 4. (A) is incorrect; (R) is incorrect

Question 5. Assertion: Electromagnetic Radiation exerts pressure.

Reason: Electromagnetic waves carry both momentum and energy.

Answer: 1. Both (A) and (R) are correct, (R) is the correct explanation of (A).

Question 6. Assertion: The EM waves of shorter wavelengths can travel longer distances than those of longer wavelengths.

Reason: The shorter the wavelength, the larger the velocity of propagation.

Answer: 3. (A) is correct; (R) is incorrect.

Question 7. Assertion: EM waves follow the Superposition principle.

Reason: Differential expression of EM wave is linear.

Answer: 1. Both (A) and (R) are correct, (R) is the correct explanation of (A).

Question 8. Assertion: Sound waves cannot travel in a vacuum, but light waves can.

Reason: Light is an electromagnetic wave – but sound is a mechanical wave.

Answer: 1. Both (A) and (R) are correct, (R) is the correct explanation of (A).

Question 9. Assertion: Microwaves are better carriers of signals than radio waves.

Reason: Electromagnetic waves do not require any medium to propagate.

Answer: 2. Both (A) and (R) are correct, (R) is not the correct explanation of (A).

Question 10. Assertion: Transverse waves are not produced in liquids and gases.

Reason: The shorter the wavelength, the larger the velocity of propagation in air.

Answer: 3. (A) is correct; (R) is incorrect.

Class 12 Physics Chapter 8 Electromagnetic Waves Short Questions And Answers

Question 1. The electric field of an electromagnetic wave is represented as E_x = E₀ sin (ωt + kz).

1. In which direction is the wave propagating?
2. In which direction does the magnetic field oscillate?

Answer:

1. Negative z direction
2. y direction

Question 2.

1. In which situation is there a displacement current but no conduction current?
2. Why are Microwaves considered suitable for radar systems used in aircraft navigation?

Answer:

1. In between the plates of a capacitor, during charging and discharging of a capacitor.
2. Microwaves have energy more than radio waves, so these can travel up to greater distances.

Question 3. Match the column

Answer: 1-C, 2-D, 3-B, 4-A, 5-E

Question 4.

1. Suppose that the earth’s atmosphere is absent, will the average temperature on the earth’s surface
   be higher or lower than what it is at present?
2. What is an electromagnetic constant?

Answer:

1. The average temperature will be lower due to the absence of the greenhouse effect.
2. All types of electromagnetic waves move with the same speed c = 3 x 10⁸ m/s in air or vacuum, so ‘c’ is called the electromagnetic constant.

Question 5.

1. The charging current for a capacitor is 0.25 A. What is the displacement current across its plates?
2. How are infrared waves produced? Write their one important use.

Answer:

1. 0.25 A
2. Infrared waves are produced by hot bodies due to molecular vibrations. These are used to
   treat muscular strain.

Question 6.

1. Which part of the electromagnetic spectrum is used for eye surgery?
2. Which part of the electromagnetic spectrum is blocked by protective welding glass?

Answer:

1. Ultra-violet rays arc used in Lasik Laser, for eye surgery.
2. Ultra-violet.

Question 7. Slate two properties of electromagnetic waves.

Answer:

1. All EM waves travel with the same speed c = 3 x 10⁸ m/s in air or vacuum.
2. EM waves have energy and momentum and these apply radiation pressure, on the surface on which they are made to fall.

Question 8.

1. The thin Ozone layer on top of the stratosphere is crucial for human survival. Why?
2. How can we show that em waves carry momentum?

Answer:

Question 9.

1. Which component of the electromagnetic wave is responsible for producing an optical effect?
2. Light can travel in a vacuum whereas sound cannot do so. Why?

Answer:

1. The Electric vector of the em wave is responsible.
2. Light is electromagnetic while sound is a mechanical wave.

Question 10.

1. For which wavelength our eyes are most sensitive?
2. Which of the electromagnetic waves is capable of penetrating layers of dust?

Answer:

1. 555nm i.e. yellow colour.
2. Infra-red.

Question 11. Electromagnetic waves of wavelengths λ₁,λ_2, and λ₃ are used in radar systems, water purifiers, and in remote switches of TV. respectively.

1. Identify the electromagnetic waves, and
2. Write one source for each of them.

Answer:

1. In radar systems ⇒ microwaves

In water purifies ⇒ UVrays

In remote switches in TV ⇒ Infrared rays

2. Microwave arc produced by special vacuum lubes (Klystrons, Magnetrons, and Gunn diodes)

• UV radiation is produced in welding arc and the sun is an important source of ultraviolet light.
• Infrared waves are produced by hot bodies and molecules.

Class 12 Physics Chapter 8 Electromagnetic Waves Long Questions And Answers

Question 1. Gamma rays are used in radiotherapy to treat cancer. They are used to spot tumors. They kill the living cells and damage malignant tumors.

(1). What is the source of gamma rays?

1. Radioactive decay of the nucleus
2. Accelerated motion of charges in conducting wire
3. Hot bodies and molecule
4. Klystron valve

Answer: 1. Radioactive decay of the nucleus

(2). How is the wavelength of gamma rays

1. Low
2. High
3. Infinite
4. Zero

Answer: 1. Low

(3). Choose the one with the correct penetrating power order of radiation.

1. Alpha > beta > gamma
2. Beta > alpha > gamma
3. Gamma > beta > alpha
4. Gamma > alpha > beta

Answer: 3. Gamma > beta > alpha

(4). What is the other use of gamma rays?

1. Used to change white topaz to blue topaz
2. Used in aircraft navigation
3. Used to kill microbes
4. Checking fractures of bone

Answer: 1. Used to change white topaz to blue topaz

Question 2. X-rays are a form of electromagnetic radiation, similar to visible light. Unlike light, however, x-rays have higher energy and can pass through most of the objects, including the body. Medical x-rays are used to generate images of tissues and structures inside the body

(1). What is the most common method of preparation of X-rays?

1. Magnetron valve
2. Vibration of atoms and molecules
3. Bombardment of metal by high-energy electrons
4. Radioactive decay of the nucleus

Answer: 3. Bombardment of metal by high-energy electrons

(2). Which of the following sets of instruments or equipment can detect X-rays

1. Photocells, photographic film
2. Thermopiles, bolometer
3. Photographic film. Geiger tube
4. Geiger lube, the human eye

Answer: 3. Photographic film. Geiger tube

(3). Where do X-rays fall on the electromagnetic spectrum?

1. Between the UV region and infrared region
2. Between gamma rays and UV region
3. Between infrared and microwaves
4. Between microwaves and radio waves

Answer: 2. Between gamma rays and UV region

(4). What is the use of rays lying beyond the X-ray region in the electromagnetic spectrum

1. Used to kill microbes
2. Used to detect heat loss in insulated systems
3. Used in standard broadcast radio and television
4. Used in oncology, to kill cancerous cells.

Answer: 4. Used in oncology, to kill cancerous cells.

Class 12 Physics Chapter 7  Alternating Current Multiple Choice Questions And Answers

Question 1. A capacitor and an inductor are connected in two different AC circuits with a bulb glowing in each circuit. The bulb glows more brightly when:

1. The number of turns in the inductor is increased
2. The separation between the plates of the capacitor is increased
3. An iron rod is introduced into the inductor
4. A dielectric is introduced into the gap between the plates of the capacitor

Answer: 4. A dielectric is introduced into the gap between the plates of the capacitor

Read and Learn More Important Questions for Class 12 Physics with Answers

Question 2. A pure inductor of 318mH and a pure resistor of 75 Ω arc connected in series to an AC source of 50 Hz. The voltage across the 75 Ω resistor is found to be 150V. The source voltage is:

1. 150 V
2. 175 V
3. 220 V
4. 250 V

Answer: 4. 250 V

⇒ \(\cos \phi=\frac{R}{Z}=\frac{V_R}{V} \text { or } V=V_R \times \frac{Z}{R}\)

∴ \(\frac{150}{75} \sqrt{75^2+\left(3.14 \times 318 \times 10^{-3}\right)^2}=2 \sqrt{5625 \times 9970}=2 \times 124.8 \simeq 2.50 \mathrm{~V}\)

Question 3. In an AC circuit, the applied voltage and resultant current are E = E₀ sin ωt and I = I₀ sin (ωt + π/2) respectively. The average power consumed in the circuit is:

1. B₀I₀
2. \(\frac{E_0 I_0}{2}\)
3. \(\frac{E_0 I_0}{\sqrt{2}}\)
4. Zero

Answer: 4. Zero

∴ \(P_{avg}=E_{rms} I_{rms} \cos 90^{\circ}=0\)

Question 4. In a series LCR circuit, at resonance, the current is equal to_____

1. \(\frac{\mathrm{V}}{\mathrm{R}}\)
2. \(\frac{\mathrm{V}}{\mathrm{x}_{\mathrm{c}}}\)
3. \(\frac{V}{X_{L}-X_C}\)
4. \(\frac{V}{\sqrt{R^2+\left(X_L+X_C\right)^2}}\)

Answer: 1. \(\frac{\mathrm{V}}{\mathrm{R}}\)

Question 5. The frequency of an AC source for which a 10 μF capacitor has a reactance of 1000 ohm is___

1. \(\frac{1000}{\pi} \mathrm{Hz}\)
2. 50 Hz
3. \(\frac{50}{\pi} \mathrm{Hz}\)
4. \(\frac{100}{\pi} \mathrm{Hz}\)

Answer: 3. \(\frac{50}{\pi} \mathrm{Hz}\)

⇒ \(X_C=\frac{1}{2 \pi \mathrm{fC}}=1000\)

∴ \(f=\frac{50}{\pi} \mathrm{Hz} \)

Question 6. Which one of the following statements is true:

1. An inductor has infinite resistance in a DC circuit.
2. An inductor and a capacitor cannot conduct in a DC circuit
3. A capacitor can conduct in a DC circuit but not an inductor.
4. An inductor can conduct in a DC circuit but not a capacitor.

Answer: 4. An inductor can conduct in a DC circuit but not a capacitor.

Question 7. In an A.C. circuit in 1 second current reduces to zero value 120 times. Hence the frequency of A.c current is ___________ Hz.

1. 50
2. 100
3. 60
4. 120

Answer: 3. 60

Question 8. What is the r.m.s. value of the current for A.C. current I = 100 cos (200 t + 45°)A.

1. 50√2 A
2. 100 A
3. 100√2 A
4. Zero

Answer: 1. 50√2 A

∴ \(I_{\mathrm{rms}}=\frac{I_0}{\sqrt{2}}=\frac{100}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=50 \sqrt{2} \mathrm{~A}\)

Question 9. In an R-C circuit when the charge on the plates of the capacitor is increasing, the energy obtained from the sources is stored in ___________.

1. Electric field
2. Magnetic field
3. Gravitational field
4. Both Magnetic field and gravitational field

Answer: 1. Electric field

Question 10. The output power in a step-up transformer is ___________

1. Greater than the input power
2. Equal to the input power
3. Maintained even during the power cut
4. Less than the input power

Answer: 4. Less than the input power

Question 11. The power factor for scries L-R A.C. circuit is ________.

1. \(\frac{\mathrm{R}}{\mathrm{X}_{L}}\)
2. \(\frac{X_L}{R}\)
3. \(\frac{R}{\sqrt{R^2+X_L^2}}\)
4. \(\frac{\sqrt{R^2+X_L^2}}{R}\)

Answer: 3. \(\frac{R}{\sqrt{R^2+X_L^2}}\)

Question 12. An alternating voltage given as V = 200 √2 sin 100 l (V) is applied to a capacitor of 5μF. The current reading of the ammeter will be equal to _________mA.

1. 80
2. 20
3. 40
4. 100

Answer: 4. 100

⇒ \(I_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{X_c}=\frac{V_0 / \sqrt{2}}{1 /\mathrm {\omega}{c}}=20\mathrm{\omega}{c}\)

∴  200 z 100 x 5 x 10^-6 = 10^-1 A = 100 x 10^-3 A = 100 mA

Question 13. The current of \(\frac{50}{\pi}\) frequency is passing through an A.C. circuit having a series combination of resistance R = 100 Ω and inductor L = 1H, then phase difference between voltage and current is __________.

1. 60°
2. 45°
3. 30°
4. 90°

Answer: 2. 45°

⇒ \(\phi=\tan ^{-1}\left(\frac{x_1}{R}\right)=\tan ^{-1}\left(\frac{2 \pi f L}{R}\right)\)

∴ \(\phi=\tan ^{-1}\left(\frac{2 \times \pi \times\frac{50}{\pi} \times 1}{100}\right)=45^{\circ}\)

Question 14. A coil of inductance L and resistance R is connected to an A.C. source of V volt. If the angular frequency of the A.C. source is equal to co rad s^-1, then the current in the circuit will be __________.

1. \(\frac{\mathrm{V}}{\mathrm{R}}\)
2. \(\frac{\mathrm{V}}{\mathrm{L}}\)
3. \(\frac{V}{R+L}\)
4. \(\frac{V}{\sqrt{R^2+\omega^2 L^2}}\)

Answer: 4. \(\frac{V}{\sqrt{R^2+\omega^2 L^2}}\)

Question 15. In an A.C. circuit current is 2A and voltage is 220 V and power is 44 W power factor is _________.

1. 0.10
2. 0.09
3. 1.80
4. 0.18

Answer: 1. 0.10

P_virtual = VI = 220 x 2 = 440 Watt

P_avg = P_virtual cos Φ

44 = 440 x cos Φ

∴ \(\cos \phi=\frac{1}{10}=0.1\)

Question 16. A 15 μF capacitor is connected to a 220, 50 Hz a.c. source. The value of capacitive reactance is ___________.

1. 424
2. 106
3. 212
4. 21.2

Answer: 3. 212

⇒ \(X_C=\frac{1}{2 \pi f C}=\frac{1}{2 \times 3.14 \times 50 \times 1.5 \times 10^{-6}}\)

∴ X_c = 212.314

Question 17. A power transmission line feeds input power at 3300 V to a step-down transformer with its primary windings having 2000 turns. What should be the number of turns in the secondary to gel output power at 330 V?

1. 200
2. 400
3. 33
4. 40

Answer: 1. 200

⇒ \(\frac{V_s}{V_p}=\frac{N_s}{N_p}\)

⇒ \(\frac{330}{3300}=\frac{N_s}{2000}\)

⇒ \(\frac{1}{10}=\frac{N_s}{2000}\)

∴ N_s = 200

Class 12 Physics Chapter 7  Alternating Current Assertion and Reason

For question numbers 1 to 7 two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (1). (2). (3) and (4) as given below.

1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is NOT the correct explanation of A
3. A is true but R is false
4. A is false and R is also false

Question 1. Assertion: In a Series LCR circuit connected to an AC source, resonance can take place.

Reason: At resonance XL = XC

Answer: 2. Both A and R are true but R is NOT the correct explanation of A

Question 2. Assertion: A transformer is used to increase or decrease AC voltage only.

Reason: A transformer works based on mutual Induction.

Answer: 1. Both A and R are true and R is the correct explanation of A

Question 3. Assertion: Average power loss in scries LC circuit is always zero.

Reason: The average value of voltage and current in A.C. is zero.

Answer: 2. Both A and R are true but R is NOT the correct explanation of A

Question 4. Assertion: The capacitor serves as a block for D.C, and offers an easy path to AC.

Reason: Capacitive reactance is inversely proportional to frequency.

Answer: 1. Both A and R are true and R is the correct explanation of A

Question 5. Assertion: When capacitive reactance is smaller than the inductive reactance in the scries LCR
circuit, voltage leads the current.

Reason: In a series LCR circuit inductive reactance is always greater than capacitive reactance.

Answer: 3. A is true but R is false

Question 6. Assertion: In the series LCR circuit, the impedance is minimal at resonance.

Reason: The currents in the inductor and capacitor arc same in the scries LCR circuit.

Answer: 2. Both A and R are true but R is NOT the correct explanation of A

Question 7. Assertion: In series LCR circuit phase difference between current and voltage is never zero.

Reason: Voltage and current are never in phase.

Answer: 4. A is false and R is also false

Class 12 Physics Chapter 7  Alternating Current Short Questions And Answers

Question 1. An alternating current I = (10 A) sin (100 πt) is passed through a resistor of 20 Ω. What is the average power consumed by the resistor over a complete cycle?

Answer:

P_avg= V_rms I_rms cos = (I_rms R) I_rms [cosΦ= 1]

So, P_avg = I²_rms R-C

⇒ \(\frac{I_{0}^2}{2} \times R\)

⇒ \(\frac{10 \times 10 \times 20}{2}\)

P_avg = 1000 watt

Question 2. Define ‘quality factor’ at resonance in series LCR circuit. What is its SI unit?

Answer:

The Q factor of the series resonant circuit is defined as the ratio of the voltage developed across the inductor or capacitor at resonance to the applied voltage, which is the voltage across R.

⇒ \(Q=\frac{I X_1}{I R}=\frac{{\omega}_0 L}{R}=\frac{{\omega}_1}{{\omega}_2-{\omega}_1}\)

It is dimensionless, hence it has no units. It represents the sharpness of resonance.

Question 3. An a.c. source of voltage V = V₀sinωt is connected to an ideal inductor. Draw graphs of voltage V and current 1 versus cot.

Answer:

Question 4. Explain why current Hows through an ideal capacitor when it is connected to an a.c. source but not when it is connected to a D.C. source in a steady state.

Answer:

When AC is connected to the capacitor, due to continuous change of polarity of the applied voltage there will be continuous change of polarity of capacitor plates. This causes the charge to flow across the capcitor.

In steady state, the capacitor acts as an open circuit as reactance offered by it to flow of dc (f = 0) is infinite, As \(X_c=\frac{1}{2 \pi f C}=\infty\)

Question 5.

An LCR series circuit is connected to an AC source. If the angular resonant frequency of the circuit is coo, will the current lead or lag or be in phase with the voltage when to ω < ω₀  and why?

1. We cannot step up the DC voltage using a transformer. Why?
2. On what principle does a metal detector work?

Answer:

1. at ω < ω₀

X_L< X_C

so current leads the voltage

2. For d.c f = 0

So. there is no mutual induction and the transformer works on the principle of mutual induction.

3. The metal detector works on the principle of resonance in AC circuits.

Question 6.

1. In an LCR series circuit connected to an AC source, the voltage and the current are in the same phase. If the capacitor is filled with a dielectric, will the current lead or lag or remain in phase with the voltage? Explain.
2. In the circuit, why is the rms value of net voltage not equal to the sum of voltage drops
   across individual elements?
3. Draw a graph showing variation of the impedance of the circuit with the frequency of the
   applied voltage.

Answer:

1. Given in question capacitor is filled with a dielectric slab, So new capacity increases as

⇒ \(C=\frac{\varepsilon_1 \varepsilon_0 A}{d}\)

So, C increase then X_C decreases

So, X_L > X_C

Current lag voltage, as the circuit is inductive.

2. Voltage across R, L, and C are at different phase angles. So. we must do vector addition of voltages then we get Net voltage.

3.

Question 7. A capacitor of unknown capacitance, a resistor of 100 Ω, and an inductor of self-inductance l, = (4/π²) henry are connected in series to an AC source of 200 V and 50 Hz. Calculate the value of the capacitance and impedance of the circuit when the current is in phase with the voltage. Calculate the power dissipated in the circuit.

Answer:

Current in phase with voltage means the angle between cmf and current is zero. This is a resonance condition. So

Inductive reactance = Capacitive Reactance

⇒ \(X_L=X_C \Rightarrow \omega L=\frac{1}{\omega C}\)

⇒ \(\mathrm{C}=\frac{1}{{\omega}^2 \mathrm{~L}} \text {, given } \mathrm{L}=4 / \pi^2 \quad f=50 \mathrm{~Hz}, \mathrm{~V}=200 \mathrm{~V}\)

So \(C=\frac{1 \times \pi^2}{(2 \pi)^2 \times 4 \times(50)^2} \Rightarrow C=25 \mu \mathrm{F}\)

Power dissipated \(P=V^2 / R=\frac{200 \times 200}{100}\)

∴ P = 400 W

Question 8. A series LCR circuit connected to a variable frequency 230 V source.

1. Determine the source frequency which drives the circuit in resonance.
2. Calculate the impedance of the circuit and amplitude of current at resonance.
3. Show that potential drop across LC combination is zero at resonating frequency.

Answer:

1. At Resonance

⇒ \(X_L=X_C \Rightarrow \omega_{\mathrm{r}} L=\frac{1}{\omega_{\mathrm{r}} \mathrm{C}}\)

⇒ \(f_r=\frac{1}{2 \pi \sqrt{L C}} \Rightarrow f_r=\frac{1}{2 \pi \sqrt{5 \times 80 \times 10^{-6}}}\)

⇒ \(f_r=\frac{1}{2 \pi \times 20 \times 10^{-3}}\)

∴ \(\mathrm{f}_{\mathrm{r}}=\frac{25}{\pi} \mathrm{Hz}\)

2. Impedance of the circuit at resonance

Z = R = Z = 40 Ω

Amplitude of current at resonance

⇒ \(V_{\mathrm{rms}}=I_{\mathrm{rm}} \mathrm{Z} \Rightarrow I_{\mathrm{rms}}=\frac{230}{40} \mathrm{A}\)

∴ \(I_{\mathrm{rms}}=\frac{23}{4} \mathrm{~A}\)

Amplitude, I₀ = √2 I_rms

∴ I₀ = 8.1 A

3. Potential drop across LC combination

V_LC = V_L-V_C

= I(X_L-X_C)

At resonance X_L = X_C => X_L– X_C = 0

V_C= 0

Question 9.

1. When an AC source is connected to an ideal capacitor, show that the average power
   supplied by the source over a complete cycle is zero.
2. A bulb is connected in series with a variable capacitor and an A.C. source as shown. What happens to the brightness of the bulb when the key is plugged in and the capacitance of the capacitor is gradually reduced?

Answer:

1. Given V = V₀ sin t

q = CV

q = CV₀ sin t

⇒ \(\frac{\mathrm{dq}}{\mathrm{dt}}=\mathrm{CV}_0 \quad \frac{\mathrm{d}}{\mathrm{dt}} \sin \omega t\)

I = CV₀ (cos ω t)

⇒ [\atex]\left.I=\frac{V_0}{\left(\begin{array}{c}\frac
{1}{\omega c}
\end{array}\right)} \cos {\omega} t \Rightarrow I=\frac{V_0}{X_c} \sin ({\omega} t+\pi / 2\right)[/latex]

Here, \(X_C=\frac{1}{{\omega} C} \text { and } \mathrm{I}_0=\frac{V_0}{X_C}\)

Average power

⇒ \(P_{a v}=\int_0^1 V I d t=\frac{V_0^2}{X_C} \int_0^1(\sin {\omega} t)(\sin {\omega} t+\pi / 2) d t=\frac{V_0^2}{X_c} \int_0^1(\sin \omega t)(\cos {\omega}t) d t\)

⇒ \(P_{\mathrm{av}}=\frac{\mathrm{V}_0^2}{2 X_C} \int_0^{\mathrm{T}} \sin (2{\omega} t) d t \quad\left\{\int_0^{\mathrm{T}} \sin (2{\omega} t) d t=0\right.\)

P_av = 0

2. When the AC source is connected, the capacitor offers capacitive reactance X_C = 1/Cω. The
current flows in the circuit and the lamp glows. On reducing C, X_C increases and currently
reduces, Therefore, the bulb’s glow reduces.

Question 10. A capacitor (C) and resistor (R) are connected in series with an AC source of voltage of frequency 50 Hz. The potential difference across C and R are 120 V and 90 V respectively, and the circuit’s current is 3 A. Calculate

1. The impedance of the circuit
2. The value of the inductance, which when connected in series with C and R will make the power factor of the circuit unity.

Answer:

Given: f = 50 Hz, I = 3A, V_C = 120 V and V_R = 90 V

1. Impedance of the circuit:

⇒ \(\mathrm{Z}=\frac{\mathrm{V}}{\mathrm{I}} \Rightarrow \mathrm{Z}=\frac{\sqrt{\mathrm{V}_{\mathrm{c}}^2+\mathrm{V}_{\mathrm{R}}^2}}{\mathrm{I}}\)

⇒ \(Z=\frac{\sqrt{(120)^2+(90)^2}}{3}\)

Z = 50

2. Power factor (cos Φ) = l. This is the condition of resonance. Let inductance (L) is connected in series with C and R. At resonance.

X_L = X_C

V_C = IX_C

120 = 3 X_C

X_C = 40

X_L = 40 Ω ⇒ ωL = 40 Ω ⇒ 2πfL = 40

π \(\mathrm{L}=\frac{40}{2 \pi f}=\frac{40}{2 \pi \times 50}=\frac{0.4}{\pi} \mathrm{H}\)

Question 11.

1. When an AC source is connected to an ideal inductor show that the average power supplied
   by the source over a complete cycle is zero.
2. A lamp is connected in series with an inductor and an AC source. What happens to the lamp’s brightness when the key is plugged in and an iron rod is inserted inside the inductor? Explain.

Answer:

1. Given

V = V₀ sinωt

⇒ \(\mathrm{V}=\mathrm{L} \frac{\mathrm{d} \mathrm{I}}{\mathrm{dt}}\) (induced emf)

⇒ \(\mathrm{dI}=\frac{\mathrm{V}}{\mathrm{L}} \mathrm{dt}\)

⇒ \(\mathrm{dI}=\frac{\mathrm{V}_0}{\mathrm{~L}} \sin \omega t \mathrm{dt}\)

By integration \(I=\frac{-V_0}{{\omega}} \cos {\omega} t\)

∴ \(I=-\frac{V_0}{\omega L} \sin \left[\frac{\pi}{2}-\omega t\right] \Rightarrow I_0 \sin \left[\omega t-\frac{\pi}{2}\right]\)

where \(I_0=\frac{V_0}{{\omega} \mathrm{L}}\)

Average power

⇒ \(P_{\mathrm{av}}=\int_0^{\mathrm{T}} V I \mathrm{dt}\)

⇒ \(-\frac{V_0^2}{\omega L} \int_0^T \sin {\omega} t \cos {\omega}t dt\)

⇒ \(-\frac{V_0^2}{2 \omega L} \int_0^{\mathrm{T}} \sin (2 \omega \mathrm{t}) \mathrm{dt} \quad\left\{\int_0^{\mathrm{T}} \sin (2 \omega t) \mathrm{d} t=0\right.\)

= 0

2. When an iron rod is inserted into the inductor, the self-inductance of the inductor increases. On increasing L. X_L increases and current reduces. Therefore glow of the bulb reduces.

Class 12 Physics Chapter 7  Alternating Current Long Questions And Answers

Question 1. When a pure resistance R, pure inductor L, and an ideal capacitor of capacitance C are connected in series to a source of alternating e.m.f., then-current at any instant through the three elements has the same amplitude and is represented as I = I₀ sin ωt.

However, the voltage across each element has a different phase relationship with the current as shown in the graph. The effective resistance of the RLC circuit is called the impedance (Z) of the circuit and the voltage leads the current by a phase angle Φ.

A resistor of 12Ω a capacitor of reactance 14Ω and a pure inductor of inductance 0.1 H are joined in series and placed across 200 V, 50 Hz a.c. supply.

(1). What is the value of inductive reactance?

Answer:

X_L = 2πfL = 2 x 3.14 x 50 x 0.1 = 31.4 Ω

(2). What is the value of impedance?

Answer:

∴ \(Z=\sqrt{R^2+\left(X_{L}-X_C\right)^2}=\sqrt{(12)^2+(31.4-14)^2}=21.13 \Omega\)

(3). What is the value of current in the circuit?

Answer:

∴ \(I=\frac{e}{Z}=\frac{200}{21.13}=9.46 \mathrm{~A}\)

(4). What is the value of the phase angle between current and voltage?

Answer:

∴ \(\tan \phi=\frac{X_L-X_C}{R}=\frac{31.4-14}{12}=1.45 \Rightarrow \phi=\tan ^{-1}(1.45)\)

Question 2. The power averaged over one full cycle of a.c. is known as average power. It is also known as true power \(P_{\mathrm{av}}=V_{\mathrm{rms}} I_{\mathrm{rms}} \cos \phi=\frac{V_0 I_0}{2} \cos \phi\)

Root mean square or simply rms watts refer to continuous power.

A circuit containing an 80 mH inductor and a 60 μF capacitor in series is connected to a 230V, 50 Hz supply. The resistance of the circuit is negligible.

(1). What is the average power transferred to the inductor?

Answer:

Zero

(2). What is the total average power absorbed by the circuit?

Answer:

This is an LC circuit so the average power absorbed by the circuit is zero.

(3). Find the value of current amplitude.

Answer:

⇒ \(I=\frac{e}{Z}=\frac{c}{X_L-X_c}\)

e = 230 V0

X_L = L = 2πfL

= 2 x 3.14 x 50 x 80 x 10^-3 = 25.120 Ω

⇒ \(X_C=\frac{1}{{\omega} C}=\frac{1}{2 \pi fC}\)

∴  \(\frac{1}{2 \times 3.14 \times 50 \times 60 \times 10^{-6}}=53 \Omega\)

So, \(I=\frac{230}{53-25.120}=8.249 \mathrm{~A}\)

I₀ = √2I_rms

= 2 x 8.249 = 11.6 A

(4). Find the rms value of current.

Answer:

⇒ \(I=\frac{e}{Z}=\frac{e}{X_L-X_C}\)

e = 230 V0

x_L = L = 2πfL

= 2 x 3.14 x 50 80 x 10^-3 = 25. 120

⇒\(X_C=\frac{1}{\omega C}=\frac{1}{2 \pi f C}\)

⇒ \(\frac{1}{2 \times 3.14 \times 50 \times 60 \times 10^{-6}}=53 \Omega\)

So, \(I=\frac{230}{53-25.120}=8.249 \mathrm{~A}\)

Question 3.

1. In a series LCR circuit connected to an a.c. source of voltage V = V_m sinωt, use phasor diagram to derive an expression for the current in the circuit. Hence obtain the expression tor the power dissipated in the circuit. Show that power dissipated at resonance is maximum.
2. In a series LR circuit, X_L = R, and the power factor of the circuit is P₁. When a capacitor with capacitance C such that X_L = X_C is put in series, the power factor becomes P₂. Calculate P₁/P₂.

Answer:

1. Suppose OA, OB, and OC represent the magnitude of phasor V_R, V_L, and V_C respectively. In the case of V_L > V_C, the resultant of (V_R) and (V_L-V_C), is represented by OE. Thus from ΔOAE

⇒ \(\mathrm{OE}=\sqrt{\mathrm{OA}^2+\mathrm{AE}^2}\)

⇒ \(V=\sqrt{V_R^2+\left(V_L-V_C\right)^2}\)

Substituting the value of VR, VL, and VC we have

⇒ \(V=\sqrt{(I R)^2+\left(I X_L-I X_C\right)^2}\)

or \(I=\frac{V}{\sqrt{(R)^2+\left(X_L-X_C\right)^2}}\)

The effective opposition offered by L, C, R to a.c. supply is called the impedance of the LCR circuit and is represented by Z.

∴ \(I=\frac{V}{Z}\)

So, comparing \(Z=\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}\)

Also from ΔOAE

⇒ \(\tan \phi=\frac{A E}{O A}=\frac{V_L-V_c}{V_R}\)

or \(\tan \phi=\left(X_L-X_C\right) / R\)

or \(\phi=\tan ^{-1} \frac{\left(X_L-X_C\right)}{R}\)

Power dissipation in LCR circuit:

The instantaneous power supplied by the source is

P = VI

⇒ \(P=\left(V_m \sin \omega t\right) \times i_m \sin (\omega t+\phi) = \frac{V_m i_m}{2}[\cos \phi-\cos (2 \omega t+\phi)]\)

[2sinAsinB = cos(A-B)-cos(A+B)]

For Average power, the second term becomes zero in the complete cycle.

So \(P_{a v}=\frac{V_m i_m}{2} \cos \phi \quad \int_0^{\mathrm{T}} \cos (2 \omega t+\phi) d t=0\)

So \(P_{\mathrm{av}}=\frac{V_{\mathrm{m}}}{\sqrt{2}} \frac{i_{\mathrm{m}}}{\sqrt{2}} \cos \phi=V_{\mathrm{rms}} i_{\mathrm{rms}} \cos \phi\)

So \(\mathrm{P}_{\mathrm{av}}=\mathrm{V}_{\mathrm{rms}} \mathrm{i}_{\mathrm{rms}} \cos \phi\)

At resonance condition, cosΦ = 1 (because Φ) = 0), R becomes the effective impedance of a circuit. So power dissipated is maximum at resonance condition

2. In series LCR circuit impedance

⇒ \(Z=\sqrt{R^2+\left(X_L-X_C\right)^2}\) and power factor p = R/Zero

case 1: In LR circuit When XL = R, So Z = (2R2)1/2

Z = √2R = Now \(P_1=\frac{R}{Z}=\frac{R}{\sqrt{2} R} \Rightarrow P_1=\frac{1}{\sqrt{2}}\)

Case 2: X_L = X_C, Z=R, So power factor P₂ becomes equal to 1

P₂ = 1

So ration \(\frac{P_1}{P_2}=\frac{1}{\sqrt{2}}: \frac{1}{1} \Rightarrow \frac{P_1}{P_2}=\frac{1}{\sqrt{2}}\)

Question 4. A 2 μF capacitor, 100 Ω resistors, and 8 H inductor are connected in series with an AC source.

1. What should be the frequency of the source such that the current drawn in the circuit is maximum, What is this frequency called?
2. If the peak value of c.m.f. of the source is 200 V, find the maximum current.
3. Draw a graph showing the variation of amplitude of circuit current with changing frequency of applied voltage in a series LCR circuit for two different values of resistance R₁ and R₂ (R₁ > R₂).
4. Define the term ‘Sharpness of Resonance’. Under what conditions, does a circuit become
   more selective?

Answer:

1. Source frequency, when current is maximum is given by

⇒ \(f=\frac{1}{2 \pi \sqrt{L C}}=\frac{1}{2 \pi \sqrt{8 \times 2 \times 10^{-6}}}\) [L = 8H and C = 2μF]

∴ \(f=\frac{1}{2 \pi \times 4 \times 10^{-3}}\)

f = 39.80 Hz

The frequency at which the current maximum, is called resonant frequency.

2. Given E₀ = 200V, R= 100Ω

∴ \(I_{\max }=\frac{E_0}{R}=\frac{200}{100}=2 \mathrm{~A}\)

3.

4. The sharpness of resonance is given by the quality factor (Q factor) of a resonant circuit. It is defined as the ratio of the voltage drop across the inductor (or capacitor) to the applied voltage. Sharper the curve, the circuit will be more selective. For resistant R₂, the circuit is more selective.

Question 5.

1. Draw a labeled diagram of a step-down transformer. State the principle and its working.
2. Express the turn ratio in terms of voltages.
3. Find the ratio of primary and secondary currents in terms of turn ratio in an ideal
   transformer.
4. How much current is drawn by the primary of a transformer connected to a 220 V supply when it delivers power to a 1 10 V- 550 W refrigerator?

Answer:

1.

Principle: The Transformer works on the principle of mutual induction, in which an EMF is induced in the secondary coil by changing the magnetic flux in the primary coil.

Working: When an alternating current source is connected to the ends of the primary coil, the current changes continuously in the primary coil, due to which magnetic flux linked with the secondary coil changes continuously. Therefore, the alternating emf of the same frequency is developed across the secondary terminals.

2. \(\frac{N_s}{N_p}=\frac{V_s}{V_p} \quad\left\{\frac{N_s}{N_p}=\right.\text { turn ratio }\)

3. For ideal transformer

Output power = Input power

V_SI_S = V_PI_P

⇒ \(\frac{V_S}{V_p}=\frac{I_p}{I_S} \quad\left\{\frac{V_s}{V_p}=\frac{N_s}{N_p}\right.\)

∴ \(\frac{N_s}{N_p}=\frac{I_p}{I_s}\)

4. Given V_P = 220 V, V_S = 1 1 0 V. P = 550 W. I_P = ?

⇒ \(I_p=\frac{\text { Power }}{\text { Primary Voltage }}=\frac{P}{V_p}\)

∴ \(\mathrm{I}_{\mathrm{P}}=\frac{550}{220}=2.5 \mathrm{~A}\)

Question 6. A device ‘X’ is connected to an AC source V = V₀ sin ωt. The variation of voltage, current, and power in one cycle is shown in the following graph:

1. Identify the device ‘X’.
2. Which of the curves A, B, and C represent the voltage, current, and power consumed in the circuit? Justify your answer.
3. How does its impedance vary with the frequency of the AC source? Show graphically.
4. Obtain an expression for the current in the circuit and its phase relation with AC voltage.

Answer:

1. Capacitor

2. Curve: A represents power because in a pure capacitive AC circuit, power consumed in one cycle is zero and the frequency of power is twice the frequency of voltage (or current). CurvcB represents voltage. Curve-C represents current because in a pure capacitive AC circuit, current leads the voltage by π/2.

3. \(Z=X_C \Rightarrow Z=\frac{1}{2 \pi f C}\)

4. We know, \(V=\frac{q}{C} \text { or } q=C V\)

Also, q = CV₀ sin t

or \(\frac{\mathrm{dq}}{\mathrm{dt}}=C \mathrm{~V}_0 \frac{\mathrm{d}}{\mathrm{dt}}(\sin {\omega} \mathrm{t})\)

I = CV₀(cos ωt)

∴ \(I=\frac{V_0}{\left(\begin{array}{c}
\frac1{\omega C}
\end{array}\right)} \cos \omega t\)

\(I=\frac{V_0}{X_c} \cos {\omega} t\) (\(x_C=\frac{1}{\omega C}\))

\(I=I_0 \sin (\omega t+\pi / 2)\)     \(\left[I_0=\frac{V_0}{X_C}\right]\)

∴ Leads by (π/2) with voltage.

Question 7. A device X is connected across an AC source of voltage V = V₀ sinωt. The current through X is given as \(I=I_0 \sin \left(\omega t +\frac{\pi}{2}\right)\)

1. Identify the device X and write the expression for its reactance.
2. Draw graphs showing the variation of voltage and current with lime over one cycle of ac, for X.
3. How does the reactance of device X vary with the frequency of the AC? Show this variation
   graphically.
4. Draw the phasor diagram for the device X.

Answer:

1. X: Capacitor

Reactance \(X_C=\frac{1}{\omega C}=\frac{1}{2 \pi f C}\)

2.

3. Reactance of the capacitor varies in inverse proportion to the frequency i.e \(x_c \propto \frac{1}{f}\)

4.

Class 12 Physics Electric Charges And Fields Multiple Choice Questions And Answers

Question 1. An electric dipole placed in a non-uniform electric field will experience:

1. Only a force
2. Only a torque
3. Both force and torque
4. Neither force nor torque

Answer: 3. Both force and torque

Read and Learn More Important Questions for Class 12 Physics with Answers

Question 2. Let N₁ be the number of electric field lines going out of an imaginary cube of side a that encloses an isolated point charge 2q and N₂, be the corresponding number for an imaginary sphere of radius a that encloses an isolated point charge 3q Then (N₁/N₂) is :

1. 1/π
2. 2/3
3. 9/4
4. 5/3

Answer: 2. 2/3

∴ \(\phi=\frac{q}{\varepsilon_0}\)

So, \( N_1 =\frac{2 q}{\varepsilon_0} \)

∴ \(N_2 =\frac{3 q}{\varepsilon_0} \Rightarrow \frac{N_1}{N_2}=\frac{2}{3}\)

Question 3. Let F₁ be the magnitude of the force between two small spheres, charged to a constant potential In free space, and F₂, be the magnitude of the force between them in a medium of dielectric constant K. Then (F₁/F₂) is:

1. \(\frac{1}{K}\)
2. K
3. K²
4. \(\frac{1}{\mathrm{~K}^2}\)

Answer: 2. k

⇒ \( F_m=\frac{F_0}{K} \)

⇒ \( F_2=\frac{F_1}{K}\)

So, \( $\frac{F_1}{F_2}=K\)

Physics Class 12 Chapter 1 Important Questions Pdf

Question 4. A charge Q is placed at the center of the line joining two charges q and The system of the three charges will be in equilibrium if Q is :

1. \(+\frac{q}{3}\)
2. \(-\frac{q}{3}\)
3. \(+\frac{q}{4}\)
4. \(-\frac{q}{4}\)

Answer: 4. \(-\frac{q}{4}\)

According to question

F_net = 0

According to q at AB

F_AC = F_AB

⇒ \(\frac{\mathrm{kqQ}}{\left(\frac{\mathrm{d}}{2}\right)^2}=\frac{\mathrm{kqq}}{\mathrm{d}^2}\)

4Q = q

⇒ \(\mathrm{Q}=\frac{\mathrm{q}}{4}\)

Q must be negative

so, \(\mathrm{Q}=\frac{\mathrm{-q}}{4}\)

Question 5. Electric flux of an electric field E through an area d\(\vec{A}\) is given by :

1. \({\vec{E}} \times \mathrm{d} \vec{A}\)
2. \(\frac{\vec{E} \times \mathrm{d} \overrightarrow{\vec{A}}}{\varepsilon_0}\)
3. \({\vec{E}} \cdot \mathrm{d}{\vec{A}}\)
4. \(\frac{\vec{E} \cdot d \vec{A}}{\varepsilon_0}\)

Answer: 3. \({\vec{E}} \cdot \mathrm{d}{\vec{A}}\)

Question 6. Two point charges +16 q and -4 q are located at x = 0, and x = L. The location of the point on the x-axis at which the resultant electric field due to these charges is zero is:

1. 8L
2. 6L
3. 4L
4. 2L

Answer: 4. 2L

Physics Class 12 Chapter 1 Important Questions Pdf

Question 7. An electric dipole of dipole moment 4 x 10^-5C-m, kept in a uniform electric field of 10^-3 NC^-1 experiences a torque of 2 x 10^-8 Nm. The angle that the dipole makes with the electric field is:

1. 30°
2. 45°
3. 60°
4. 90°

Answer: 1. 30°

τ = pE sin θ

Question 8. Three identical charges arc placed at the x-axis from left to right with adjacent charges separated by a distance d. The magnitude of the force on a charge from its nearest neighbor charge is F. Let \(\hat{i}\) be the unit vector along + x-axis. then the net force on each charge from left to right is :

1. \((2 F \hat{i},-2 F \hat{i}, 2 F \hat{i})\)
2. \((\mathrm{F} \hat{\imath}, 0, \mathrm{~F} \hat{\imath})\)
3. \((-5 / 4 F \hat{\imath}, 0,+5 / 4 F \hat{\imath})\)
4. \((2 F \hat{\imath}, 0,2 F \hat{\imath})\)

Answer: 3. \((-5 / 4 F \hat{\imath}, 0,+5 / 4 F \hat{\imath})\)

Question 9. A lest charge of 1.6 x 10^-19C is moving with a velocity \({\vec{v}}=(4 \hat{\mathrm{i}}+3 \hat{\mathrm{k}}) \mathrm{ms}^{-1}\) in a magnetic field \({\vec{B}}=(3 \hat{\mathrm{k}}+4 \hat{\mathrm{i}}) \mathrm{T}\) The force on this lest charge is:

1. \(24 \hat{j} \mathrm{~N}\)
2. \(-24 \hat{i} \mathrm{~N}\)
3. \(24 \hat{k} \mathrm{~N}\)
4. 0

Answer: 4. 0

Question 10. If a charge is moved against a coulomb force of an electric field, then the

1. The intensity of the electric field increases
2. The intensity of the electric field decreases
3. Work is done by the electric field
4. Work is done by the external source

Answer: 4. Work is done by the external source

Question 11. A charge Q is located at the center of a circle of radius r. The work done in moving a test charge q₀ from point A to point B (at opposite ends of diameter AB) to complete a semicircle is \(\left[k=\frac{1}{4 \pi \varepsilon_0}\right]\)

1. \(k \frac{q_0 Q}{r}\)
2. \(k \frac{q_0 Q}{r^2}\)
3. kq₀Qr
4. Zero

Answer: 4. Zero

Physics Class 12 Chapter 1 Important Questions Pdf

Question 12. Two charged spheres A and B having their radii in ratio 1: 2 are connected with a conducting wire, the ratio of their surface charge densities (σ_A/ σ_B) will be:

1. \(\frac{1}{2}\)
2. 2
3. \(\frac{1}{4}\)
4. 4

Answer: 2. 2

Question 13. The force acting between two point charges kept at a certain distance is 5 N. Now the magnitudes of charges arc doubled and the distance between them is halved, the force acting between them is _______ N.

1. 5
2. 20
3. 40
4. 80

Answer: 4. 80

Now, \(F=\frac{k(2 q)(2 q)}{\left(\begin{array}{l}
r \\
2
\end{array}\right)^2}\) = 16 x 5 = 80 N

Question 14. When an electron and a proton are placed in an electric field _________.

1. The electric forces acting on them are equal in magnitude as well as direction.
2. Only the magnitudes of forces are the same.
3. Accelerations produced in them are the same
4. The magnitudes of accelerations produced in them are the same

Answer: 2. Only the magnitudes of forces are the same.

Question 15. Two spheres carrying charges q arc hanging from, the same point of suspension with the threads of length 2 m, in space free from gravity. The distance between them will be, ______ m.

1. 0
2. 1.0
3. 4.0
4. 2.0

Answer: 3. 4.0

Question 16. When two spheres having 2Q and -Q are placed at a certain distance. The force acting between them is F. Now these are connected by a conducting wire and again separated from each other. How much, force will act between them if the separation, now is the same as before?

1. F
2. \(\frac{F}{2}\)
3. \(\frac{F}{4}\)
4. \(\frac{F}{8}\)

Answer: 4. \(\frac{F}{8}\)

⇒ \(\mathrm{F}=\frac{2 \mathrm{k} \mathrm{Q}^2}{\mathrm{r}^2}\) →(1)

These are connected after the separation

⇒ \(\mathrm{F}^{\prime}=\frac{\mathrm{k}\left(\frac{\mathrm{Q}}{2}\right)\left(\frac{\mathrm{Q}}{2}\right)}{\mathrm{r}^2}=\frac{\mathrm{kQ}^2}{4 \mathrm{r}^2}\) →(2)

Equation (2)÷(1)

∴ \(\frac{F^{\prime}}{F}=\frac{1}{8} \Rightarrow F^{\prime}=\frac{F}{8}\)

Physics Class 12 Chapter 1 Important Questions Pdf

Question 17. When a 10 μC charge is enclosed by a closed surface, the flux passing through the surface is Φ. Now another -5μC charge is placed inside the same closed surface, then the flux passing through the surface is ________.

1. 2Φ
2. Φ/2
3. Φ
4. Zero

Answer: 2. Φ/2

Question 18. An electric dipole is placed in a uniform electric field. The resultant force acting on it is _________.

1. Always zero
2. Never zero
3. Depends on the relative position
4. Depends upon the dipole moment

Answer: 1. Always zero

Question 19. Electric field due to a dipole at a large distance (r) falls off as ________.

1. \(\frac{1}{r}\)
2. \(\frac{1}{r^2}\)
3. \(\frac{1}{r^3}\)
4. \(\frac{1}{r^4}\)

Answer: 3. \(\frac{1}{r^3}\)

⇒ \(E=\frac{2 k p}{r^3}\) (On axis)

∴ \(\mathrm{E}=\frac{\mathrm{kp}}{\mathrm{r}^3}\) (on equator for large distance)

Question 20. The resultant force and resultant torque acting on an electric dipole kept in a uniform electric Held (θ ≠ 0° or 180°) are \(\vec{F}\) and \(\vec{\tau}\) then:

1. \({\vec{F}} \neq 0 ; \quad \vec{\tau}=0 \)
2. \({\vec{F}}=0 ; \quad \vec{\tau} \neq 0\)
3. \({\vec{F}}=0 ; \vec{\tau}=0\)
4. \(\dot{\vec{F}} \neq 0 ; \vec{\tau} \neq 0\)

Answer: 2. \({\vec{F}}=0 ; \quad \vec{\tau} \neq 0\)

Question 21. The liquid drop of mass ‘m’ has a charge ‘q’. What should be the magnitude of electric field E to balance this drop?

1. \(\frac{E}{m}\)
2. \(\frac{\mathrm{mg}}{\mathrm{q}}\)
3. mgq
4. \(\frac{\mathrm{mq}}{\mathrm{g}}\)

Answer: 2. \(\frac{\mathrm{mg}}{\mathrm{q}}\)

∴ mg = qE

Question 22. The number of electric field lines that emerged from 1 mC charge is _____.

1. 1.13×10²
2. 9×10⁹
3. 1.13×10¹¹
4. 9×10^-9

Answer: 1. 1.13×10²

∴ \(\phi=\frac{\mathrm{q}}{\varepsilon_0}=\frac{1 \mathrm{mC}}{8.85 \times 10^{-12}}=1.13 \times 10^8 \frac{\mathrm{N}}{\mathrm{C}} \times \mathrm{m}^2\)

Physics Class 12 Chapter 1 Important Questions Pdf

Question 23. A charge Q is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled, the outward electric flux will:

1. Increase four times
2. Be reduced to half
3. Remains The same
4. Be doubled

Answer: 3. Remains The same

Physics Class 12 Chapter 1 Important Questions Assertion And Reason

For questions numbers 1 to 4 two statements are given-one labeled Assertion (A) and the other labeled Reason (R). Select the correct answer to these questions from the codes (1), (2), (3), and (4) as given below.

1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is NOT the correct explanation of A
3. A is true but R is false
4. A is false and R is also false

Question 1. Assertion: In electrostatics, electrostatic field lines can never be closed loops.

Reason: The number of electric field lines originating from or terminating on a charge is
proportional to the magnitude of the charge.

Answer: 2. A is true but R is false

Question 2. Assertion: Under electrostatic conditions net electric field inside a solid conductor will be zero.

Reason: Under electrostatics conditions, there will be no free electrons inside a conductor.

Answer: 3. Both A and R are true but R is NOT the correct explanation of A

Question 3. Assertion: Gauss law shows diversion when inverse square law is not obeyed.

Reason: Gauss law is a consequence of conservation of charge.

Answer: 3. A is true but R is false

Question 4. Assertion: The electrostatic force between two charges is a nonconservative force.

Reason: Electric force between two charges proportional to the square of distance between the two.

Answer: 4. A is false and R is also false

Question 5. Assertion: Electrostatic field lines are perpendicular to the surface of the conductor.

Reason: The surface of a conductor is equipotential.

Answer: 1. Both A and R are true and R is the correct explanation of A

Physics Class 12 Chapter 1 Short Questions And Answers

Question 1. Derive an expression for the work done in rotating a dipole from the angle θ₀ to θ₁, in a uniform electric field.

Answer:

As we know, when a dipole is placed in a uniform electric field, the net force on the dipole is zero but it experiences a torque, which can be given as, \(\vec{\tau}=\vec{p} \times \vec{E}\)

This torque rotates the dipole unless it is placed parallel or anti-parallel to the external field. If we apply an external and opposite torque, it neutralizes the effect of this torque given by τ_ext and it rotates the dipole from the angle θ₀ to an angle θ₁ at an infinitesimal angular speed without any angular acceleration.

The amount of work done by the external torque can be given by

∴ \(\mathrm{W}=\int_{0_{0}}^{0_1} \tau_{\mathrm{cxt}} \mathrm{d} 0=\int_{0_0}^{0_1} p E \sin 0 \mathrm{~d} 0=p \mathrm{E}\left(\cos \theta_0-\cos \theta_1\right)\)

Question 2.

1. Draw the pattern of electric field lines due to an electric dipole.
2. Write any two properties of electric field lines.

Answer:

1. 

2. Field lines of the electrostatic field have the following properties:

• Never intersect each other.
• Electrostatic field lines never form closed loops.

Question 3. A system has two charges q_A = 2.5 x 10^-7C and q_B = – 2.5 x 10^-7C located at points A : (0, 0, -15 cm) and B : (0, 0, +15 cm), respectively. What is the total charge and electric dipole moment of the system?

Answer:

Total charge = 2.5 x 10^-7– 2.5 x 10^-7 = 0

Electric dipole moment is \(\overrightarrow{\mathrm{p}}=\mathrm{q}(2 \overrightarrow{\mathrm{a}})\)

= 2.5×10^-7x (0.15+0.15) C-m

= 7.5×10⁸ C-m

The direction of the dipole moment is along the Z-axis.

Question 4. Find the expression for torque experienced by an electric dipole in a uniform electric field.

Answer:

Torque on an electric dipole in a uniform electric field

We consider a dipole with charges +q and -q which are at a distance d away from each other. Let it be placed in a uniform electric field of strength E such that the axis of the dipole forms an angle 0 with the electric field.

The force on the charges is

F_+q = +qE → towards the direction of the electric field

F_-q = -qE → opposite to the direction of the electric field

Since the magnitudes of forces are equal and they are separated by a distance d,

The torque on the dipole is given by :

Torque (τ) = Force x perpendicular distance between both forces

τ = F.d sin 0

or τ = qEdsinO

So τ = pE .sin 0 (p = qd)

or \(\vec{\tau}=\vec{p} \times \vec{E}\) [in vector form]

Question 5.

1. Define electric flux Write its SI Unit.
2. How does the electric flux due to a point charge enclosed by a spherical Gaussian surface get affected when its radius is increased?

Answer:

1. Electric flux is proportional to the number of electric field lines passing through a given area.

It is numerically equal to the dot product of the electric field and area vector.

Φ = A (Ecosθ)

⇒ \(\phi={\vec{E}} \cdot {\vec{A}}\)

The electric flux through an area is the dot product of the magnitude of \(\vec{E}\) and \(\vec{A}\).

The S.I. unit of ‘electric flux’ is N-m²C^-1 or V-m.

2. Wc knows that flux through the Gaussian surface is given by \(\phi=q / \varepsilon_0\)

As flux is independent of radius, it is not affected by changing the radius.

Question 6.

1. A uniformly charged large plane sheet has charge density \(\sigma=\left(\frac{1}{18 \pi}\right) \times 10^{-15} \mathrm{C} / \mathrm{m}^2\). Find the electric field at point A which is 50 cm from the sheet. Consider a straight line with three points P. Q and R, placed 50 cm from the charged sheet on the right side as shown in the figure. At which of these points, does the magnitude of the electric field due to the sheet remain the same as that at point A and why?

2. Two small identical conducting spheres carrying charge 10 μC and- 20μC when separated by a distance r, experience a force F each. If they are brought in contact and then separated to a distance of \(\frac{r}{2}\), what is the new force between them in terms of F?

Answer:

1. \(E=\frac{\sigma}{2 \varepsilon_0}=\frac{1}{18 \pi} \times \frac{10^{-15}}{2 \times 1} \times 4 \pi \mathrm{k}\)  (\(k=\frac{1}{4 \pi \varepsilon_0}\))

= \(\frac{1}{9} \times 10^{-15} \times 9 \times 10^9\)

= 10^-6 V/m

Point →Q, Because at 50 cm, the charge sheet acts as a finite sheet, and thus the magnitude
remains the same towards the middle region of the planar sheet.

2.

According to Coulomb’s Law

⇒ \(\mathrm{F}=\frac{\mathrm{k}(10 \mu \mathrm{C})(20 \mu \mathrm{C})}{\mathrm{r}^2}\)

⇒ \(\mathrm{F}=\frac{2 \times 9 \times 10^9 \times 100 \times 10^{-12}}{\mathrm{r}^2}\)

⇒ \(F=\frac{1.8}{r^2}\) →(1)

After contact

⇒ \(F^{\prime}=\frac{k(-.5 \mu C)(-.5 \mu C)}{\left(\begin{array}{l}
r{\prime} \\
2\end{array}\right)^2}\)

⇒ \(F^{\prime}=\frac{4 \times 9 \times 10^{11} \times 2.5 \times 10^{-12}}{r^2}\)

∴ \(F^{\prime}=\frac{0.9}{r^2}\) →(2)

So, \(F^{\prime}=\frac{F}{2}\)

Question 7. A charge is distributed uniformly over a ring of radius ‘a’. Obtain an expression for the electric intensity \(\vec{E}\) at a point on the axis of the ring. Hence shows that for points at a large distance from the ring, it behaves like a point charge.

Answer:

Suppose that the ring is placed with its plane perpendicular to the x-axis. as shown in the above diagram. We consider a small element dl of the ring. So the charge dq on the element dl is

⇒ \(\mathrm{dq}=\frac{\mathrm{q}}{2 \pi \mathrm{a}} \mathrm{d} l\) [∵ \(\lambda=\frac{\mathrm{q}}{2 \pi \mathrm{a}}\)] = charge per unit length

∴ The magnitude of the field de has two components:

1. The axial component ⇒ dEcosθ
2. The perpendicular component ⇒ dEsinθ

Since the perpendicular components of any two diametrically opposite elements are equal and opposite, they all cancel out in pairs. Only the axial components will add up to produce the resultant field at point P, which is given by,

⇒ \(\mathrm{E}=\int_0^{2 \pi \mathrm{a}} \mathrm{d} \mathrm{E} \cos \theta\)

⇒ \(\mathrm{E}=\int_0^{2 \pi \mathrm{a}} \frac{\mathrm{kq}}{2 \pi \mathrm{a}} \frac{\mathrm{d}
l}{\mathrm{r}^2}\left(\frac{\mathrm{x}}{\mathrm{r}}\right)=\frac{\mathrm{kqx}}{2 \pi \mathrm{ar}^3} \int_0^{2 \pi \mathrm{a}} \mathrm{d} l\) [∵ \(\cos \theta=\frac{x}{r}\)]

∴ \(\mathrm{E}=\frac{\mathrm{kqx}}{2 \pi \mathrm{ar}^3}[l]_0^{2 \pi \mathrm{a}}=\frac{\mathrm{kqx}}{2 \pi \mathrm{a}} \frac{1}{\left(\mathrm{x}^2+\mathrm{a}^2\right)^{3 / 2}}(2 \pi \mathrm{a}) \quad\left[\mathrm{as}^2=\mathrm{x}^2+\mathrm{a}^2\right]\)

or \(\mathrm{E}=\frac{\mathrm{kqx}}{\left(\mathrm{x}^2+\mathrm{a}^2\right)^{3 / 2}}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{qx}}{\left(\mathrm{x}^2+\mathrm{a}^2\right)^{3 / 2}}\)

Special case: For points at large distance from the ring x >> a

∴ \(E=\frac{k q}{x^2}=\frac{1}{4 \pi \varepsilon_0} \frac{q}{x^2}\)

Tins are the same as the field due to a point charge, indicating that for far-off axial points, the charged ring behaves as a point charge.

Question 8. Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0 x 10^-22 C/m². What is electric field intensity:

1. In the outer region of the first plate.
2. In the outer region of the second plate, and
3. Between the plates?

Answer:

Where. E_P = Electric field due to Plate P

E_Q = Electric field due to Plate Q

1. In the outer region of the first plate

⇒ \(E_1=E_P-E_Q=\frac{\sigma_P-\sigma_Q}{2 \varepsilon_0}\)

= \(\frac{17 \times 10^{-22}-17 \times 10^{-22}}{2 \varepsilon_0}\) = E₁ = 0 (i.e Electric field is zero)

2. Similarly, the electric field is zero in this case also E₁₁ = E_P– E_Q = 0

3. Between the plates

⇒ \(\mathrm{E}_{\mathrm{3}}=\mathrm{E}_{\mathrm{P}}+\mathrm{E}_{\mathrm{Q}}=\frac{\sigma_{\mathrm{P}}+\sigma_{\mathrm{Q}}}{2 \varepsilon_0}\)

= \(\frac{17 \times 10^{-22}+17 \times 10^{-22}}{2 \varepsilon_0}\)

∴ \(\mathrm{E}_{3}=\frac{34 \times 10^{-22}}{2 \times 8.854 \times 10^{-12}}=1.92 \times 10^{-10} \mathrm{NC}^{-1}\)

Question 9. Derive an expression for the electric field intensity at a point on the equatorial line of an electric dipole of dipole moment \(\vec{P}\) and length 2a. What is the direction of this field?

Answer:

We consider a dipole consisting of -q and +q separated by a distance 2a. Let P be a point on the equatorial line.

⇒ \({\vec{E}}_{\mathrm{A}}=\frac{\mathrm{l}}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{(\mathrm{AP})^2} \text { along } \vec{PA}\)

⇒ \(E_A=\frac{1}{4 \pi \varepsilon_0} \frac{q}{\left(r^2+a^2\right)}\)

⇒ \({\vec{E}}_{\mathrm{B}}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{(\mathrm{BP})^2} \text { along } \vec{BP}\)

∴ \(E_B=\frac{1}{4 \pi \varepsilon_0} \frac{q}{\left(r^2+a^2\right)}\)

The resultant intensity is the vector sum of the intensities along PA and BP. E_A and E_B can be resolved into vertical and horizontal components. The vertical components of E_A and E_B cancel each other as they are equal and oppositely directed. So the horizontal components add up to the resultant field.

E = E_A cos θ + E_B cos θ

E = 2E_Acosθ , as E_A = E_B

Substituting, \(\cos \theta=\frac{a}{\left(r^2+a^2\right)^{\frac{1}{2}}}\) in the above equation

⇒ \(E=2 E_A \cos \theta=\frac{2}{4 \pi \varepsilon_0} \frac{q}{\left(r^2+a^2\right)} \frac{a}{\left(r^2+a^2\right)^{\frac{1}{2}}}\)

∴ \(\mathrm{E}=\frac{\mathrm{kp}}{\left(\mathrm{r}^2+\mathrm{a}^2\right)^{\frac{3}{2}}}\)along \(\overrightarrow{\mathrm{BA}}\) (As p+qx2a)

As a special case,

If a²<<r² then, \(E=\frac{k p}{r^3}\) along \(\overrightarrow{\mathrm{BA}}\)

Electric field intensity at an axial point is twice the electric field intensity on the equatorial line.

The direction of the field will be against the direction of the dipole moment.

Question 10. Four point charges Q, q, Q, and q are placed at the corners of a square of side ‘a’ as shown in the figure.

Find the resultant electric force on a charge Q

Answer:

Let us find the force on the charge Q at the point C. Force due to the other charge Q

∴ \(\mathrm{F}_1=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Q}^2}{(\mathrm{a} \sqrt{2})^2}=\frac{1}{4 \pi \varepsilon_0}\left(\frac{\mathrm{Q}^2}{2 \mathrm{a}^2}\right)\) (along AC)

Force due to the charge q placed at B

⇒ \(\mathrm{F}_2=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{qQ}}{\mathrm{a}^2} \text { along } \mathrm{BC}\)

Force due to the charge q placed at D

⇒ \(F_3=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{qQ}}{\mathrm{a}^2} \text { along } \mathrm{DC}\) along DC

Resultant of these two equal forces F₂ and F₃

⇒ \(\mathrm{F}_{23}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{qQ}(\sqrt{2})}{\mathrm{a}^2} \text { (along } \mathrm{AC} \text { ) }\)

∴ The net force on charge Q (at point C)

⇒ \(F=F_1+F_{23}=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{\mathrm{a}^2}\left[\frac{Q}{2}+\sqrt{2} \mathrm{q}\right]\)

This force is directed along the AC

(For the charge Q, at the point A, the force will have the same magnitude but will be directed along CA)

Physics Class 12 Chapter 1 Important Long Questions And Answers

Question 1. When an electric dipole is placed in a uniform electric field, its two charges experience equal and opposite forces, which cancel each other and hence net force on the electric dipole in a uniform electric field is zero. However, these forces are not collinear, so they give rise to some torque on the dipole. Since the net force on an electric dipole in a uniform electric field is zero. So no work is done in moving the electric dipole in a uniform electric field. However, some work is done in rotating the dipole against the torque acting on it.

1. The dipole moment of a dipole in a uniform external field \(\vec{E}\) is \(\vec{p}\). Write the expression of torque acting on the dipole.
2. An electric dipole consists of two opposite charges, each of magnitude 1.0 μC separated by a distance of 2.0 cm. The dipole is placed in an external field of 10⁵ NC^-1. Find the value of max Torque.
3. Write the value of angle θ, when τ is minimum.
4. When an electric dipole is held at an angle θ (θ ≠ 0° or 180°) in a uniform electric field Write the value of net force \(\vec{F}\) and torque \(\vec{\tau}\)?

Answer:

1. \(\vec{\tau}=\vec{p} \times \vec{E}\)

2. \(\tau=P E \sin 90^{\circ}=p E=1 \times 10^{-6} \times 2 \times 10^{-2} \times 10^5=10^{-3} \mathrm{~N} . \mathrm{m}\)

3. 0° or 180° or nπ

4. F = 0, τ ≠ 0

Question 2. Concept of Electric field

An electric field is an elegant way of characterizing the electrical environment of a system of charges. An electric field at a point in the space around a system of charges tells you the force a unit-positive test charge would experience if placed at that point (without disturbing the system). The electric field is a characteristic of the system of charges and is independent of the last charge that you place at a point to determine the field

1. Write one property of electric field lines
2. Define electric field intensity.
3. The SI unit of the electric field is_______
4. A proton of mass ‘m’ placed in the electric field region remains stationary in the air. What is the magnitude of the electric field?

Answer:

1. The electric field line starts from +ve charge and ends at -ve charge.
2. It is defined as the electric force experienced per unit positive test charge is known as electric field intensity.
3. N/C and V/m
4. mg = eE
   E = mg/c

Question 3.

1. Use Gauss’ law to derive the expression for the electric field \((\vec{E})\) due to a straight uniformly charged infinite line of charge density λ C/m.
2. Draw a graph to show the variation of E with perpendicular distance r from the line of
   charge.
3. Find the work done in bringing a charge q at a perpendicular distance from co-long charged
   wire r₁, to r₂, (r₂ > r₁).

Answer:

1. To calculate the electric field, imagine a cylindrical Gaussian surface, since the field is everywhere radial, flux through two ends of the cylindrical Gaussian surface is zero. At the cylindrical part of the surface electric field E is normal to the surface at every point and its magnitude is constant. Therefore flux through the Gaussian surface.

= Flux through the curved cylindrical part of the surface.

= E x 2πl → (1)

Applying Gauss’s Law

⇒ \(\text { Flux } \phi=\frac{q_{\text {enclosed }}}{\varepsilon_0}\)

Total charge enclosed

= Linear charge density x l = xl

∴ \(\phi=\frac{\lambda l}{\varepsilon_0}\) → (2)

Using Equations (1) and (2)

∴ \(\mathrm{E} \times 2 \pi \mathrm{r} l=\frac{\lambda l}{\varepsilon_0} \Rightarrow \overrightarrow{\mathrm{E}}=\frac{\lambda}{2 \pi \varepsilon_0 \mathrm{r}} \hat{\mathrm{n}}\)

(where \(\hat{n}\) is a unit vector normal to the line charge)

2. The required graph is as shown :

3. Work done in moving a charge q with displacement ‘dr’

⇒ \(\mathrm{dW}=\overrightarrow{\mathrm{F}} \cdot \mathrm{d} \overrightarrow{\mathrm{r}}\)

⇒ \(\mathrm{dW}=\mathrm{q} \overrightarrow{\mathrm{E}} \cdot \mathrm{d} \overrightarrow{\mathrm{r}}=\mathrm{q} \mathrm{Edr} \cos θ\)

⇒ \(\mathrm{dW}=\mathrm{q} \times \frac{\lambda}{2 \pi \varepsilon_0 \mathrm{r}} \mathrm{dr} \)

Work done in moving the given charge from r₁ to r₂ (r₂ > r₁)

⇒ \(\int_{r_1}^{r_2} d W=\int_{r_1}^{r_2} \frac{\lambda q d r}{2 \pi \varepsilon_0 r}\)

⇒ \(\mathrm{W}=\frac{\lambda \mathrm{q}}{2 \pi \varepsilon_0}\left[\log _{\mathrm{e}} \mathrm{r}_2-\log _{\mathrm{e}} \mathrm{r}_1\right]\)

∴ \(\mathrm{W}=\frac{\lambda \mathrm{q}}{2 \pi \varepsilon_0}\left[\log _{\mathrm{e}} \frac{\mathrm{r}_2}{\mathrm{r}_1}\right]\)

Question 4. Define electric flux. Is it a scalar or a vector quantity?

1. A point charge q is at a distance of d/2 directly above the center of a square of side d, as shown in the figure. Use Gauss’ law to obtain the expression for the electric flux through the square.

2. If the point charge is now moved to a distance ‘d’ from the center of the square and the side of the square is doubled, explain how the electric flux will be affected.

Answer:

1. Electric flux through a given surface is defined as the dot product of the electric field and area vector over that surface.

Alternatively \(\phi=\int \vec{E} \cdot d \vec{S}\)

Also, accept

Electric flux, through a surface, equals the surface integral of the electric field over a closed surface. It is a scalar quantity.

Constructing a cube of side ‘d’ so that charge ‘q’ is placed within this cube (Gaussian surface)

According to Gauss’ law the Electric flux \(\phi=\frac{\text { charge enclosed }}{\varepsilon_0}=\frac{\mathrm{q}}{\varepsilon_0}\)

This is the total flux. through all the six faces of the cube.

Hence electric flux through the square \(\frac{1}{6} \times \frac{q}{\varepsilon_0}=\frac{q}{6 \varepsilon_0}\)

2. If the charge is moved to distance d and the side of the square is doubled even then the total charge enclosed in it will remain the same. Hence the total flux will remain the same as before.

Question 5.

1. Derive an expression for the electric field E due to a dipole of length ‘2a’ at a point distant r from the center of the dipole on the axial line.
2. Draw a graph of E versus r for r >> a.
3. If this dipole were kept in a uniform external electric field E₀, with the help of a diagram represent the position of the dipole in stable and unstable equilibrium and write the
   expressions for the torque acting on the dipole in both cases.

Answer:

1. Let’s consider a dipole system,

Here, AO = OB = a

OP = r

BP = r-a

AP = r + a

Elec, field (\(\overrightarrow{\mathrm{E}}_{\mathrm{B}}\)), due to the charge at point ‘B’ being towards ‘P’

Elec, field (\(\overrightarrow{\mathrm{E}}_{\mathrm{A}}\)), due to the charge at point ‘A’ being opposite to ‘P’

Now, according to the superposition principle,

⇒ \(\overrightarrow{\mathrm{E}}_{\mathrm{p}}=\overrightarrow{\mathrm{E}}_{\mathrm{ix} \mathrm{axial}}=\overrightarrow{\mathrm{E}}_{\Lambda}+\overrightarrow{\mathrm{E}}_{\mathrm{B}}\)

E_P = E_B-E_A

⇒ \(E_p=\frac{-k q}{(r+a)^2}+\frac{k q}{(r-a)^2}=k q\left[\frac{1}{(r-a)^2}-\frac{1}{(r+a)^2}\right]=k q\left[\frac{(r+a+r-a)(r+a-r+a)}{(r-a)^2(r+a)^2}\right]\)

=\(\frac{k q(2 r)(2 a)}{\left(r^2-a^2\right)^2}\)

⇒ \(E_{a x i a l}=\frac{2 r k(2 a)(q)}{\left(r^2-a^2\right)^2}\) [∵ \(|\overrightarrow{\mathrm{p}}|=2 \mathrm{a} \times \mathrm{q}\)] = \(\frac{2 \mathrm{kpr}}{r^{4}}\)

if a₂ << r₂ then \(\dot{\vec{E}}_{\mathrm{axial}}=\frac{2 \mathrm{k} {\vec{p}}}{\mathrm{r}^3}\)

It will be directed in the direction of electric dipole moment, \(\vec{P}\).

2.

3. (1). Stable equilibrium

Torque (τ) = pEsinθ

= pE x sinθ° = 0 (vsinθ° = 0)

(2). Unstable equilibrium

Torque (τ) = pE sinθ

= pE sin 180°

= pE x 0 = 0 (∵ sin 180° = 0)

Question 6.

1. Use Gauss’ theorem to find the electric field due to a uniformly charged infinitely large plane thin sheet with surface charge density σ.
2. An infinitely large thin plane sheet has a uniform surface charge density +σ. Obtain the expression for the amount of work done in bringing a point charge q from infinity to a point distant r, in front of the charged plane sheet.

Answer:

1.

As shown in the figure, considering a cylindrical Gaussian surface of cross-section A

Flux through the curved surface :

∴ \(\phi=\int \overrightarrow{\vec{E}} \cdot \mathrm{d} {\vec{S}}=\int \mathrm{Eds} \cos 90^{\circ}=0\)

At the points on the curved surface, the field vector E and area vector dS make an angle of 90° with each other. Therefore, curved surfaces do not contribute to the flux.

Flux through end caps :

∴ \(\phi=\oint \vec{E} \cdot \mathrm{dS}=\oint \mathrm{EdS} \cos 0^{\circ}=\mathrm{EA}\)

Hence, the total flux through the closed surface is :

Φ = Flux through both end caps + flux through curved surface

or Φ = EA + EA + 0 = 2EA → (1)

Now according to Gauss’ law for electrostatics

⇒ \(\phi=q / \varepsilon_0\) → (2)

Comparing equations (1) and (2), we get

2EA = q/ε₀

E = q/2ε₀A → (3)

The area of the sheet enclosed in the Gaussian cylinder is also A. Therefore, the charge contained in the cylinder, q = σA as a (surface charge density) = q/A.

Substituting this value of q in equation (3), we get

E = σA/2ε₀A

or E = σ/2ε₀

This is the relation for the electric field due to an infinite plane sheet of charge. The field is uniform and does not depend on the distance from the plane sheet of charge.

2. \(V=\frac{W}{c}=\int_{\infty}^r \vec{E} \cdot d \vec{r}\)

⇒ \(\mathrm{W}=\mathrm{c} \int_{\infty}^{\mathrm{r}}(-\mathrm{Edr})\)

⇒ \(W=-q \int_{\infty}^r \frac{\sigma}{2 c_0} d r\)

∴ \(W=\frac{q \sigma}{2 \varepsilon_0}(\infty-r)\) W = ∞