NEET Physics Chapter 3 Gravitation MCQs with Answers

Gravitation Chapter NEET Physics Questions PDF download

Question 1. Why do planets revolve around the sun?
Answer:

Planets revolve around the sun:

The planet revolve around the sun and not straight into it is because of gravitational motion.

Question 2. What is Universal law of gravitation?
Answer:

Universal law of gravitation:

According to Newton, the gravitational force acting between two bodies is directly proportional to the product of their masses and inversely proportional to the square of distance between their centres.

Question 3. Define gravitational constant.
Answer:

Gravitational constant:

Gravitational Constant ‘G’ is numerically equal to the gravitational force of attraction between two masses, each of mass 1 kg placed at a distance of 1 m.

Physics Gravitation NEET previous year questions

Question 4. Describe the importance of Universal law of gravitation.
Answer:

The importance of Universal law of gravitation:

Universal law of gravitation is important because it is:

The force that binds us to the earth.

The motion of moon around earth.

The motion of planets around the Sun.

The tides due to moon and sun.

Question 5. A Saturn year is 29.5 times the earth year. How far is Saturn from the sun (M) if the earth is 1.5 × 10⁸ km away from the sun?
Answer:

Given:

A Saturn year is 29.5 times the earth year.

it is given that

T_S = 29.5 T_e

R_e = 1.5 × 10¹¹ m

Now, according to Kepler’s third law

T_S²/T_e² = R_s³/R_e³

R_S = R_e (T_S/T_e)^2/3 = 1.5 × 10¹¹ ((29.5T_e)/T_e)^2/3

= 1.43 × 10¹² m

= 1.43 × 10⁹ km

NEET Physics Gravitation problems with step-by-step solution

Question 6. Define centre of gravity.
Answer:

Centre of gravity:

Each particle or portion of a body exp

 

 

 

 

eriences the force of gravity. The net effect of all these forces is equivalent to the effect of a single force mg acting through a point called centre of gravity of the body.

Question 7. A sphere of mass 40 kg is attached by a second sphere of mass 15 kg when their centres are 20 cm  apart, with a force of 0.1 milligram weight. Calculate the value of gravitational constant.
Answer:

Given:

A sphere of mass 40 kg is attached by a second sphere of mass 15 kg when their centres are 20 cm  apart, with a force of 0.1 milligram weight.

Here, m₁ = 40 kg, m₂ = 15 kg

R = 20 cm = 20/100 m = 2 × 10^-1 m

F = 0.1 milligram weight = 0.1 × 10^-3 gram weight

F = 10^-4 × 10^-3 kg wt

F = 10^-7 × 9.8 N (1 kg wt = 9.8 N)

F = \(G \frac{m_1 m_2}{r^2}\)

G = \(\frac{F \times r^2}{m_1 \times m_2}\)

G = 10^-7 × 9.8 × (2 × 10^-1)²/(40 × 15)

G = 6.53 × 10^-11 N m² kg^-2

Question 8. Find the gravitational force between the two protons kept at a separation of 1  femtometer (1 femtometre = 10^-15 m). The mass of a proton is 1.67 × 10^-27 kg.
Answer:

Gravitational force is given by:

F = \(G \frac{m_1 m_2}{r^2}\)

= (6.67 × 10^-11 Nm² kg^-2) × (1.67 × 10^-27 kg)²/(10^-15 m)²

= 1.86 × 10^-34 N

Question 9. Define acceleration due to gravity. Write its symbol and its value.
Answer:

Acceleration due to gravity:

The rate at which the velocity of a freely falling object increases is called acceleration due to gravity. It is denoted by g and its value is 9.8 m/s2.

Question 10. What is the relationship between acceleration due to gravity (g) and Universal gravitational constant?
Answer:

Relationship between acceleration due to gravity (g) and Universal gravitational constant

Acceleration due to gravity (g) is given by

g = GM/R²

Where M = mass of the planet

R = radius of the planet

Question 11. Define weight. Write its SI unit.
Answer:

Weight:

Weight of an object is the force with which earth attracts it towards its centre. SI unit of weight is Newton.

Question 12. The mass of an object is 120 kg on the surface of the earth. What would be its weight when  measured on the surface of the moon? What would be its mass on moon? (Take g = 10 m/s²)
Answer:

Given:

The mass of an object is 120 kg on the surface of the earth.

Mass of an object on the earth, m_e = 120 kg

Acceleration due to gravity on the earth, g_e = 10 m/s²

Weight of the object on the earth, We = ?

W_e = m_e × g_e

By putting the value we get,

W_e = 120 kg × 10 ms^-2

W_e = 1200 N

We know that the weight on the surface of moon is = (1/6) × its weight on the earth

W_e = 1/6 × 1200 N

W_e = 200 N

Question 13. An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the  magnitude of escape velocity from the earth.

1. Determine the height of satellite above the earth’s surface.
2. If satellite is stopped suddenly in the orbit and allowed to fall freely to the earth, find the speed  with which it hits the surface of the earth.
   Radius of earth = 6.4 × 10⁶ m

Answer:

(1) Let M be the mass of the earth and R be its radius then escape velocity of earth is given by the equation

ve = \(\sqrt{\frac{2 G M}{R}}\)

For satellite of mass m and orbital radius r = R + h where h is the height of the satellite above the earth’s surface moving with speed v, we thus have

\(\frac{m v^2}{r}=\frac{G M m}{r^2}\)

v = \(\sqrt{\frac{G M}{(R+h)}}\)

or,

It is given that

v = \(\frac{v_t}{2}\)

From equation (1) and (2)

\(\frac{G M}{(R+h)}=\frac{G M}{2 R}\)

From this we calculate that R + h = 2R or h = R

(2) Total initial energy of the satellite at height h above the surface of earth is

E_i = K.E. at height h + PE at height h

E_i = \(=0-\frac{G M m}{(R+h)}=-\frac{G M m}{R+h}\)

As the satellite is stopped in its orbit, KE at height h = 0. Let be the speed with which the satellite hits the surface of the earth. The final energy at the surface of the earth is

E_f = K.E. at h = 0 + P.E. at h = 0 = \(\frac{m v^2}{2}-\frac{G M m}{R}\)

From principle of conservation of energy Ef = Ei i.e.,

\(\frac{1}{2} m v_0^2-\frac{G M m}{R}=-\frac{G M m}{(R+h)}\)

From above equation we can calculate v0 which is

v₀ = \(\sqrt{G M / R}\)

Now acceleration due to gravity on the surface of earth is given by

g = \(\frac{G M}{R^2} \text { or } \frac{G M}{R}=g R\)

Putting this in equation (3) and calculating, we get

v₀ = 7.92 × 10³ m/s

Question 14. Define one Pascal.
Answer:

Pascal:

One Pascal is the pressure experienced by the surface when 1 N force acts on 1 m2 area of the body.

Question 15. Give characteristics of pressure inside fluids.
Answer:

Characteristics of pressure inside fluids:

A fluid contained in a vessel at the same level exerts equal pressure in all directions and on the walls of the container.

Question 16. What are the factors affecting the pressure at a point in a liquid?
Answer:

The pressure at a point inside the liquid depends on the following factors:

• Depth of the point below the free surface.
• Density of liquid
• Acceleration due to gravity.

Question 17. Describe the laws of liquid pressure.
Answer:

The various laws of liquid pressure are:

• Pressure at a point inside liquid increases with depth from the free surface.
• In a stationary liquid, pressure is same at all point in a horizontal plane.
• Pressure is same in all directions about a point in liquid.
• Pressure at same depth is different in different liquids. It increases with increase in density of the liquid.

Question 18. What do you mean by the term buoyancy?
Answer:

Buoyancy:

When an object is partially or completely immersed in a liquid an upward force acts on it. This upward force is known as buoyancy.

Question 19. A solid body of mass 150 g and volume 250 cm³ is put in water. Will the body float or sink?
Answer:

Here, mass of the body, M = 150 g

Volume of body, V = 250 cm³

Density of body,

\(d=\frac{M}{V}=\frac{150 \mathrm{~g}}{250 \mathrm{~cm}^3}=0.6 \mathrm{~g} / \mathrm{cm}^3\)

Since the density of the body is less than the density of water which is 1 g/cm3 the body will float on water.

Question 20. What do you mean by Archimedes’ principle?
Answer:

Archimedes’ Principle:

When an object is immersed in a liquid partially or completely, it experiences an upthrust which is equal to the weight of the liquid displaced by it.

Question 21. State 2 uses of Archimedes’ principle.
Answer:

The uses of Archimedes’ principle are:

• It is used in designing submarines and ships.
• It is used in lactometers, which is used to determine the purity of milk.

The mass of an empty bucket of capacity 10 litres is 1 kg. Find its mass when ­completely filled with a liquid relative density 0.8.

Answer:

Here, mass of empty bucket,

m = 1 kg

Volume of bucket,

V = 10 litres = 10 × 10^-3 m³ = 10^-2m³

Relative density of liquid,

R.D. = 0.8

R.D. = density of liquid/density of water

0.8 = d/1000 kg/m³

d = 0.8 × 1000 kg/m³

d = 800 kg/m³

Mass of liquid in bucket,

M = V × d

M = (10^-2m³) × (800 kg/m³)

M = 8 kg

Total mass in bucket = m + M = 1 kg + 8 kg = 9 kg

Question 22. A ball is thrown vertically upwards and rises to a height of 10 m. Calculate the velocity with which the object was thrown upwards.
Answer. 14 m/s

Question 23. What is the magnitude of the gravitational force between the earth and a 1 kg object kept on its surface? (Mass of the earth is 6 × 10²⁴ kg and radius of the earth is 6.4 × 10⁶ m).
Answer. 10N

Question 24. Gravitational force on the surface of the moon is only as strong as gravitational force on the earth. What is the weight in Newton of a 10 kg object on the moon and on the earth?
Answer. Weight of body on earth = 100N, Weight of body on moon = 17N

Question 25. A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s², find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?
Answer. Height = 80m, Total distance travelled = 160m, Total displacement = 0

Question 26. A ball thrown up vertically returns to the thrower after 6 s. Find (1) the velocity with which it was thrown up, (2) the maximum height it reaches, and (3) its position after 4 s.
Answer.

(1) Velocity = 30m/s

(2) 45m

(3) 40m above ground