CBSE Solutions For Class 10 Mathematics Chapter 13 Volume And Surface Area Of Solids

Class 10 Maths Volume And Surface Area Of Solids

Question 1. A tent of cloth is Cylindrical upto I’m height and Conical above it of the Same radius of base. If the diameter of the tent is 6m and the slant height of the Conical part is 5m, find the cloth required to make this tent.
Solution:

Diameter of base 2r = 6m

r = \(\frac{6}{2}\) = 3m

Height of Cylindrical path h = 1m

The slant height of Conical part l = 5m

Cloth required in tent = 2πrh +πrl

= πr(2h+1)

⇒ \(\frac{22}{7}\) x 3 (2×1 +5)

⇒ \(\frac{22}{7}\) x 3(10)

= 66m²

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Question 2. The volume and Surface area of a Solid hemisphere are numerically equal. What is the diameter of the hemisphere?
Solution:

We have,

Volume of hemisphere = Surface area of hemisphere

⇒ \(\frac{2}{3}\) = 3πr²

⇒ \(\frac{2}{3}\) r = 3

2r = 9

Hence, the diameter of the hemisphere = 9 units.

Question 3. 2 Cubes each of volume 64 cm3 are joined end to end. Find the Surface area of the resulting Cuboid.
Solution:

Given,

volume of cube = 64 cm³

(Side)³ = 64

(Side)³ = 43

Side = 4cm

Side of cube = 4cm

A Cuboid is formed by joining two Cubes together.

∴ For Cuboid

length l = 4+4=8cm,

breadth b = 4cm

height b = 4cm

Now, the total surface area of the cuboid.

= 2(l.b+b.h+l.h)

2(8×4+ 4×4+8×4)

= 2(32+16+32)

2(80)

= 160 cm²

Question 4. From a Solid Cylinder whose height is 2.4cm and diameter is 1.4 cm, a Conical Cavity of the Same height and diameter is hollowed out. Find the total Surface area of the remaining Solid to the nearest Cm².
Solution:

Diameter of Cylinder 2r = 1.4cm

r = 0.7cm

∴ Radius of Cylinder = radius of cone = r= 0.70m

Height of Cylinder = height of cone

h = 2.40m

If the slant height of a cone is l, then

l² = h²+r² = (2.4)² + (0.7)²

5.76 +0.49 = 6.25

1 = \(\sqrt{6.25}\) = 2.5cm

The surface area of the remaining solid = area of the base of the cylinder + curved Surface of the cylinder + curved surface of the cone

πr² + 2πrh + πrl

=πr (r+2h+1)

= \(\frac{22}{7}\) × 0.7 (0.7+2×2.4+2.5)

= \(\frac{22}{7}\) × 0.7 × (5.86) 223×0.7

= \(\frac{22}{7}\): × 4.102

= 17.6cm²

Question 5. The radius and height of a solid right Circular Cone are in the ratio of 5:12. If its volume is 314 cm³, find its total Surface area. [Take π = 3.14]
Solution:

Let the radius of Cone = 5x

∴ Height of Cone =12x

l² = (5x)²+ (12x)²

l² = 25x² +144x²

l² = 169x²

l = \(\sqrt{169 x^2}\)

l = 13x

It is given that volume = 314 cm²

∴ \(\frac{1}{3}\)π(5x)2 (12x) = 314

⇒ \(\frac{1}{3}\) × 3.14 x 25×12 × x3 = 314

⇒  x³ = \(\frac{314 \times 3}{3.14 \times 25 \times 12}\) = 1

∴ x = 1cm

∴ Radius r = 5×1 = 5cm

Height h = 12×1 = 12cm

and slant height 1 = 13×1 = 13cm

Now, total surface area of Cone = πr(l+r) = 3.14 × 5(13+5) = 3.14×5×18

= 282.60cm²

Hence, the total surface area of cone is 282.60cm²

Question 6. The Curved Surface area of a Cone of height 8m is 188.4m². Find the volume of Cone.
Solution:

πrl = 188.4

⇒ rl = \(\frac{188.4}{3.14}\) = 60

r²l² = 3600

r² (h² +r²)=3600

r² (64+r²)=3600

r⁴+64r² = 3600 = 0

(r² +100) (r² – 36)=0

∴ r² = -100 or r²=36

r = 6

(∵ r² = -100 is not possible)

∴ Volume of a cone = \(\frac{1}{3} \pi r^2 h=\frac{1}{3} \times 3.14 \times 36 \times 8=301.44 \mathrm{~m}^3\)

Question 7. A Conical tent is required to accommodate 157 persons, each person must have 2m² of space on the ground and 15m³ of air to breathe. Find the height of the tent” Also Calculate the slant height.
Solution:

1 person needs 2m² of Space.

∴ 157 Persons needs 2 × 157 m² of Space on the ground.

πr² = 2 × 157

∴ r² = \(\frac{2 \times 157}{3.14}\)

r² = 100

r = 10m

Also, 1 person needs 15m3 of air.

∴ 157 Persons need 15×157 m2 of air.

∴ \(\frac{1}{3}\) πr2h = 15×157

⇒ h = \(\frac{15 \times 157 \times 3}{3.14 \times 100}\) = 22.5m

∴ l² = h² + r² = (225)² + (10)² = 606.25

∴ l = \(\sqrt{606.25}\) = 24.62m

Question 8. Three Cubes of metal whose edges are in the ratio 3:4:5 are melted down into a single cube whose diagonal is \(12 \sqrt{3}\) cm. Find the edges of the three cubes.
Solution:

The ratio in the edges = 3:4:5

Let edges be 3x, 4x and 5x respectively.

∴ Volumes of three cubes will be 27x³, 64x³, and 125x³ in cm³ respectively.

Now, the Sum of the Volumes of these three Cubes = 27x³ + 64x³+125x³

216x³ cm³

Let the edge of the new cube be a cm.

∴ Diagonal of new Cube = \(a \sqrt{3}\) cm

∴ \(a \sqrt{3}=12 \sqrt{3}\)

⇒ a = 12

∴ Volume of new Cube = (12)³ = 1728 cm³

Now by the given Condition

216x³ = 1728

x³ = 8

x = 2

∴ Edge of 1 Cube = 3×2 = 6cm

Edge of 2 Cube = 4×2 = 8 Cm

Edge of 3 Cube = 5×2 = 10cm

Question 9. A Solid is in the shape of a Cone Standing on a hemisphere with both their radii being equal to Icm and the height of the cone is equal to its radius. Find the volume of the Solid in terms of π.
Solution:

Rodius of hemisphere = radius of Cone = r = 1 cm

Height of Cone h = radius of Cone = 1 cm

Volume of hemisphere = \(\frac{2}{3}\) πr3

volume of Come = \(\frac{1}{3}\) πr2h

∴ Volume of Solid = Volume of hemisphere + volume of Come

⇒ \(\frac{2}{3} \pi r^3+\frac{1}{3} \pi r^2 h\)

⇒ \(\frac{2}{3} \pi(1)^3+\frac{1}{3} \pi(1)^2(1)\)

⇒ \(\frac{2}{3} \pi+\frac{1}{3} \pi\)

= π cm³

Question 10. A granary is in the shape of a Cuboid of a size 8m x 6m x 3m. If a bag of grain occupies a space of 0.65 m³, how many bags Can be stored in the granary?
Solution:

The Size of the granary is 8m x 6m x 3m.

∴ volume of granary =8×6×3 = 144m³

the volume of one bag of grain = 0.65m³

∴ The number of bags that can be stored in the granary

= \(\frac{\text { volume of granary }}{\text { volume of each bag }}\)

= \(\frac{144}{0.65}\)

= 221.54 or 221 bags.

Question 11. A cylindrical bucket 28cm in diameter and 72cm high is full of water. the water is emptied into a rectangular tank 66 cm long and 28cm wide. Find the height of the water level in the tank.
solution:

Let the height of the water level in the tank = xm, then according to problem

πr²h = l×b×x

Or \(\frac{22}{7} \times 14 \times 14 \times 72=66 \times 28 \times x\)

Or \(x=\frac{\frac{22}{7} \times 14 \times 14 \times 72}{66 \times 28}\)

x = 24 Cm

Question 12. A Solid Spherical ball of Iron with a radius of 6cm is melted and recast into three Solid Spherical balls. The radii of the two balls are 3cm and 4cm respectively, determining the diameter of the third ball.
Solution:

Let the radius of the third ball = r cm

∴ The volume of three balls formed = volume of the ball melted

⇒ \(\frac{4}{3} \pi(3)^3+\frac{4}{3} \pi(4)^3+\frac{4}{3} \pi(r)^3=\frac{4}{3} \pi(6)^3\)

⇒ 27 +64 +r³ = 216

⇒ r³ = 125, i.e., r = 5cm

The diameter of the third ball = 2×5cm = 10cm

Question 13. A Semicircle of radius 17.5cm is rotated about its diameter. Find the Curved Surface of So generated Solid.
Solution:

The Solid generated by a circle rotated about its diameter is a Sphere

Now, a radius of Sphere r = 17.5 cm

and its Curved Surface = 4π²

= \(4 \times \frac{22}{7} \times 17.5 \times 17.5\)

= 3850 cm²

Question 14. A sphere of radius 6cm is melted and recast into a Cone of height 6cm. Find the radius of the Cone.
Solution:

Radius of Sphere = 6cm

∴ Volume of Sphere = \(\frac{4}{3} \pi(6)^3\) = 288π cm³

Let radius of Cone = r

Height of Cone = 6cm

∴ Volume of Cone = \(\frac{1}{3} \pi r^2 h=\frac{1}{3} \pi r^2 \times 6=2 \pi r^2\)

Given that, Volume of Cone = volume of a sphere

2πr² = 288π

r² = 144

r = 12

Therefore, a radius of Cone = 12cm

Question 15. A metallic Cylinder of diameter 16 cm and height 9cm is melted and recast Into Sphere of diameter 6cm. How many such Spheres can be formed?
Solution:

For the Cylinder,

Radius = \(\frac{16}{2}\) = 8cm

Height = 9cm

∴ Volume of Cylinder = π (8)²(9) = 576 π cm³

Diameter of Sphere = 6cm

∴ Radius of Sphere = \(\frac{6}{2}\) = 3 cm

Now, Volume of one Sphere = \(\frac{4}{3} \pi(3)^3\) = 36π cm³

∴ Number of Spheres formed = \(=\frac{\text { volume of Cylinder }}{\text { volume of one Sphere }}\)

⇒ \(\frac{576 \pi}{36 \pi}\)

= 16

Question 16. The volume of a sphere is 288π Cm³, 27 Small Spheres Can be formed with this Sphere. Find the radius of the Small Sphere.
Solution:

Volume of 27 Small Spheres = Volume of One big Sphere = 288π

⇒ Volume of 1 Small Sphere = \(\frac{288}{27} \pi\)

⇒ \(\frac{4}{3} \pi r^3=\frac{22}{3} \pi\)

⇒ r³ = 8

⇒ r = 2cm

Question 17. A metallic Sphere of radius 4.2cm is melted and recast into the Shape of a Cylinder of radius 6cm, Find the height of the cylinder.
Solution:

Radius of Sphere R = 4.2cm

Radius of Cylinder r = 6cm

Let the height of the Cylinder = h

Now, the volume of the cylinder = Volume of the Sphere

⇒ \(\pi r^2 h=\frac{4}{3} \pi R^3\)

⇒ h = \(\frac{4 R^3}{3 r^2}\)

⇒ \(\frac{4 \times 4.2 \times 4.2 \times 4.2}{3 \times 6 \times 6} \mathrm{~cm}\)

= 2.744 сm

∴ Height of Cylinder = 2.744cm

Question 18. Metallic Spheres of radii 6cm, 8cm, and 10 cm, respectively, are melted to form a Single Solid Sphere. Find the radius of the resulting Sphere.
Solution:

Let r₁ = 6cm r₂ =8cm and r₃ = 10cm

let the radius of a bigger Solid Sphere = R

The volume of bigger Solid Volume = Sum of volumes of three given Spheres

⇒ \(\frac{4}{3} \pi R^3=\frac{4}{3} \pi r_1^3+\frac{4}{3} \pi r_2^3+\frac{4}{3} \pi r_3^3\)

⇒ \(R^3=r_1^3+r_2^3+r_3^3\)

⇒ \(R^3=6^3+8^3+10^3\)

⇒ R³ = 216 +512 +1000

⇒ R³ = 1728

⇒ R² = 12³

⇒ R = 12cm

∴ Radius of new Solid Sphere = 12 cm

Question 19. A 20m deep well with a diameter 7m is dug and the earth from digging is evenly Spread out to form a platform 22m of the platform.
Solution:

Radius of well, r = \(\frac{7}{2} m\)

and depth h = 20m

Let the height of the platform be H meter.

∴ The volume of platform = Volume of well

⇒ 22 x 14 x H = πr²h

⇒ \(22 \times 14 \times H=\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 20\)

⇒ H = \(\frac{7 \times 5}{14}\)

⇒ H = \(\frac{35}{14}\)

⇒ H = 2.5mn

Height of platform = 2.5m

Question 20. A Cone is divided into two parts by drawing a plane through the mid-point of its axis, parallel to its base, Compare the volume of the two parts.
Solution:

We Can Solve this using Similarity

Let r and h be the radius and height of a Cone OAB

Let OE = \(\frac{h}{2}\)

As OED and OFB are Similar

∴ \(\frac{OE}{O F}=\frac{ED}{F B}\)

⇒ \(\frac{h / 2}{h}=\frac{ED}{r}\)

⇒ ED = \(\frac{r}{2}\)

Now volume of Cone OCD = \(\frac{1}{3} \pi r^2 h=\frac{1}{3} \times \pi \times\left(\frac{r}{2}\right)^2 \times \frac{h}{2}\)

⇒ \(\frac{\pi r^2 h}{24}\)

and volume of Cone DAB = \(\frac{1}{3} \times \pi x r^2 \times h\) = \(\frac{\pi r^2 h}{3}\)

∴ \(\frac{\text { volume of Part } O C D}{\text { volume of Part } C D A B}\)

⇒ \(\frac{\frac{\pi r^2 h}{24}}{\frac{\pi r^2 h}{3}-\frac{\pi r^2 h}{24}}\)

⇒ \(\frac{\frac{1}{24}}{\frac{1}{3}-\frac{1}{24}}\)

⇒ \(\frac{\frac{1}{24}}{\frac{1}{3}-\frac{1}{24}}\)

⇒ \(\frac{\frac{1}{24}}{\frac{8-1}{24}}\)

⇒ \(\frac{1}{7}\)

Question 21. A well of diameter 3m is dug lum deep. The earth taken out of it has been Spread evenly all around it in the Shape of a Circular ring of width um to form an embankment. Find the height of the embankment.
Solution:

Diameter of well 2r = 3m

r = \(\frac{2}{3}\) = 1.5m

and depth h = 14m

∴ Volume of earth taken out from well = πr²h

⇒ \(\frac{22}{7} \times 1.5 \times 1.5 \times 14=99 \mathrm{~m}^3\)

Now, the outer radius of the well, R = 1.5+4 = 5.5m

∴ Area of the ring of platform = π(R²– r²)

⇒ \(\frac{22}{7}\left[(5.5)^2-(1.5)^2\right]\)

⇒ \(\frac{22}{7}[30.25-2.25]\)

⇒ \(\frac{22}{7} \times 7 \times 4=88 \mathrm{~m}^2\)

let the height of the embankment = H

∴ 88 x H=99

⇒ \(H=\frac{99}{88}=\frac{9}{8}\)

= 1.125m

∴ Height of embankment = 1.125m