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# Class 9 RD Sharma Solutions- Chapter 21 Surface Area and Volume of a Sphere – Exercise 21.2 | Set 1

• Last Updated : 13 Jan, 2021

### Question 1. Find the volume of a sphere whose radius is :

(i) 2 cm (ii) 3.5 cm (iii) 10.5 cm.

Solution:

As we know that,

Volume of a sphere = 4/3πr3 Cubic Units Where r is radius of a sphere

(i) Given that, Radius = 2 cm

put in formula and we get,

Volume = 4/3 × 22/7 × (2)3 = 33.52

Volume = 33.52 cm3

(ii) Given that, Radius = 3.5cm

putting value in formula and we get,

Volume = 4/3×22/7×(3.5)3 = 179.666

Volume = 179.666 cm3

(iii) Given that, Radius = 10.5 cm

putting this value in formula and we get,

Volume = 4/3×22/7×(10.5)3 = 4851

Volume = 4851 cm3

### Question 2. Find the volume of a sphere whose diameter is :

(i) 14 cm (ii) 3.5 dm (iii) 2.1 m

Solution:

As we know that,

Volume of a sphere = 4/3πr3 Cubic Units Where r is radius of a sphere

(i) Given that, diameter = 14 cm

So, radius = diameter / 2 = 14/2 = 7cm

putting these value in formula and we get,

Volume = 4/3×22/7×(7)3 = 1437.33

Volume = 1437.33 cm3

(ii) Given that,

Diameter = 3.5 dm

So, radius = diameter/2 = 3.5/2 = 1.75 dm

putting these value in formula and we get,

Volume = 4/3×22/7×(1.75)3 = 22.46

Volume = 22.46 dm3

(iii) Given that,

Diameter = 2.1 m

So, radius = diameter/2 = 2.1/2 = 1.05 m

putting these value in formula and we get,

Volume = 4/3×22/7×(1.05)3 = 4.851

Volume = 4.851 m3

### Question 3. A hemispherical tank has the inner radius of 2.8 m. Find its capacity in liters.

Solution:

Given that,

Radius of hemispherical tank is 2.8 m

Capacity of hemispherical tank is 2/3 πr3 = 2/3×22/7×(2.8)3 m3 = 45.997 m3

[As we know that 1m3 = 1000 liters]

Therefore, capacity in liters = 45997 liters

### Question 4. A hemispherical bowl is made of steel 0.25 cm thick. The inside radius of the bowl is 5 cm. Find the volume of steel used in making the bowl.

Solution:

Given that,

Inner radius of a hemispherical bowl is 5 cm

Outer radius of a hemispherical bowl is 5 cm + 0.25 cm = 5.25 cm

As we know that,

Volume of steel used = Outer volume – Inner volume

= 2/3×π×((5.25)3−(5)3) = 2/3×22/7×((5.25)3−(5)3) = 41.282

Hence Volume of steel used is 41.282 cm3

### Question 5. How many bullets can be made out of a cube of lead, whose edge measures 22 cm, each bullet being 2 cm in diameter?

Solution:

Given that,

Edge of a cube = 22 cm,

Diameter of bullet = 2 cm,

So, radius of bullet(r) = 1 cm,

Volume of the cube = (side)3 = (22)3 cm3 = 10648 cm3 and,

Volume of each bullet which will be in spherical in shape = 4/3πr3

= 4/3 × 22/7 × (1)3 = 4/3 × 22/7

= 88/21 cm3

As we know that,

Number of Bullets = (Volume of Cube) / (Volume of Bullet)

= 10648 / (88/21) = 2541

Hence, 2541 bullets can be made.

Solution:

Given that,

Volume of laddoo having radius 5 cm (V1) = 4/3×22/7×(5)3 (Using Volume of Sphere formula)

= 11000/21 cm3

= 4/3×22/7×(2.5)3 = 1375/21 cm3

Hence, Number of laddoos of radius 2.5 cm that can be made are = V1/V2 = 11000/1375 = 8

### Question 7. A spherical ball of lead 3 cm in diameter is melted and recast into three spherical balls. If the diameters of two balls be 3/2cm and 2 cm, find the diameter of the third ball.

Solution:

Given that,

Volume of lead ball with radius 3/2 cm = 4/3πr3 = 4/3×π×(3/2)3

Lets,

Diameter of first ball (d1) = 3/2cm,

Radius of first ball (r1) = 3/4 cm,

Diameter of second ball (d2) = 2 cm,

Radius of second ball (r2) = 2/2 cm = 1 cm,

Diameter of third ball (d3) = d,

Radius of third ball (r3) = d/2 cm,

As we know that,

Volume of lead ball = 4/3πr13 + 4/3πr23 + 4/3πr33

Volume of lead ball = 4/3π(3/4)3 + 4/3π(2/2)3 + 4/3π(d/2)3

4/3π(3/2)3 = 4/3π[(3/4)3 + (2/2)3 + (d/2)3]

27/8 = 27/64 + 1 + d3/8

d3 = (125 x 8) / 64

d = 10 / 4

Hence, d = 2.5 cm

### Question 8. A sphere of radius 5 cm is immersed in water filled in a cylinder, the level of water rises 5/3 cm. Find the radius of the cylinder.

Solution:

Given that,

Height of water rised = 5/3cm,

Let us assume that radius of Cylinder is r cm,

As we know that Volume of Sphere = 4/3πr3

= 4/3 × π × (5)3

As we know that, Volume of water rised in cylinder = πr2h

Therefore,

Volume of water rises in cylinder = Volume of sphere

πr2h = 4/3πr3

r2 × 5/3 = 4/3 × π × (5)^3

r2 × 5/3 = 4/3 × 22/7 × 125

r2 = 20 × 5

r = √100

r = 10 cm

Hence the radius of cylinder is 10 cm.

### Question 9. If the radius of a sphere is doubled, what is the ratio of the volume of the first sphere to that of the second sphere?

Solution:

Let us assume that v1 and v2 be the volumes of the first and second sphere respectively,

Radius of the first sphere = r,

Radius of the second sphere = 2r

therefore (Volume of first sphere) / (Volume of second sphere)

= 4/3πr3 / 4/3π(2r)3 = 1 / 8

Hence the ratio is 1 : 8

### Question 10. A cone and a hemisphere have equal bases and equal volumes. Find the ratio of their heights.

Solution:

Given that,

Volume of Cone = Volume of Hemisphere

1/3πr2h = 2/3πr3

r2h = 2r3

h = 2r

h/r = 1/1 × 2 = 2

Hence, Ratio of their heights is 2 : 1

### Question 11. A vessel in the form of a hemispherical bowl is full of water. Its contents are emptied in a right circular cylinder. The internal radii of the bowl and the cylinder are 3.5 cm and 7 cm respectively. Find the height to which the water will rise in the cylinder.

Solution:

Given that,

Volume of water in the hemispherical bowl = Volume of water in the cylinder

Let h be the height to which water rises in the cylinder

Inner radii of the bowl = r1 = 3.5 cm

Inner radii of the bowl = r2 = 7 cm

2/3π(r13 )= π(r22)h

h = 2r13 / 3r22

h = 2(3.5)3 / 3(72)

h = 7 / 12 cm

Hence the height to which the water will rise in the cylinder is 7/12 cm.

### Question 12. A cylinder whose height is two thirds of its diameter has the same volume as a sphere of radius 4 cm. Calculate the radius of the base of the cylinder.

Solution:

Given that,

Height of the cylinder = 2/3 diameter

We know that

h = 2/3 × 2r = 4/3r

Volume of Cylinder = Volume of Sphere

πr2h = 4/3πr3

π × r2 × (4/3r) = 4/3π(4)3

(r)3 = (4)3

r = 4 cm

Hence, the radius of the base of the Cylinder is 4 cm.

### Question 13. A vessel in the form of a hemispherical bowl is full of water. The contents are emptied into a cylinder. The internal radii of the bowl and cylinder are respectively 6 cm and 4 cm. Find the height of water in the cylinder.

Solution:

Given that,

Volume of water in Hemispherical bowl = Volume of Cylinder

2/3πr13 = πr22h

h = 2x(6)3 / 3x(4)2

h = 9 cm

Hence the height of water in the cylinder is 9 cm.

### Question 14. A cylindrical tub of radius 16 cm contains water to a depth of 30 cm. A spherical iron ball is dropped into the tub and thus level of water is raised by 9 cm. What is the radius of the ball?

Solution:

Given that,

Radius of the cylinder = 16 cm,

Let’s r be the radius of the iron ball

Then,

Volume of iron ball = Volume of water raised in the hub

4/3 x π x r3 = π x (r)2 x h

4/3 x r3 = (16)2 x 9

r^3 = 1728 = (12)^3

Hence radius of ball is 12 cm.

### Question 15. A cylinder of radius 12 cm contains water to a depth of 20 cm. A spherical iron ball is dropped into the cylinder and thus the level of water is raised by 6.75 cm. Find the radius of the ball. (Use = 227).

Solution:

Given that,

Radius of the cylinder = r1 = 12cm,

Raised in raised = r2 = 6.75 cm,

Volume of water raised = Volume of the sphere

π x (r1)2 x h = 4/3 x π x (r2)3

12 x 12 x 6.75 = 4/3 x (r2)3

= (r2)3 = (12 x 12 x 6.75 x 3) / 4

= r2 = 9 cm

Hence radius of Sphere is 9 cm.

### Question 16. The diameter of a copper sphere is 18 cm. The sphere is melted and is drawn into a long wire of uniform circular cross-section. If the length of the wire is 108 m, find its diameter.

Solution:

Given that,

Diameter of a copper sphere = 18 cm,

Radius of the sphere = 9 cm,

Length of the wire = 108 m = 10800 cm,

Volume of cylinder = Volume of sphere

π x (r1)^2 x h = 4/3 x π x (r2)^3

= (r1)^2 x 10800 = 4/3 x 9 x 9 x 9

= (r1)^2 = 0.009

= r1 = 0.3 cm

Hence Diameter is 0.6 cm.

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