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

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.

Question 6. A shopkeeper has one laddoo of radius 5 cm. With the same material, how many laddoos of radius 2.5 cm 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

Also, Volume of laddoo having radius 2.5 cm (V2) = 4/3πr³

= 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,

Radius of sphere = 5cm,

Height of water raised = 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 raised 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

Diameter = 2(radius)

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.