Read RD Sharma Solutions Class 9 Chapter 21 Surface Area and Volume of A Sphere below, students should study RD Sharma class 9 Mathematics available on Studiestoday.com with solved questions and answers. These chapter wise answers for class 9 Mathematics have been prepared by teacher of Grade 9. These RD Sharma class 9 Solutions have been designed as per the latest NCERT syllabus for class 9 and if practiced thoroughly can help you to score good marks in standard 9 Mathematics class tests and examinations

**Exercise 21.1**

**Question 1: Find the surface area of a sphere of radius:**

**(i) 10.5 cm (ii) 5.6 cm (iii) 14 cm**

**Solution:**

(i) It is give that,

Radius = 10.5 cm

Surface area of a sphere = 4πr

^{2}Surface area = 4 × 22/7 × (10.5)

^{2}Surface area = 4 × 22/7 × 10.5 × 10.5

Surface area = 4 × 22 × 1.5 × 10.5

Surface area = 1386

Surface area is 1386 cm

^{2}(ii) It is given that,

Radius = 5.6 cm

Surface area = 4 × 22/7 × (5.6)

^{2}Surface area = 4 × 22/7 × 5.6 × 5.6

Surface area = 4 × 22 × 0.8 × 5.6

Surface area = 394.24

Surface area is 394.24 cm

^{2}(iii) It is given that,

Radius = 14 cm

Surface area = 4 × 22/7 × (14)

^{2}Surface area = 4 × 22/7 × 14 × 14

Surface area = 4 × 22 × 2 × 14

Surface area = 2464

Surface area is 2464 cm

^{2}**Question 2: Find the surface area of a sphere of diameter:**

**(i) 14 cm (ii) 21 cm (iii) 3.5 cm**

**Solution:**

Where, r = radius of a sphere

(i) Diameter= 14 cm

Radius = diameter/2

Radius = 14/2 cm

Radius = 7 cm

Surface area of a sphere = 4πr

^{2}Surface area = 4 × 22/7 × (7)

^{2}Surface area = 4 × 22/7 × 7 × 7

Surface area = 4 × 22 × 7

Surface area = 616

Surface area is 616 cm

^{2}(ii) Diameter = 21cm

Radius = diameter/2

Radius = 21/2 cm

Radius = 10.5 cm

Surface area of a sphere = 4πr

^{2}Surface area = 4 × 22/7 × (10.5)

^{2}Surface area = 4 × 22/7 × 10.5 × 10.5

Surface area = 4 × 22 × 1.5 × 10.5

Surface area = 1386

Surface area is 1386 cm

^{2}(iii) Diameter= 3.5cm

Radius = diameter/2

Radius = 3.5/2 cm

Radius = 1.75 cm

Surface area of a sphere = 4πr

^{2}Surface area = 4 × 22/7 × (1.75)

^{2}Surface area = 4 × 22/7 × 1.75 × 1.75

Surface area = 4 × 22 × 0.25 × 1.75

Surface area = 38.5

Surface area is 38.5 cm

^{2}**Question 3: Find the total surface area of a hemisphere and a solid hemisphere each of radius 10 cm. (π = 3.14)**

**Solution:**

It is given that, the Radius of hemisphere is 10cm.

Surface area of the hemisphere = 2πr

^{2}Surface area of the hemisphere = 2 × 3.14 × (10)

^{2 }Surface area of the hemisphere = 2 × 3.14 × 10 × 10

Surface area of the hemisphere = 628cm

^{2}Surface area of solid hemisphere = 3πr

^{2}Surface area of solid hemisphere = 3 × 3.14 × (10)

^{2}Surface area of solid hemisphere = 3 × 3.14 × 10 × 10

Surface area of solid hemisphere = 942cm

^{2}**Question 4: The surface area of a sphere is 5544 cm**

^{2}, find its diameter.**Solution:**

It is given that,

Surface area of a sphere is 5544cm

^{2}Surface area of a sphere = 4πr

^{2}4πr2 = 5544

4 × 22/7 × (r)

^{2}= 5544r

^{2}= (5544 × 7)/(22 × 4)r

^{2}= (5544 × 7)/88r

^{2}= (5544 × 7)/88r

^{2}= 63×7r

^{2}= 441r = 21cm

Diameter = 2(radius)

Diameter = 2(21)

Diameter = 42cm

Hence, the diameter is surface area of a sphere is 42cm.

**Question 5: A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin plating it on the inside at the rate of Rs.4 per 100 cm**

^{2}.**Solution:**

It is given that,

Inner diameter of hemispherical bowl = 10.5 cm

Radius = Diameter/2

Radius = 10.5/2cm

Radius = 5.25 cm

Surface area of hemispherical bowl = 2πr

^{2}Surface area of hemispherical bowl = 2 × 3.14 × (5.25)

^{2}Surface area of hemispherical bowl = 173.25

Surface area of hemispherical bowl is 173.25cm

^{2}It is also given that,

Cost of tin plating 100cm

^{2}area is Rs.4Cost of tin plating 173.25cm

^{2}area = Rs. 4 × 173.25/100 = Rs. 6.93Hence, cost of tin plating the inner side of hemispherical bowl is Rs.6.93.

**Question 6: The dome of a building is in the form of a hemisphere. Its radius is 63 dm. Find the cost of painting it at the rate of Rs. 2 per sq m.**

**Solution:**

It is given that, Radius of hemispherical = 63dm

Change in meters,

Radius of hemispherical dome = 63/10

Radius of hemispherical dome = 6.3m

Inner surface area = 2πr

^{2}Inner surface area = 2 × 3.14 × (6.3)

^{2}Inner surface area = 2 × 3.14 × 6.3 × 6.3

Inner surface area = 249.48

Hence, the inner surface area of dome is 249.48 m

^{2}It is also given that,

Cost of painting 1m

^{2}= Rs.2Cost of painting 249.48 m

^{2}Cost of painting = Rs. 249.48 × 2

Cost of painting = Rs. 498.96

Hence, the cost of painting is Rs. 498.96.

**Exercise 21.2**

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

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

**Solution:**

(i) It is given that,

Radius = 2 cm

Volume of a sphere = 4/3πr3

Volume of a sphere = 4/3 × 22/7 × (2)

^{3}Volume of a sphere = 33.52

Volume of a sphere = 33.52 cm

^{3}(ii) It is given that,

Radius = 3.5cm

Volume of a sphere = 4/3πr

^{3 }Volume of a sphere = 4/3 × 22/7 × (3.5)

^{3}Volume of a sphere = 4/3 × 22/7 × 3.5 × 3.5

Volume of a sphere = 179.666

Volume of a sphere = 179.666 cm

^{3}(iii) It is given that,

Radius = 10.5 cm

Volume of a sphere = 4/3πr

^{3 }Volume of a sphere = 4/3 × 22/7 × (10.5)

^{3}Volume of a sphere = 4/3 × 22/7 × 10.5 × 10.5

Volume of a sphere = 4/3 × 22 × 1.5 × 10.5

Volume of a sphere = 4851

Volume of a sphere = 4851 cm

^{3}**Question 2: Find the volume of a sphere whose diameter is:**

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

**Solution:**

(i) It is given that,

Diameter = 14 cm

Radius = diameter/2 = 14/2 = 7cm

Volume of a sphere = 4/3πr

^{3 }Volume of a sphere = 4/3 × 22/7 × (7)

^{3}Volume of a sphere = 4/3 × 22/7 × 7 × 7 × 7

Volume of a sphere = 1437.33

Volume of a sphere = 1437.33 cm

^{3}(ii) It is given that,

Diameter = 3.5dm

Radius = diameter/2

Radius = 3.5/2

Radius = 1.75dm

Volume of a sphere = 4/3πr

^{3}Volume of a sphere = 4/3 × 22/7 × (1.75)

^{3}Volume of a sphere = 4/3 × 22/7 × 1.75 × 1.75 × 1.75

Volume of a sphere = 22.46

Volume of a sphere = 22.46 dm

^{3}(iii) It is given that,

Diameter = 2.1m

Radius = diameter/2

Radius = 2.1/2

Radius = 1.05 m

Volume of a sphere = 4/3πr

^{3}Volume of a sphere = 4/3 × 22/7 × (1.05)

^{3}Volume of a sphere = 4/3 × 22/7 × 1.05 × 1.05 × 1.05

Volume of a sphere = 4 × 22 × 0.05 × 1.05 × 1.05

Volume of a sphere = 4.851

Volume of a sphere = 4.851 m

^{3}**Question 3: A hemispherical tank has the inner radius of 2.8 m. Find its capacity in litre.**

**Solution:**

It is given that, the radius of hemispherical tank is 2.8 m

Capacity of hemispherical tank = 2/3 πr

^{3}Capacity of hemispherical tank = 2/3 × 22/7 × (2.8)

^{3 }Capacity of hemispherical tank = 2/3 × 22/7 × 2.8 × 2.8 × 2.8

Capacity of hemispherical tank = 2/3 × 22 × 0.4 × 2.8 × 2.8

Capacity of hemispherical tank = 45.997 m

^{3}Change in litres

1m

^{3}= 1000 litresCapacity in litres = 45.997 × 1000

Capacity in litres = 45997 litres

Hence, the capacity in litres is 45997litres.

**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:**

It is given that, the inner radius of a hemispherical bowl is 5cm

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

Volume of steel used = Outer volume – Inner volume

Volume of hemisphere = 2/3 πR

^{3}- 2/3 πr^{3 }Volume of hemisphere = 2/3 π(R

^{3}- r^{3})Volume of steel used = 2/3 × π × ((5.25)

^{3}− (5)^{3})Volume of steel used = 2/3 × 22/7 × (144.70 − 125)

Volume of steel used = 2/3 × 22/7 × (19.70)

Volume of steel used = 41.282cm

^{3}Hence, the volume of steel used in making the bowl is 41.282 cm

^{3}**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:**

Edge of a cube = 22cm

Diameter of bullet = 2cm

Radius of bullet (r) = 2/2 cm = 1cm

Volume of the cube = (side)

^{3 }Volume of the cube = (22)3 cm

^{3 }Volume of the cube = 10648 cm

^{3}Volume of sphere = 4/3πr3

Volume of sphere = 4/3 × 22/7 × (1)3 cm

^{3}Volume of sphere = 4/3 × 22/7 cm

^{3}Volume of sphere = 88/21 cm

^{3}Number of bullets =( (Volume of cube) )/((Volume of bullet))

Number of bullets = (10648/88)/21

Number of bullets = (10648 × 21)/88

Number of bullets = 2541

Hence, 2541 bullets can be made out of a cube of lead.

**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:**

Volume of laddoo (of Radius 5 cm) = 4/3πr

^{3}Volume of laddoo (of Radius 5 cm) = 4/3 × 22/7 × (5)

^{3}Volume of laddoo (of Radius 5 cm) = 4/3 × 22/7 × 5 × 5 × 5

Volume of laddoo (of Radius 5 cm) = 4/3 × 22/7 × 125

Volume of laddoo (of Radius 5cm) = 11000/21 cm

^{3}Volume of laddoo (of Radius 2.5) = 4/3πr

^{3}Volume of laddoo (of Radius 2.5) = 4/3 × 22/7 × (2.5)

^{3 }Volume of laddoo (of Radius 2.5) = 4/3 × 22/7 × 2.5 × 2.5 × 2.5

Volume of laddoo (of Radius 2.5) = 4/3 × 22/7 × 15.625

Volume of laddoo (of Radius 2.5) = 1375/21 cm

^{3}Number of laddoos of radius 2.5 cm that can be made with laddoos of radius 5cm

Number of laddoos = (Volume of Laddoo of R=5cm)/(Volume of Laddoo of R=2.5cm)

Number of laddoos = (11000/21)/(1375/21)

Number of laddoos = 11000/1375

Number of laddoos = 8

Hence, the Number of laddoos is 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:**

It is given that,

Diameter of spherical ball = 3cm

Radius of spherical ball = 3/2cm

Volume of lead ball = 4/3πr

^{3}Volume of lead ball = 4/3 × π × (3/2)

^{3}It is also given that,

Diameter of first ball = 3/2cm

Radius of first ball = 3/4 cm

Diameter of second ball = 2 cm

Radius of second ball = 2/2 cm = 1 cm

Let us assumed that,

Diameter of third ball = d

Radius of third ball = d/2 cm

We know that,

Volume of Lead ball = Volume of First Ball + Volume of Second Ball + Volume of Third Ball

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

**Solution:**

It is given that, Radius of sphere is 5 cm

Let ‘r’ be the radius of cylinder

Volume of sphere = 4/3πr

^{3}Volume of sphere = 4/3 × π × (5)

^{3}Volume of sphere = 4/3 × π × 5 × 5 × 5

Volume of sphere = 4/3 × π × 125

It is also given that,

Height (h) of water rises = 5/3 cm

Volume of water rises in cylinder = πr

^{2}hVolume of water rises in cylinder = Volume of sphere

πr

^{2}h = 4/3πr^{3}π × r

^{2}× 5/3 = 4/3 × π × 125r

^{2}= 4/3 × 125 × 3/5r

^{2}= 4 × 25r

^{2}= 100r = 10

Hence, the radius of the 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 assumed that, the radius of the first sphere = r

Radius of the second sphere = 2r

Volume of sphere = 4/3 πr

^{3}**Question 10: A cone and a hemisphere have equal bases and equal volumes. Find the ratio of their heights.**

**Solution:**

It is given that,

Volume of the cone = Volume of the hemisphere

1/3 πr

^{2}h = 2/3 πr^{3}r

^{2}h = 2r^{3}h = (2r

^{3})/r^{2}h = 2r

Height and radius of a hemisphere is equal.

h/r = 2/1

h : r = 2 : 1

Hence, the 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:**

It is given that,

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

Inner radius of the bowl (r) = 3.5cm

Inner radius of cylinder (R) = 7cm

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

2/3 π(r)

^{3}= π(R)^{2}h2/3 π(3.5)

^{3}= π(7)^{2}h2/3×3.5×3.5×3.5 = 7×7×7×h

(2 × 3.5 × 3.5 × 3.5)/(3 × 7 × 7 × 7) = h

(2 × 0.5 × 0.5 × 0.5)/3 = h

0.25/3=h

0.0833…. cm = h

Hence, the height to which water rises in the cylinder is 0.0833cm.

**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:**

It is given that,

Radius of a sphere = 4cm

Let the radius of cylinder be r

Height of the cylinder = 2/3 diameter

h = 2/3 × (2r)

h = 4r/3

Volume of the cylinder = Volume of the sphere

πr

^{2}h = 4/3πR^{3}π × r

^{2}× (4r/3) = 4/3 π (4)^{3}r

^{2}× (r) = (4)^{3}r

^{3}= (4)^{3}r

^{3}= 4×4×4r = 4

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:**

It is given that,

Radius of a bowl (r

_{1}) = 6cmRadius of a cylinder (r

_{2}) = 4cmLet

Height of a cylinder = h

Volume of water in hemispherical bowl = Volume of cylinder

2/3πR

^{3}= πr^{2}h2/3 × π × (6)

^{3}= π × (4)^{2}× h2 × 2 × 6 × 6 = 4 × 4 × 4 × h

(2 × 2 × 6 × 6)/(4 × 4 × 4) = h

3 × 3 = h

h = 9

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

**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:**

It is given that,

Radius of the cylinder (R) = 16cm

Let us assumed that,

Radius of the iron ball = r

Height = 9 cm

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

4/3πr

^{3}= πR^{2}h4/3× π × r

^{3}= π × (16)^{2}× 94/3 × r

^{3}= 16 × 16 × 9r

^{3}= (16 × 16 × 9 × 3)/4r

^{3}= 16 × 4 × 9 × 3r

^{3}= 1728r = 12

Hence, the radius of the ball is 12cm.

**Exercise VSAQs ..........................**

**Question 1: Find the surface area of a sphere of radius 14 cm.**

**Solution:**

It is given that,

Radius of a sphere = 14 cm

Surface area of a sphere = 4πr

^{2}Surface area of a sphere = 4 × 22/7 × 14 × 14

Surface area of a sphere = 2464 cm

^{2}**Question 2: Find the total surface area of a hemisphere of radius 10 cm.**

**Solution:**

It is given that,

Radius of a hemisphere = 10 cm

TSA of a hemisphere = 3πr

^{2}TSA of a hemisphere = 3 × (22/7) × 10 × 10

TSA of a hemisphere = 942.85cm

^{2}**Question 3: Find the radius of a sphere whose surface area is 154 cm**

^{2}.**Solution:**

It is given that,

Surface area of a sphere = 154cm

^{2}Surface area of a sphere = 4πr

^{2}4πr

^{2}= 1544 × 22/7 × r

^{2}= 154r

^{2}= (154 × 7)/(22 × 4)r

^{2}= (7 × 7)/4r

^{2}= 49/4r

^{2}= √(49/4)r = 7/2

r = 3.5

Hence, the radius of a sphere is 3.5 cm.

**Question 4: The hollow sphere, in which the circus motor cyclist performs his stunts, has a diameter of 7 m. Find the area available to the motorcyclist for riding.**

**Solution:**

It is given that,

Diameter of hollow sphere = 7 m

Radius of hollow sphere = 7/2m = 3.5 cm

Surface area of a sphere = 4πr

^{2}Surface area of a sphere = 4 × 22/7 × 3.5 × 3.5

Surface area of a sphere = 4 × 22 × 0.5 × 3.5

Surface area of a sphere = 154 m

^{2}Hence, the area available to the motorcyclist for riding is 154 m

^{2}.**Question 5: Find the volume of a sphere whose surface area is 154 cm**

^{2}.**Solution:**

It is given that,

Surface area of a sphere = 154 cm

^{2}Surface area of a sphere = 4πr

^{2}4πr

^{2}= 1544 × 22/7 × r

^{2}= 154r

^{2}= (154 × 7)/(22 × 4)r

^{2}= (7 × 7)/4r

^{2}= 49/4r

^{2}= √(49/4)r = 7/2

r = 3.5

Radius = 3.5 cm

Volume of sphere = 4/3πr

^{3}Volume of sphere = 4/3 × 22/7 × 3.5 × 3.5 × 3.5

Volume of sphere = 4/3 × 22 × 0.5 × 3.5 × 3.5

Volume of sphere = 179.66

Hence, the volume of sphere is 179.66 cm

^{3}.