# 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: **

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As we know that,

Volume of a sphere = 4/3πr

^{3}Cubic Units Where r is radius of a sphere

(i)Given that, Radius = 2 cmput in formula and we get,

Volume = 4/3 × 22/7 × (2)

^{3}= 33.52

Volume = 33.52 cm^{3}

(ii)Given that, Radius = 3.5cmputting value in formula and we get,

Volume = 4/3×22/7×(3.5)

^{3}= 179.666

Volume = 179.666 cm^{3}

(iii)Given that, Radius = 10.5 cmputting this value in formula and we get,

Volume = 4/3×22/7×(10.5)

^{3}= 4851

Volume = 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:**

As we know that,

Volume of a sphere = 4/3πr

^{3}Cubic Units Where r is radius of a sphere

(i)Given that, diameter = 14 cmSo, 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 cm^{3}

(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 dm^{3}

(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 m^{3}

**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 πr

^{3}= 2/3×22/7×(2.8)^{3}m^{3}= 45.997 m^{3}[As we know that 1m

^{3}= 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 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:**

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}cm^{3 }= 10648 cm^{3}and,Volume of each bullet which will be in spherical in shape = 4/3πr

^{3}= 4/3 × 22/7 × (1)

^{3}= 4/3 × 22/7= 88/21 cm

^{3}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 cm

^{3}Also, Volume of laddoo having radius 2.5 cm (V2) = 4/3πr

^{3}= 4/3×22/7×(2.5)

^{3}= 1375/21 cm^{3}

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 = 4/3πr

^{3}= 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 = 4/3πr

_{1}^{3}+ 4/3πr_{2}^{3}+ 4/3πr_{3}^{3}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 rised = 5/3cm,

Let us assume that radius of Cylinder is r cm,

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

^{3}= 4/3 × π × (5)

^{3}As we know that, Volume of water rised in cylinder = πr

^{2}hTherefore,

Volume of water rises in cylinder = Volume of sphere

πr

^{2}h = 4/3πr^{3}r

^{2}× 5/3 = 4/3 × π × (5)^3r

^{2 }× 5/3 = 4/3 × 22/7 × 125r

^{2 }= 20 × 5r = √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)

= 4/3πr

^{3}/ 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

1/3πr

^{2}h = 2/3πr^{3}r

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

_{1}^{3 })= π(r_{2}^{2})hh = 2r1

^{3 }/ 3r2^{2}h = 2(3.5)

^{3}/ 3(7^{2})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

πr

^{2}h = 4/3πr^{3}π × r

^{2}× (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

2/3πr1

^{3}= πr2^{2}hh = 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 r

^{3}= π x (r)^{2}x h4/3 x r

^{3}= (16)^{2}x 9r^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= r

^{2 }= 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.