The document RD Sharma Solutions -Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.

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**Q1. Find the volume of a sphere whose radius is : (i) 2 cm (ii) 3.5 cm (iii) 10.5 cm.**

**Sol.**

**(i) ** Radius(r) = 2cm

Therefore volume = Ï€r^{3}

= Ã—Ã—(2)^{3}

= 33.52cm^{3}

**(ii)** Radius(r) = 3.5cm

Therefore volume = Ï€r^{3}

= Ã—Ã—(3.5)^{3} = 179.666cm^{3}

**(iii)** Radius(r) = 10.5cm

Therefore volume = 4^{3}Ï€r^{3}

= 4851cm^{3}

**Q2. Find the volume of a sphere whose diameter is : (i) 14 cm (ii) 3.5 dm (iii) 2.1 m**

**Sol.**

**(i) ** Diameter = 14cm, Radius(r) = = 7cm

Therefore volume = Ï€r^{3}

= 1437.33cm^{3}

**(ii)** Diameter = 3.5dm, Radius(r) = = 1.75dm

Therefore volume = Ï€r^{3}

= 22.46dm^{3}

**(iii)** Diameter = 2.1m, Radius(r) = = 1.05m

Therefore volume = Ï€r^{3}

= 4.851m^{3}

**Q3. A hemispherical tank has the inner radius of 2.8 m. Find its capacity in liters.**

**Sol.**

Radius of the tank = 2.8m

Therefore Capacity = Ï€r^{3}

= 45.994m^{3}

1m^{3} = 1000l

Therefore capacity in litres = 45994 litres

**Q4. 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.**

**Sol.**

Inner radius = 5cm

Outer radius = 5 + 0.25 = 5.25

Volume of steel used = Outer volume - Inner volume

= 41.282cm^{3}

**Q5. How many bullets can be made out of a cube of lead, whose edge measures 22 cm, each bullet being 2 cm in diameter?**

**Sol.**

Cube edge = 22cm

Therefore volume of the cube = (22)^{3} = 10648cm^{3}

And,

Volume of each bullet = Ï€r^{3 }=

Number of bullets = = 2541

**Q6. A shopkeeper has one laddoo of radius 5 cm. With the same material, how many laddoos of radius 2.5 cm can be made?**

**Sol.**

Volume of laddoo having radius = 5cm

i.e Volume(V_{1}) = Ï€r^{3}

Also Volume of laddoo having radius 2.5cm

i.e Volume(V_{2}) = Ï€r^{3}

Therefore number of laddoos = = 8

**Q7. A spherical ball of lead 3 cm in diameter is melted and recast into three spherical balls. If the diameters of two balls be 32cm and 2 cm, find the diameter of the third ball.**

**Sol.**

Volume of lead ball = Ï€r^{3}

Diameter of first balld_{1 } = cm

Radius of first ball r_{1} cm

Diameter of second ball d_{2} = 2cm

Radius of second ball r_{2} = cm = 1cm

Diameter of third balld_{3} = d

Radius of third ball r_{3} = cm

Volume of lead ball =

Volume of lead ball =

d =

d = 2.5cm

** **

**Q8. A sphere of radius 5 cm is immersed in water filled in a cylinder, the level of water rises ****cm. Find the radius of the cylinder.**

**Sol.**

Radius of cylinder = r

Radius of sphere = 5cm

Volume of sphere = Ï€r^{3}

= Ã—Ï€Ã—(5)^{3}

Height of water rised = cm

Volume of water rised in cylinder = Ï€r^{2}h

Therefore, Volume of water rises in cylinder = Volume of sphere

Let r be the radius of the cylinder

Ï€r^{2}h = Ï€r^{3}

r^{2} = 20Ã—5

r =

r = 10cm

**Q9. 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?**

**Sol.**

Let v_{1} and v_{2} be the volumes of the first and second sphere respectively

Radius of the first sphere = r

Radius of the second sphere = 2r

Therefore,

**Q10. A cone and a hemisphere have equal bases and equal volumes. Find the ratio of their heights.**

**Sol.**

Given that

Volume of the cone = Volume of the hemisphere

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

h = 2r

= 2

Therefore

Ratio of their heights = 2 : 1

**Q11. 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.**

**Sol.**

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 = r_{1 } = 3.5cm

Inner radii of the bowl = r_{2} = 7cm

** **

**Q12. 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.**

**Sol.**

Given that

Height of the cylinder = diameter

We know that

Diameter = 2(radius)

Volume of the cylinder = Volume of the sphere

Ï€r^{2}h = Ï€r^{3}

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

r = 4cm

**Q13. 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.**

**Sol.**

It is given that

Volume of water in hemispherical bowl = Volume of cylinder

h = 9cm

**Q14. 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?**

**Sol.**

Let r be the radius of the iron ball

Radius of the cylinder = 16cm

Then,

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

= Ï€r^{2}h

= (16)^{2}Ã—9

r^{3 }=

r^{3} = 1728

r = 12cm

Therefore radius of the ball = 12cm.

**Q15. 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 = ****).**

**Sol.**

Given that :

Radius of the cylinder = 12cm = r_{1}

Raised in raised = 6.75 cm = r_{2}

Volume of water raised = Volume of the sphere

= 12Ã—12Ã—6.75 =

= 729

= r_{2} = 9cm

Radius of the sphere is 9cm

**Q16. 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.**

**Sol.**

Given that diameter of a copper sphere = 18cm

Radius of the sphere = 9cm

Length of the wire = 108m = 10800cm

Volume of cylinder = Volume of sphere

Ã—9Ã—9Ã—9

= 0.009

r_{1} = 0.3cm

Therefore Diameter = 2Ã—0.3 = 0.6cm