## RD Sharma Solutions Class 9 Chapter 21 Exercise 21.2

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

**Sol.**

**(i)Â ** Radius(r)= 2cm

Therefore volume=\(\frac{4}{3}\pi r^{3}\)

=\(\frac{4}{3}\times\frac{22}{7}\times\left ( 2 \right )^{3}\)

=33.52\(cm^{3}\)

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

Therefore volume=\(\frac{4}{3}\pi r^{3}\)

=\(\frac{4}{3}\times\frac{22}{7}\times\left ( 3.5 \right )^{3}\)

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

Therefore volume=\(\frac{4}{3}\pi r^{3}\)

=\(\frac{4}{3}\times\frac{22}{7}\times\left ( 10.5 \right )^{3}\)

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

**Sol.**

**(i)Â ** Diameter=14cm, Radius(r)= \(\frac{14}{2}\)

Therefore volume=\(\frac{4}{3}\pi r^{3}\)

=\(\frac{4}{3}\times\frac{22}{7}\times\left ( 7 \right )^{3}\)

**(ii)** Diameter=3.5dm, Radius(r)= \(\frac{3.5}{2}\)

Therefore volume=\(\frac{4}{3}\pi r^{3}\)

=\(\frac{4}{3}\times\frac{22}{7}\times\left ( 1.75 \right )^{3}\)

=22.46\(dm^{3}\)

**(iii)** Diameter=2.1m, Radius(r)= \(\frac{2.1}{2}\)

Therefore volume=\(\frac{4}{3}\pi r^{3}\)

=\(\frac{4}{3}\times\frac{22}{7}\times\left ( 1.05 \right )^{3}\)

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

**Sol.**

Radius of the tank= 2.8m

Therefore Capacity= \(\frac{2}{3}\pi r^{3}\)

=\(\frac{2}{3}\times\frac{22}{7}\times\left ( 2.8 \right )^{3}\)

\(1m^{3}=1000l\)

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

= \(\frac{2}{3}\times\pi\times\left ( \left ( 5.25 \right )^{3}-\left ( 5 \right )^{3} \right )\)

= \(\frac{2}{3}\times\frac{22}{7}\times\left ( \left ( 5.25 \right )^{3}-\left ( 5 \right )^{3} \right )\)

= 41.282\(cm^{3}\)

**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= \(\left ( 22 \right )^{3}\)

And,

Volume of each bullet=\(\frac{4}{3}\pi r^{3}\\ =\frac{4}{3}\times\frac{22}{7}\times\left(1 \right )^{3}\\ =\frac{4}{3}\times\frac{22}{7}\\ =\frac{88}{21}cm^{3}\)

Number of bullets=\(\frac{Volume\;of\;cube}{Volume\;of\;bullet}\\ =\frac{10648}{\frac{88}{21}}\\ =2541\)

**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\(\left ( V_{1} \right )\)

\(\left ( V_{1} \right )\)

\(\left ( V_{1} \right )\)

Also Volume of laddoo having radius 2.5cm

i.e Volume\(\left ( V_{2} \right )\)

\(\left ( V_{2} \right )\)

\(\left ( V_{2} \right )\)

Therefore number of laddoos=\(\frac{V_{1}}{V_{2}}\)

**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 \(\frac{3}{2}\)cm and 2 cm, find the diameter of the third ball.**

**Sol.**

Volume of lead ball= \(\frac{4}{3}\pi r^{3}\)

=\(\frac{4}{3}\times\frac{22}{7}\times\left ( \frac{3}{2} \right )^{3}\)

Diameter of first ball\(d_{1}\)

Radius of first ball\(r_{1}\)

Diameter of second ball\(d_{2}\)

Radius of second ball\(r_{2}\)

Diameter of third ball\(d_{3}\)

Radius of third ball\(r_{3}\)

Volume of lead ball=\(\frac{4}{3}\pi r_{1}^{3}+\frac{4}{3}\pi r_{2}^{3}+\frac{4}{3}\pi r_{3}^{3}\)

Volume of lead ball= \(\frac{4}{3}\times\pi\times\left ( \frac{3}{4} \right )^{3}\)

\(\frac{4}{3}\times\frac{22}{7}\times\left ( \frac{3}{2} \right )^{3}\)

\(\frac{4}{3}\pi\left [ \left ( \frac{3}{2} \right )^{3} \right ]=\frac{4}{3}\pi\left [ \left ( \frac{3}{4} \right )^{3}+\left ( \frac{2}{2} \right )^{3}+\left ( \frac{d}{2} \right )^{3} \right ]\)

\(\frac{27}{8}=\frac{27}{64}+1+\frac{d^{3}}{8}\)

\(d^{3}=8\left [ \frac{27}{8}-\frac{27}{64}-1 \right ]\)

\(\frac{d^{3}}{8}=\frac{125}{64}\)

\(\frac{d}{2}=\frac{5}{4}\)

\(d=\frac{10}{4}\)

d=2.5cm

**Â **

**Q8. A sphere of radius 5 cm is immersed in water filled in a cylinder, the level of water rises \(\frac{5}{3}\)cm. Find the radius of the cylinder.**

**Sol.**

Radius of cylinder=r

Radius of sphere= 5cm

Volume of sphere=\(\frac{4}{3}\pi r^{3}\)

=\(\frac{4}{3}\times\pi \times\left ( 5 \right)^{3}\)

Height of water rised= \(\frac{5}{3}\)

Volume of water rised in cylinder=\(\pi r^{2}h\)

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

Let r be the radius of the cylinder

\(\pi r^{2}h=\frac{4}{3}\pi r^{3}\)

\(r^{2}\times\frac{5}{3}=\frac{4}{3}\times\pi\times \left ( 5 \right )^{3}\)

\(r^{2}\times\frac{5}{3}=\frac{4}{3}\times\frac{22}{7}\times 125\)

\(r^{2}=20\times5\)

\(r=\sqrt{100}\)

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 \(v_{1}\)

Radius of the first sphere = r

Radius of the second sphere = 2r

Therefore, \(\frac{Volume of first sphere}{Volume of second sphere}\)

= \(\frac{1}{8}\)

**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

\(\frac{1}{3}\pi r^{2}h=\frac{2}{3}\pi r^{3}\)

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

h=2r

\(\frac{h}{r}=\frac{1}{1}\times 2=2\)

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= \(r_{1}\)

Inner radii of the bowl= \(r_{2}\)

\(\frac{2}{3}\pi r_{1}^{3}=\pi r_{2}^{2}h\)

\(h=\frac{2r_{1}^{3}}{3r_{2}^{2}}\)

\(h=\frac{2\left (3.5 \right )^{3} }{3\left (7^{2} \right )}\)

\(h=\frac{7}{12}cm\)

**Â **

**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= \(\frac{2}{3}diameter\)

We know that

Diameter= 2(radius)

\(h=\frac{2}{3}\times 2r=\frac{4}{3}r\)

Volume of the cylinder=Volume of the sphere

\(\pi r^{2}h=\frac{4}{3}\pi r^{3}\)

\(\pi\times r^{2}\times \left ( \frac{4}{3}r \right )=\frac{4}{3}\pi\left ( 4 \right )^{3}\)

\(\left ( r \right )^{3}=\left ( 4 \right )^{3}\)

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

\(\frac{2}{3}\pi r_{1}^{3}=\pi r_{2}^{2}h\)

\(\frac{2}{3}\pi\left ( 6 \right )^{3}=\pi\left ( 4 \right )^{2}h\)

\(h=\frac{2}{3}\times \frac{6\times 6\times 6}{4\times 4}\)

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

\(\frac{4}{3}\pi r^{3}=\pi r^{2}h\)

\(\frac{4}{3}r^{3}=\left ( 16 \right )^{2}\times 9\)

\(r^{3}=\frac{27\times 16\times 16}{4}\)

\(r^{3}=1728\)

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 = \(\frac{22}{7}\)). **

**Sol.**

Given that:

Radius of the cylinder = 12cm = \( r_{1}\)

Raised inÂ raised = 6.75 cm = \(r_{2}\)

Volume of water raised = Volume of the sphere

= \(\pi r_{1}^{2}h=\frac{4}{3}\pi r_{2}^{3}\)

= \(12\times12\times6.75=\frac{4}{3} r_{2}^{3}\)

= \(r_{2}^{3} = \frac{12\times12\times6.75\times3}{4}\)

= \(r_{2}^{3}\)

= \(r_{2}\)

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. **

**Â **

**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

\(\pi r_{1}^{2}h=\frac{4}{3}\pi r_{2}^{3}\\ r_{1}^{2}\times10800=\frac{4}{3}\times9\times9\times9\\ r_{1}^{2}=0.009\\ r_{1}=0.3cm\)

Therefore \(Diameter=2\times0.3=0.6cm\)

**Â **

**Q17. A cylindrical jar of radius 6 cm contains oil. Iron spheres each of radius 1.5 cm are immersed in the oil. How many spheres are necessary to raise the level of the oil by two centimeters?**

**Sol.**

Given that,

Radius of the cylinder jar=6cm=\(r_{1}\)

Level to be rised=2cm

Radius of each iron sphere=1.5cm=\(r_{2}\)

\( Number\;of\;sphere=\frac{Volume\;of\;cylinder}{Volume\;of\;sphere}\\ =\frac{\pi r_{1}^{2}h}{\frac{4}{3}\pi r_{2}^{3} }\\ =\frac{r_{1}^{2}h}{\frac{4}{3} r_{2}^{3}}\\ =\frac{6\times6\times2}{\frac{4}{3}\times1.5\times1.5\times1.5}\\ Number\;of\;sphere=16\)

**Q18. A measuring jar of internal diameter 10 cm is partially filled with water. Four equal spherical balls of diameter 2 cm each are dropped in it and they sink down in water completely. What will be the change in the level of water in the jar?**

**Sol.**

Given that,

Diameter of jar=10cm

Radius of jar=5cm

Let the level of water be raised by h

Diameter of the spherical bowl=2cm

Radius of the ball=1cm

Volume of jar=4(Volume of spherical ball)

\(\pi r_{1}^{2}h=4\left(\frac{4}{3}\pi r_{2}^{3} \right )\\ r_{1}^{2}h=4\left(\frac{4}{3} r_{2}^{3} \right )\\ 5\times 5\times h=4\times\frac{4}{3}r_{2}^{3}\\ 5\times 5\times h=4\times\frac{4}{3}\times 1\times 1\times1\\ h=\frac{4\times4\times1}{3\times5\times5}\\ h=\frac{16}{75}cm\\ Height of water in jar=\frac{16}{75}cm\)

**Q19. The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 0.2 cm. Find the length of the wire.**

**Sol.**

Given that

Diameter of sphere=6cm

\(Radius\;of\;sphere=\frac{d}{2}=\frac{6}{2}=3cm=r_{1}\\ Diameter\;of\;the\;wire=0.2cm\\ Radius\;of\;the\;wire=0.1cm=r_{2}\\\)

Volume of sphere=Volume of wire

\(\frac{4}{3}\pi r_{1}^{3}=\pi r_{2}^{2}h\\ =\frac{4}{3}\times3\times3\times3=0.1\times0.1\times h\\ h=\frac{4\times3\times3}{0.1\times0.1}\\ h=3600cm\\ h=36m\\\)

Therefore length of wire=36m

**Q20. The radius of the internal and external surfaces of a hollow spherical shell are 3 cm and 5 cm respectively. If it is melted and recast into a solid cylinder of height \(2 2/3\)cm. Find the diameter of the cylinder.**

**Sol.**

Given that,

Internal radius of the sphere=3cm=\(r_{1}\)

External radius of the sphere=5cm=\(r_{2}\)

Height of the cylinder=\(\frac{8}{3}cm=h\)

Volume of the spherical shell = Volume of cylinder

\(\frac{4}{3}\pi\left(r_{2}^{3}-r_{1}^{3} \right )=\pi r_{3}^{2}h\\ \frac{4}{3}\left(5^{3}-3^{3} \right )=\frac{8}{3}r_{3}^{2}\\ r_{3}^{2}=\frac{4\times98\times3}{3\times8}\\ r_{3}=\sqrt{49}\\ r_{3}=7cm\)

Therefore diameter of the cylinder=2(radius)=14cm

**Q21. A hemisphere of the lead of radius 7 cm is cast into a right circular cone of height 49 cm. Find the radius of the base.**

**Sol.**

Given

Radius of the hemisphere=Volume of cone

\(\frac{2}{3}\pi r_{1}^{3}=\frac{1}{3}\pi r_{2}^{2}h\\ \frac{2}{3}\times 7^{3}=\frac{1}{3}r_{2}^{2}\times49\\ r_{2}^{2}=\frac{2\times 7\times 7\times 7\times 3}{3\times 49} \\ r_{2}^{2}=\frac{2058}{147}\\ r_{2}=3.47cm\)

Therefore radius of the base=3.74cm

**Q22. A hollow sphere of internal and external radii 2cm and 4 cm respectively is melted into a cone of base radius 4cm. Find the height and slant height of the cone.**

**Sol.**

Given that

Hollow sphere external radii=\(r_{2}\)

Internal radii=\(r_{1}\)

Cone base radius(R)=4cm

Height=h

Volume of cone=Volume of sphere

\(\frac{1}{3}\pi r^{2}h=\frac{4}{3}\pi\left(r_{2}^{2}-r_{1}^{2} \right )\\ \;4^{2}h=4\left(4^{3}-2^{3} \right )\\ \;h=\frac{4\times56}{16}\\ \;h=14cm\\ Slant\;height\left(l \right )=\sqrt{r^{2}+h^{2}}\\ Slant\;height\left(l \right )=\sqrt{r_{2}^{2}+h^{2}}\\ l=\sqrt{4^{2}+\left(14^{2} \right )}\\ l=\sqrt{16+196}\\ l=\sqrt{212}\\ l=14.56cm\)

**Q23. A metallic sphere of radius 10.5 cm is melted and thus recast into small cones, each of radius 3.5 cm and height 3 cm. Find how many cones are obtained.**

**Sol.**

Given that

Metallic sphere of radius=10.5cm

Cone radius=3.5cm

Height of radius=3cm

Let the number of cones obtained be x

\(v_{s}=x\times v_{cone}\\ \frac{4}{3}\pi r^{3}=x\times \frac{1}{3}\pi r^{2}h\\ x=\frac{4\times 10.5\times 10.5\times 10.5}{3.5\times 3.5\times 3}\\ x=126\)

Therefore number of cones=126

**Â **

**Q24. A cone and a hemisphere have equal bases and equal volumes. Find the ratio of their heights.**

**Sol.**

Given that

A cone and a hemisphere have equal bases and volumes

\(v_{cone}=v_{hemisphere}\\ \frac{1}{3}\pi r^{2}h=\frac{2}{3}\pi r^{3}\\ r^{2}h=2 r^{3}\\ h=2r\\ \frac{h}{r}=\frac{2}{1}\\ h:r=2:1\)

Therefore the ratio is 2:1

**Q25. A cone, a hemisphere, and a cylinder stand on equal bases and have the same height. Show that their volumes are in the ratio 1 : 2 : 3.**

**Sol.**

Given that

A cone, a hemisphere and a cylinder stand on one equal bases and have the same weight

We know that

\(v_{cone}:v_{hemisphere}:v_{cylinder}\\ \frac{1}{3}\pi r^{2}h:\frac{2}{3}\pi r^{3}:\pi r^{2}h\\ multiplying\;by\;3\\ \pi r^{2}h:2\pi r^{3}:3\pi r^{2}h\\ \pi r^{3}:2\pi r^{3}:3\pi r^{3}\left(âˆµ r=h\;and\;r^{2}h=r^{3} \right )\\ 1:2:3\\ Therefore\;the\;ratio\;is\;1:2:3\)

**Q26. A cylindrical tub of radius 12 cm contains water to a depth of 20 cm. A spherical form ball is dropped into the tub and thus the level of water is raised by 6.75 cm. What is the radius of the ball?**

**Sol.**

Radius of cylindrical tub=12cm

Depth=20cm

Let r be the radius of the ball

Then

Volume of the ball= Volume of water raised

\(\frac{4}{3}\pi r^{3}=\pi r^{2}h\\ r^{3}=\frac{3.14\times\left(12 \right )^{2}\times 6.75\times 3}{4}\\ r^{3}=729\\ r=\sqrt[3]{729}\\ r=9cm\)

Therefore radius of the ball=9cm

**Q27. The largest sphere is carved out of a cube of side 10.5 cm. Find the volume of the sphere.**

**Sol.**

Side of cube=10.5cm

Volume of sphere=v

Diameter of the largest sphere=10.5cm

2r=10.5

r=5.25cm

\(Volume\;of\;sphere=\frac{4}{3}\pi r^{3}=\frac{4}{3}\times\frac{22}{7}\times5.25\times 5.25 \times 5.25\\ v=\frac{11\times 441}{8}cm^{3}\\ v=606.375cm^{3}\)

**Q28. A sphere, a cylinder, and a cone have the same diameter. The height of the cylinder and also the cone are equal to the diameter of the sphere. Find the ratio of their volumes.**

**Sol.**

Let r be the common radius

Height of the cone=height of the cylinder=2r

Let

\(v_{1}\)

\(v_{1}\)

\(v_{1}\)

Now

\(v_{1}:v_{2}:v_{3}=\frac{4}{3}\pi r^{3}:2\pi r^{3}:\frac{2}{3}\pi r^{3}\\ =4:6:2\\ =2:3:1\)

**Q29. A cube of side 4 cm contains a sphere touching its side. Find the volume of the gap in between.**

**Sol.**

It is given that

Cube side=4cm

Volume of cube=\(\left(4cm \right )^{3}\)

Diameter of the sphere= Length of the side of the cube=4cm

Therefore radius of the sphere=2cm

Volume of the sphere=\(\frac{4}{3}\pi r^{3}\)

Volume of gap=Volume of cube-Volume of sphere

=64\(cm^{3}\)

**Q30. A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.**

**Sol.**

Given that,

Inner radius of the hemispherical tank=1m=\(r_{1}\)

Thickness of the hemispherical tank=1cm=0.01m

Outer radius of hemispherical tank=(1+0.01)=1.01m==\(r_{2}\)

Volume of iron used to make the tank=\(\frac{2}{3}\pi\left(r_{2}^{3}-r_{1}^{3} \right )\\ =\frac{2}{3}\times\frac{22}{7}\left[\left(1.01 \right )^{3}-1^{3} \right ]\\ =\frac{44}{21}\left[\left(1.0303 \right )-1 \right ]m^{3}\\ =0.06348m^{3}\)

**Â **

**Q31. A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (\(mm^{3}\)) is needed to fill this capsule?**

**Sol.**

Given that

Diameter of capsule=3.5mm

Radius=\(\frac{3.5}{2}\)

Volume of spherical sphere=\(\frac{4}{3}\pi r^{3}\\ =\frac{4}{3}\times\frac{22}{7}\times\left(1.75 \right )^{3}\\ =22.458mm^{3}\)

Therefore 22.46\(mm^{3}\)

**Â **

**Q32. The diameter of the moon is approximately one-fourth of the diameter of the earth. What is the earth the volume of the moon?**

**Sol.**

Diameter of moon= \(\frac{1}{4}th\)

Let the diameter of earth be d, so radius=\(\frac{d}{2}\)

Then diameter of moon=\(\frac{d}{4}\)

Radius=\(\frac{\frac{d}{2}}{4}\)

Volume of moon=\(\frac{4}{3}\pi r^{3}=\frac{4}{3}\pi \left(\frac{d}{8} \right )^{3}=\frac{4}{3}\times \frac{1}{512}\pi d^{3}\)

Volume of earth=\(\frac{4}{3}\pi r^{3}=\frac{4}{3}\pi \left(\frac{d}{2} \right )^{3}=\frac{4}{3}\times \frac{1}{8}\pi d^{3}\)

\(\frac{Volume\;of\;moon}{Volume\;of\;earth}=\frac{\frac{4}{3}\times \frac{1}{512}\pi d^{3}}{\frac{4}{3}\times \frac{1}{8}\pi d^{3}}\)

= \(\frac{1}{64}\)

Thus the volume of the moon is \(\frac{1}{64}\)