# RD Sharma Solutions Class 9 Surface Area And Volume Of Sphere Exercise 21.2

## RD Sharma Solutions Class 9 Chapter 21 Exercise 21.2

### RD Sharma Class 9 Solutions Chapter 21 Ex 21.2 Free Download

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

Sol.

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

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

=33.52$cm^{3}$

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

=$\frac{4}{3}\times\frac{22}{7}\times\left ( 3.5 \right )^{3}$ = 179.666$cm^{3}$

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

=$\frac{4}{3}\times\frac{22}{7}\times\left ( 10.5 \right )^{3}$ = 4851$cm^{3}$

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)= $\frac{14}{2}$= 7cm

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

=$\frac{4}{3}\times\frac{22}{7}\times\left ( 7 \right )^{3}$ = 1437.33$cm^{3}$

(ii) Diameter=3.5dm, Radius(r)= $\frac{3.5}{2}$= 1.75dm

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)= $\frac{2.1}{2}$= 1.05m

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

=$\frac{4}{3}\times\frac{22}{7}\times\left ( 1.05 \right )^{3}$ = 4.851$m^{3}$

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

Sol.

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

=$\frac{2}{3}\times\frac{22}{7}\times\left ( 2.8 \right )^{3}$ = 45.994$m^{3}$

$1m^{3}=1000l$

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.

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= $\left ( 22 \right )^{3}$ = 10648$cm^{3}$

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=$\frac{Volume\;of\;cube}{Volume\;of\;bullet}\\ =\frac{10648}{\frac{88}{21}}\\ =2541$

Sol.

i.e Volume$\left ( V_{1} \right )$ =$\frac{4}{3}\pi r^{3}$

$\left ( V_{1} \right )$=$\frac{4}{3}\times\frac{22}{7}\times\left ( 5 \right )^{3}$

$\left ( V_{1} \right )$= $\frac{11000}{21}$ $cm^{3}$

i.e Volume$\left ( V_{2} \right )$ =$\frac{4}{3}\pi r^{3}$

$\left ( V_{2} \right )$=$\frac{4}{3}\times\frac{22}{7}\times\left ( 2.5 \right )^{3}$

$\left ( V_{2} \right )$= $\frac{1375}{21}$ $cm^{3}$

Therefore number of laddoos=$\frac{V_{1}}{V_{2}}$= $\frac{11000}{1375}$= 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 $\frac{3}{2}$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$d_{1}$ = $\frac{3}{2}$cm

Radius of first ball$r_{1}$ =$\frac{\frac{3}{2}}{2}$ =$\frac{3}{4}$cm

Diameter of second ball$d_{2}$ = 2cm

Radius of second ball$r_{2}$ = $\frac{2}{2}$cm = 1cm

Diameter of third ball$d_{3}$ = d

Radius of third ball$r_{3}$ = $\frac{d}{2}$cm

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

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

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

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

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

$\frac{d^{3}}{8}=\frac{125}{64}$

$\frac{d}{2}=\frac{5}{4}$

$d=\frac{10}{4}$

d=2.5cm

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.

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

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

Height of water rised= $\frac{5}{3}$cm

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 $v_{1}$ 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, $\frac{Volume of first sphere}{Volume of second sphere}$= $\frac{\frac{4}{3}\pi r^{3}}{\frac{4}{3}\pi \left ( 2r \right )^{3}}$

= $\frac{1}{8}$

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= $r_{1}$ =3.5cm

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

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

$h=\frac{2r_{1}^{3}}{3r_{2}^{2}}$

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

$h=\frac{7}{12}cm$

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

$h=\frac{2}{3}\times 2r=\frac{4}{3}r$

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

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

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 in  raised = 6.75 cm = $r_{2}$

Volume of water raised = Volume of the sphere

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

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

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

= $r_{2}^{3}$ = 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

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 $Diameter=2\times0.3=0.6cm$

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

Let the level of water be raised by h

Diameter of the spherical bowl=2cm

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

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

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

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

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=$r_{2}$=4cm

Internal radii=$r_{1}$=2cm

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

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

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

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

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.

Depth=20cm

Let r be the radius of the ball

Then

Volume of the ball= Volume of water raised

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

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

$v_{1}$=Volume of sphere=$\frac{4}{3}\pi r^{3}$

$v_{1}$=Volume of cylinder=$\pi r^{2}h$= $\pi r^{2}\times 2r$

$v_{1}$=Volume of cone=$\frac{1}{3}\pi r^{2}h$= $\frac{1}{3}\pi r^{3}$

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=$\left(4cm \right )^{3}$=64$cm^{3}$

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

Volume of the sphere=$\frac{4}{3}\pi r^{3}$= $\frac{4}{3}\times\frac{22}{7}\times\left(2 \right )^{3}=33.52cm^{3}\\$

Volume of gap=Volume of cube-Volume of sphere

=64$cm^{3}$-33.52$cm^{3}$=30.48$cm^{3}$

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=$\frac{2}{3}\pi\left(r_{2}^{3}-r_{1}^{3} \right )\\ =\frac{2}{3}\times\frac{22}{7}\left[\left(1.01 \right )^{3}-1^{3} \right ]\\ =\frac{44}{21}\left[\left(1.0303 \right )-1 \right ]m^{3}\\ =0.06348m^{3}$.

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=$\frac{3.5}{2}$=1.75mm

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$mm^{3}$ of medicine is required

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= $\frac{1}{4}th$diameter of earth

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

Then diameter of moon=$\frac{d}{4}$

Radius=$\frac{\frac{d}{2}}{4}$= $\frac{d}{8}$

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 $\frac{1}{64}$ of volume of earth

#### Practise This Question

Find the sum of roots and product of roots respectively for the polynomial :
2x2+x5=0