ML Aggarwal Solution Class 10 Chapter 17 Mensuration Exercise 17.3

 Exercise 17.3

Question 1

Find the surface area of a sphere of radius :

(i) 14 cm

(ii) 10.5 cm

Sol :

(i) Radius (r) = 14 cm

Surface area $=4 \pi r^{2}=4 \times \frac{22}{7} \times 14 \times 14 \mathrm{~cm}^{2}$

$=2964 \mathrm{~cm}^{2}$


(ii) Radius (r)=10.5 cm

$\therefore$ Surface area $=4 \pi r^{2}$

$=4 \times \frac{22}{7} \times 10.5 \times 10.5 \mathrm{~cm}^{2}=1386 \mathrm{~cm}^{2}$


Question 2

Find the volume of a sphere of radius :

(i) 0.63 m

(ii) 11.2 cm

Sol :

(i) Radius (r) = 0.63 m

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

$=\frac{4}{3} \times \frac{22}{7} \times \frac{63}{100} \times \frac{63}{100} \times \frac{63}{100} \mathrm{~m}^{3}$

$=\frac{1047816}{1000000}=1.0478=1.05 \mathrm{~m}^{3}($ approx $)$


(ii) Radius $(r)=11.2 \mathrm{~cm}$

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

$=\frac{4}{3} \times \frac{22}{7} \times 11.2 \times 11.2 \times 11.2 \mathrm{~cm}^{3}$

$=\frac{17661.952}{3}=5887.32 \mathrm{~cm}^{3}($ approx $)$


Question 3

Find the surface area of a sphere of diameter: (i) 21 cm (ii) 3.5 cm

Sol :

(i) Diameter = 21 cm

Radius $(r)=\frac{21}{2} \mathrm{~cm}$

$\therefore$ Surface area $=4 \pi r^{2}$

$=4 \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} \mathrm{~cm}^{2}=1386 \mathrm{~cm}^{2}$


(ii) Diameter=3.5 cm

$\therefore$ Radius $(r)=\frac{3.5}{2}=1.75 \mathrm{~cm}$

$\therefore$ Surface area $=4 \pi r^{2}$

$=4 \times \frac{22}{7} \times 1.75 \times 1.75 \mathrm{~cm}^{2}=38.5 \mathrm{~cm}^{2}$


Question 4

A shot-put is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g per $cm^3$, find the mass of the shot-put.

Sol :

Radius of the metallic shot-put = 4.9 cm

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

$=\frac{4}{3} \times \frac{22}{7} \times 4.9 \times 4.9 \times 4.9 \mathrm{~cm}^{3}$

$=\frac{1479.016}{3}=493 \mathrm{~cm}^{3}$

Density of metal $=7.8 \mathrm{~g}$ per $\mathrm{cm}^{3}$ Total mass $=493 \times 7.8 \mathrm{~g}=3845.4 \mathrm{~g}$

$=3.845 \mathrm{~kg}=3.85 \mathrm{~kg}$ (approx)


Question 5

Find the diameter of a sphere whose surface area is 154 $\mathrm{cm}^{2}$.

Sol :

Surface area of a sphere = 154 $\mathrm{cm}^{2}$

$\therefore \operatorname{Radius}(r)=\sqrt{\frac{\text { Surface area }}{4 \pi}}=\sqrt{\frac{154 \times 7}{4 \times 22}} \mathrm{~cm}$

$=\sqrt{\frac{49}{4}}=\frac{7}{2} \mathrm{~cm}$

$\therefore$ Diameter $=2 r=2 \times \frac{7}{2} \mathrm{~cm}=7 \mathrm{~cm}$


Question 6

Find:

(i) the curved surface area.

(ii) the total surface area of a hemisphere of radius 21 cm.

Sol :

Radius of a hemisphere = 21 cm

(i) Curved surface area $=2 \pi r^{2}$

$=2 \times \frac{22}{7} \times 21 \times 21 \mathrm{~cm}^{2}=2772 \mathrm{~cm}^{2}$

(ii) Total surface area $=3 \pi r^{2}$

$=3 \times \frac{22}{7} \times 21 \times 21 \mathrm{~cm}^{2}=4158 \mathrm{~cm}^{2}$


Question 7

A hemispherical brass bowl has inner- diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 $\mathrm{cm}^{2}$

Sol :

The inner diameter of hemispherical bowl = 10.5 cm

Radius $(r)=\frac{10.5}{2}=5.25 \mathrm{~cm}$

$\therefore$ Inner curved surface area $=2 \pi r^{2}$

$=2 \times \frac{22}{7} \times \frac{10.5}{2} \times \frac{10.5}{2} \mathrm{~cm}^{2}$

$=\frac{693}{4} \mathrm{~cm}^{2}=173.25 \mathrm{~cm}^{2}$

Rate of tin-plating $=₹ 16$ per $100 \mathrm{~cm}^{2}$

$\therefore$ Total cost $=₹ \frac{693 \times 16}{4 \times 100}$

$=₹ \frac{2772}{100}=₹ 27.72$


Question 8

The radius of a spherical balloon increases from 7 cm to 14 cm as air is jumped into it. Find the ratio of the surface areas of the balloon in two cases.

Sol :
Original radius of balloon = 7 cm
Radius after filling the air in it = 14 cm
The surface area of balloon, the original position
$=4 \pi r^{2}$
$=4 \times \frac{22}{7} \times 7 \times 7=616 \mathrm{~cm}^{2}$

Surface area after filling air $=4 \pi r^{2}$

$=4 \times \frac{22}{7} \times 14 \times 14 \mathrm{~cm}^{2}=2464 \mathrm{~cm}^{2}$

Ratio between the two position 

=616: 2464=1: 4


Question 9

A sphere and a cube have the same surface. Show that the ratio of the volume of the sphere to that of the cube is √6 : √π

Sol :

Let the edge of a cube = a

Surface area $=6 \mathrm{a}^{2}$

and surface area of sphere $=6 a^{2}$

$\therefore$ Radius $(r)=\sqrt{\frac{6 a^{2}}{4 \pi}}=\sqrt{\frac{3}{2 \pi}} a$

and volume of cube $=a^{3}$

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

$=\frac{4}{3} \pi\left[\sqrt{\frac{3}{2 \pi}} a\right]^{3}$

$=\frac{4}{3} \pi \times \frac{3 \sqrt{3} a^{3}}{2 \sqrt{2} \pi \times \sqrt{\pi}}=\frac{2 \sqrt{3} a^{3}}{\sqrt{2} \sqrt{\pi}}$

$\therefore$ Ratio $=\frac{2 \sqrt{3} a^{3}}{\sqrt{2} \sqrt{\pi}}: a^{3}$

$=\sqrt{2} \sqrt{3}: \sqrt{\pi}=\sqrt{6}: \sqrt{\pi}$


Question 10

(a) If the ratio of the radii of two sphere is 3 : 7, find :

(i) the ratio of their volumes.

(ii) the ratio of their surface areas.

(b) If the ratio of the volumes of the two sphere is 125 : 64, find the ratio of their surface areas.

Sol :

(a) Ratio in radii of two spheres = 3 : 7

Let radius of the first sphere = 3x

and radius of the second sphere = 7x


(i) $\therefore$ Volume of the first sphere $=\frac{4}{3} \pi(3 x)^{3}$
and of second sphere $=\frac{4}{3} \pi(7 x)^{3}$
Ratio $=\frac{4}{3} \pi 27 x^{3}: \frac{4}{3} \pi 343 x^{3}$
=27: 343

(ii) Surface area of the first sphere $=4 \pi r^{2}$
$=4 \pi(3 x)^{2}$
and of second sphere $=4 \pi(7 x)^{2}$

$\therefore$ Ratio $=4 \pi(3 x)^{2} ; 4 \pi(7 x)^{2}$

$=9 x^{2}: 49 x^{2}=9: 49$


(b) Let the volume of sphere be $\mathrm{V}_{1}$ and $\mathrm{V}_{2}$ 

According to question,

$\Rightarrow \frac{V_{1}}{V_{2}}=\frac{125}{64}$

$\Rightarrow \frac{\frac{4}{3} \pi r_{1}^{3}}{\frac{4}{3} \pi r_{2}^{3}}=\frac{125}{64}$

$\Rightarrow \frac{\pi r_{1}^{3}}{\pi r_{2}^{3}}=\frac{125}{64}$

$ \Rightarrow\left(\frac{r_{1}}{r_{2}}\right)^{3}=\left(\frac{5}{4}\right)^{3}$

$\therefore r_{1}=5$ and $r_{2}=4$

Now, total surface $=\frac{4}{3} \pi r^{2}$

According to question,

$\frac{\frac{4}{3} \pi r_{1}^{2}}{\frac{4}{3} \pi r_{2}^{2}}=\frac{\left(r_{1}\right)^{2}}{\left(r_{2}\right)^{2}}=\frac{(5)^{2}}{(4)^{2}}=\frac{25}{16}$

$\therefore$ Ratio of their surface area be 25: 16


Question 11

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

Sol :

Side of a cube = 4 cm

Volume (side)³ = 4 × 4 × 4 = 64 cm³

Diameter of sphere contained by this cube is d = 4 cm

$\therefore$ Radius $(r)=\frac{4}{2}=2 \mathrm{~cm}$

Volume $=\frac{4}{3} \pi r^{3}$
$=\frac{4}{3} \pi \times 2 \times 2 \times 2=\frac{32}{3} \pi \mathrm{cm}^{3}$
$=\frac{32}{3} \times \frac{22}{7} \mathrm{~cm}^{3}=\frac{704}{21} \mathrm{~cm}^{3}$

$\therefore$ Volume of space between them

$=64-\frac{704}{21}$

$=\frac{1344-704}{21}=\frac{640}{21} \mathrm{~cm}^{3}$

$=30.48 \mathrm{~cm}^{3}($ approx $)$


Question 12

Find the volume of a sphere whose surface area is 154 cm².

Sol :

Given that

Surface area of a sphere = 154 cm²

$\therefore$ Radius $(r)=\sqrt{\frac{154}{4 \pi}}=\sqrt{\frac{154 \times 7}{4 \times 22}} \mathrm{~cm}$

$=\sqrt{\frac{49}{4}}=\frac{7}{2} \mathrm{~cm}$

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

$=\frac{4}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \frac{7}{2} \mathrm{~cm}^{3}$

$=\frac{539}{3}=179 \frac{2}{3} \mathrm{~cm}^{3}$


Question 13

If the volume of a sphere is $179 \frac{2}{3} \mathrm{~cm}^{3}$ find its radius and the surface area.

Sol :

Given that

Volume of a sphere $=179 \frac{2}{3} \mathrm{~cm}^{3}$

$=\frac{539}{3} \mathrm{~cm}^{3}$

$\therefore$ Radius $=\left[\frac{539}{3} \times \frac{3}{4 \pi}\right]^{\frac{1}{3}}$

$=\left[\frac{539 \times 3 \times 7}{3 \times 4 \times 22}\right]^{\frac{1}{3}}=\left[\frac{343}{8}\right]^{\frac{1}{3}}$

$=\left[\left(\frac{7}{2}\right)^{3}\right]^{\frac{1}{3}}=\frac{7}{2} \mathrm{~cm}=3.5 \mathrm{~cm}$

$\therefore$ Surface area $=4 \pi r^{2}$

$=4 \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \mathrm{~cm}^{2}=154 \mathrm{~cm}^{2}$


Question 14

A hemispherical bowl has a radius of 3.5 cm. What would be the volume of water it would contain?

Sol :

Radius of a hemispherical bowl (r) = 3.5 cm

$=\frac{7}{2} \mathrm{~cm}$

$\therefore$ Volume of water in it $=\frac{2}{2} \pi r^{2}$

$=\frac{2}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \frac{7}{2} \mathrm{~cm}^{3}$

$=\frac{539}{6} \mathrm{~cm}^{3}=89 \frac{5}{6} \mathrm{~cm}^{3}$


Question 15

The water for a factory is stored in a hemispherical tank whose internal diameter is 14 m. The tank contains 50 kilolitres of water. Water is pumped into the tank to fill to its capacity. Find the volume of water pumped into the tank.

Sol :

Internal diameter of a hemispherical tank (r) = 14 m

Radius of the tank $=\frac{14}{2}=7 \mathrm{~m}$

Water stored in it = 50 kilolitres of water

$\therefore$ Volume $=\frac{2}{3} \pi r^{3}=\frac{2}{3} \times \frac{22}{7} \times 7 \times 7 \times 7 \mathrm{~m}^{3}$

$=718.66 \mathrm{~m}^{3}$ or 718.66 kilolitre water

$\therefore$ More water to be pumped in $=718.66-50=668.66$ kilolitres


Question 16

The surface area of a solid sphere is 1256 cm². It is cut into two hemispheres. Find the total surface area and the volume of a hemisphere. Take π = 3.14.

Sol :

Surface area of a solid sphere = 1256 cm²

By cutting it into two hemisphere,

Curved surface area of each hemisphere

$=\frac{1256}{2}=628 \mathrm{~cm}^{2}$

Now radius $=\sqrt{\frac{\text { Surface area }}{4 \pi}}$
$=\sqrt{\frac{1256}{4 \times 3.14}} \mathrm{~cm}=\sqrt{\frac{314 \times 100}{314}}$
$=\sqrt{100}=10 \mathrm{~cm}$

Now total surface area of hemisphere $=3 \pi r^{2}$
$=3 \times 3.14 \times 10 \times 10 \mathrm{~cm}^{2}=942 \mathrm{~cm}^{2}$

and volume of hemisphere $=\frac{2}{3} \pi r^{3}$

$=\frac{2}{3}(3014) \times(10 \times 10 \times 10) \mathrm{cm}^{3}$

$=\frac{2}{3} \times 3140=\frac{6280}{3} \mathrm{~cm}^{3}=2093 \frac{1}{3} \mathrm{~cm}^{3}$


Question 17

Write whether the following statements are true or false. Justify your answer :

(i) The volume of a sphere is equal to two-third of the volume of a cylinder whose height and diameter are equal to the diameter of the sphere.

(ii) The volume of the largest right circular cone that can be fitted in a cube whose edge is 2r equals the volume of a hemisphere of radius r.

(iii) A cone, a hemisphere and a cylinder stand on equal bases and have the same height. The ratio of their volumes is 1 : 2 : 3.

Sol :

(i) The volume of a sphere is equal to the two third of the volume of a cylinder

whose height and diameter are equal to the diameter of the sphere.

True, as $\frac{4}{3} \pi r^{3}=\frac{2}{3} \pi r^{2} h$

$=\frac{2}{3} \pi r^{2} \times 2 r=\frac{4}{3} \pi r^{3}$


(ii) The volume of the longest right circular cone that can be filled in a cube whose edge is 2r equal to the volume of a hemisphere of radius r

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

$\Rightarrow \frac{1}{3} \pi r^{2} \times 2 r=\frac{2}{3} \pi r^{3}$

$ \Rightarrow \frac{2}{3} \pi r^{3}=\frac{2}{3} \pi r^{3}$


(iii) A cone, a hemisphere and a cylinder stand on equal bases and have the same height.

The ratio of their volumes is 1 : 2 : 3

As $\frac{1}{3} \pi r^{2} \times r: \frac{2}{3} \pi r^{3}: \pi r^{2} \cdot r$

$\Rightarrow \frac{1}{3} \pi r^{3}: \frac{2}{3} \pi r^{3}: \pi r^{3}$

= 1 : 2 : 3

∴It is true

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