1 Marks Question

Question 11 Mark

How many litres of milk can a hemispherical bowl of diameter $10.5 \ cm$ hold?

Answer

Diameter $= 10.5 \ cm.$
$\therefore$ Radius $(r) =\frac{10.5} {2}$
$=5.25 \ cm$
$\therefore$ Amount of milk = ${2\over3}\pi r^3$
$={2\over3}\times{22\over7}\times(5.25)^3\ cm^3$
$= 303 \ cm^3 $
$= 0.303 \ l$

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Question 21 Mark

Find the amount of water displaced by a solid spherical ball of diameter 0.21 m.

Answer

Diameter = 0.21 m
∴ Radius (r) = ${0.21}\over2$m = 0.105 m
∴ Amount of water displaced = ${4\over3}\pi r^3$
$={4\over3}\times{22\over7}\times(0.105)^3=0.004851\ m^3$

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Question 31 Mark

Find the amount of water displaced by a solid spherical ball of diameter 28 cm.

Answer

Diameter = 28 cm
∴ Radius (r) =$28\over2$ cm = 14 cm
∴ Amount of water displaced = ${4\over3}\pi r^3$
$={4\over3}\times{22\over7}\times(14)^3={34496\over3}\ cm^3=11498{2\over3}\ cm^3$

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Question 41 Mark

Find the volume of a sphere whose radius is 0.63 m.

Answer

r = 0.63 m
∴ volume =${4\over3}\pi r^3$
$={4\over3}\times{22\over7}\times(0.63)^3=1.05\ m^3$

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Question 51 Mark

Find the volume of a sphere whose radius is 7 cm.

Answer

r = 7 cm
∴ Volume = ${4\over3}\pi r^3$
$={4\over3}\times{22\over7}\times(7)^3={4312\over3}\ cm^3=1437{1\over3}\ cm^3$

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Question 61 Mark

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

Answer

Diameter of the capsule $= 3.5 \ mm$
$\therefore$ Radius of the capsule $(r) = 3.5\over2\ mm = 1.75 \ mm$
$\therefore$ Capacity of the capsule = ${4\over3}\pi r^3$
$={4\over3}\times{22\over7}\times(1.75)^3\ mm^3$
$= 22.46 \ mm^3$
$\therefore 22.46 \ mm^3$ of medicine is needed to fill this capsule.

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Question 71 Mark

Find the volume of the right circular cone with radius $3.5$ and height $12 \ cm.$

Answer

$r = 3.5 \ cm, h = 12 \ cm.$
$\therefore$ Volume of the right circular cone $={1\over3}\pi r^2h$
$={1\over3}\times{22\over7}\times(3.5)^2\times12 = 154 \ cm^3$

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Question 81 Mark

Find the volume of the right circular cone with radius $6 \ cm$ and height $7 \ cm.$

Answer

$r = 6 \ cm, h = 7 \ cm$
$\therefore $Volume of the right circular cone $= {1\over3}\pi r^2h$
$={\frac{1}{3}}\times{\frac{22}{7}}\times(6)^2\times7$
$= 264 \ cm^3$

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Question 91 Mark

A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.

Answer

Inner radius of bowl $\left( r \right)$ = 5 cm
Thickness of steel $\left( t \right)$ = 0.25 cm
$\therefore $ Outer radius of bowl (R) = $r+t$ = 5 +0.25 = 5.25 cm
$\therefore $ Outer curved surface area of bowl = $2\pi {{\text{R}}^{2}}$= $2\times \frac{22}{7}\times 5.25\times 5.25$
= $2\times \frac{22}{7}\times \frac{21}{4}\times \frac{21}{4}$
= $\frac{693}{4}$
$=\text{ }173.25\text{ }c{{m}^{2}}$

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Question 101 Mark

Find the total surface area of a hemisphere of radius $10 \ cm.$

Answer

$r = 10 \ cm$
Total surface area of the hemisphere $= 3\pi r^2$
$= 3 \times 3.14 \times (10)^2 $
$= 942 \ cm^2$

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Question 111 Mark

Find the surface area of a sphere of diameter $3.5 \ cm.$

Answer

Diameter $= 3.5 \ cm$
$\therefore$ Radius $(r) = \frac{3.5}{2} \ cm $
$= 1.75 \ cm$
Surface area of sphere $={4\pi r^2}$
$= 4\times \frac{22}{7} \times (1.75)^2$
$= 38.5 \ cm^2$

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Question 121 Mark

Find the surface area of a sphere of diameter $21 \ cm.$

Answer

Diameter $= 21 \ cm$
$\therefore$ Radius $(r) = \frac{21}{2} \ cm$
Surface area of sphere $={4\pi r^2}$
= $4\times{\frac{22}{7}r}\times({\frac{21}{2}})^2$
$= 4\times{\frac{22}{7}}\times{\frac{21}{2}}\times{\frac{21}{2}}$
$=1386 \ cm^2$

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Question 131 Mark

Find the surface area of a sphere of diameter $14 \ cm.$

Answer

Diameter $= 14 \ cm$
$\therefore$ Radius $(r) = \frac{14}{2} \ cm $
$= 7 \ cm$
Surface area of sphere $ = {4\pi r^2}$
$= 4\times \frac{22}{7} \times (7)^2$
$ = 616 \ cm^2$

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Question 141 Mark

Find the surface area of a sphere of radius $14 \ cm.$

Answer

$r = 14 \ cm$
Surface area of a sphere $=4\pi r^2$
$=4\times{22\over7}\times(14)^2$
$=4\times22\times17\times2$
$= 2464 \ cm^2$

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Question 151 Mark

Find the surface area of a sphere of radius 5.6 cm.

Answer

r = 5.6 cm
Surface area of a sphere = $4\pi {r^2}$
$ = 4 \times \frac{{22}}{7} \times {(5.6)^2} = 394.24\;c{m^2}$

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Question 161 Mark

Find the surface area of a sphere of radius $10.5 \ cm.$

Answer

$r = 10.5 \ cm$
Surface area of a sphere $=4\pi r^2$
$= 4\times{22\over7}\times(10.5)^2$
$= 4\times22\times10.5\times1.5$
$= 1386 \ cm^2$

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Question 171 Mark

Curved surface area of a cone is $308 \ cm^2$ and its slant height is $14 \ cm.$ Find total surface area of the cone.

Answer

Total surface area of cone $=$ Curved surface Area of cone $+$ Area of base
$= \pi rl + \pi r^2$
$= [308 + \frac { 22 } { 7 } \times (7)^2] \ cm^2$
$= 462 \ cm^2$
Thus, the total surface area of the cone is $462 \ cm^2.$

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Question 181 Mark

Curved surface area of a cone is $308 \ cm^2$ and its slant height is $14 \ cm.$ Find the radius of the base

Answer

Slant height of cone $= 14 \ cm$
Let radius of circular end of cone be $r.$
Curved surface area of cone = $\pi rl$
$308 \ cm^2 = (\frac { 22 } { 7 } \times r \times 14) \ cm$
$\Rightarrow r = \left( \frac { 308 } { 44 } \right) cm = 7 \ cm$
Thus, the radius of circular end of the cone is $7 \ cm.$

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Question 191 Mark

The diameter of the base of a cone is $10.5 \ cm$ and its slant height is $10 \ cm.$ Find its curved surface area.

Answer

Diameter $= 10.5 \ cm$
$\Rightarrow$ Radius $(r) =\frac{10.5}{2} = \frac{21}{4} \ cm$
Slant height of cone $(l) = 10 \ cm$
Now we have, Curved surface area of cone
$= \pi rl = \frac{22}{7}\times \frac{21}{4}\times 10$
$= 165 \ cm^2$

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Question 201 Mark

Find the surface area and total surface area of a hemisphere of radius $21 \ cm.$

Answer

We know that the surface area $S$ and total surfaces area $S_1 $of $a$ hemisphere of radius $r$ are given by
$S = 2\pi r ^ { 2 }$ and, $S_1 = 3\pi r ^ { 2 }$ respectively.
Here, $r = 21 \ cm$
$\therefore S = 2 \times \frac { 22 } { 7 } \times 21 \times 21 \ cm^2$ and,
$S_1 = 3 \times \frac { 22 } { 7 } \times 21 \times 21 \ cm^2$
$\Rightarrow S = 2772 \ cm^2$ and,
$S_1 = 4158 \ cm^2$

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Question 211 Mark

Find the surface area of a sphere of radius $7 \ cm.$

Answer

We known that the surface area $S$ of a sphere of radius $r$ is given by
$S = 4 \pi r ^ { 2 }$
Here, $r = 7 \ cm$
$\therefore S = 4 \times \frac { 22 } { 7 } \times 7 \times 7 \ cm^{2 }$
$= 616 \ cm^2$

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Question 221 Mark

Find the curved surface area of a right circular cone whose slant height is $10 \ cm$ and base radius is $7 \ cm.$

Answer

We know that,
Curved surface area of cone $=\pi r l$
$= \frac {22}7 \times 7 \times 10 \ cm^2$
$= 220 \ cm^2$

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Question 231 Mark

Savitri had to make a model of a cylindrical Kaleidoscope for her science project. She wanted to use chart paper to make the curved surface of the Kaleidoscope. What should be the area of chart paper required by her, if she wanted to make a Kaleidoscope of length $25 \ cm$ with a $3.5 \ cm$ radius? You may take $\pi=\frac{22}{7}$.

Answer

We have given that,
$r =$ Radius of the base of the cylindrical Kaleidoscope $= 3.5 \ cm$
$h =$ Height (length) of Kaleidscope $= 25 \ cm$
Therefore, Area of the chart paper required $=$ Curved surface area of the Kaleidoscope
$= 2 \pi rh = 2 \times \frac{22}{7} \times 3.5 \times 25 \ cm^2 $
$= 550 \ cm^2$

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Question 241 Mark

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

Answer

The volume of water the bowl contain = $\frac{2}{3}\pi {{r}^{3}}$
Radius = $3.5cm$
Then volume = $\frac{2}{3}\times \frac{22}{7}\times {{\left( 3.5 \right)}^{3}}$
$=\frac{2}{3}\times \frac{22}{7}\times 3.5\times 3.5\times 3.5$
$=\frac{2}{3}\times \frac{22}{7}\times \frac{35}{10}\times \frac{35}{10}\times \frac{35}{10}$
$=89.8c{{m}^{3}}$

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Question 251 Mark

Find the volume of a sphere of radius $11.2 \ cm.$

Answer

We know that volume of sphere= $\frac 43 \pi r^3$
$= \frac 43 \times \frac {22}7 \times 11.2 \times 11.2 \times 11.2 \ cm^3 $
$= 5887.32 \ cm^3$

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Question 261 Mark

The height and the slant height of a cone are $21 \ cm$ and $28 \ cm$ respectively. Find the volume of the cone.

Answer

We know that,$ l^2 = r^2 + h^2,$
Therefore , we have
$r=\sqrt{l^{2}-h^{2}}$
$=\sqrt{28^{2}-21^{2}} \mathrm{ cm}$
$=7 \sqrt{7} \mathrm{ cm}$
So, volume of the cone $= \frac{1}{3} \pi r^{2} h$
$=\frac{1}{3} \times \frac{22}{7} \times 7 \sqrt{7} \times 7 \sqrt{7} \times 21 \mathrm{ cm}^{3}$
$= 7546 \ cm^3$

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Question 271 Mark

A child playing with building blocks, which are of the shape of the cubes, has built a structure as shown in the given figure. If the edge of each cube is $3 \ cm,$ find the volume of the structure built by the child.

Answer

Volume of each cube $=$ edge $\times$ edge $\times$ edge
$= 3 \times 3 \times 3 \ cm^3 $
$= 27 \ cm^3$
Number of cubes in the structure $= 15$
Therefore, volume of the structure $= 27 \times 15 \ cm^3 $
$= 405 \ cm^3$

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