3 Mark Question

Question 13 Marks

The length of a road roller is 2.1 m and its diameter is 1.4 m. For levelling a ground 500 rotations of the road roller were required. How much area of ground was levelled by the road roller? Find the cost of levelling at the rate of ₹ 7 per sq.m.
Given: For road roller,
diameter (d) = 1.4 m,
length (h) = 2.1 m
number of rotations required for levelling the ground = 500,
rate of levelling = ₹ 7 per sq. m.
To find: Area of ground leveled by the road roller and cost of levelling

Answer

i. Since, area of ground levelled in 1 rotation of road roller = curved surface area of road roller
∴Curved surface area of the road roller = 2πrh
= πdh …[∵ d = 2r]
$=\frac{22}{7} \times 1.4 \times 2.1$
= 22 x 0.2 x 2.1
= 9.24 sq.m.

ii. Area of ground levelled in 1 rotation = 9.24 sq.m.
∴Area of ground levelled in 500 rotations = 9.24 x 500
= 4620 sq.m.

iii. Rate of levelling ₹ 7 per sq.m.
∴Total cost = Area of ground levelled x Rate of levelling
= 4620 x 7
= ₹32340
∴ The road roller levels 4620 sq.m. land in 500 rotation, and the cost of levelling is ₹32340.

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Question 23 Marks

If diameter of a road roller is $0.9\ m$ and its length is $1.4\ m,$ how much area of a field will be pressed in its $500$ rotations?$\left(\pi=\frac{22}{7}\right)$
Given: For road roller,
diameter $(d) = 0.9\ m,$ length $(h) = 1.4\ m$
To find: Area of a field pressed in $500$ rotations

Answer

i. Since, area of field pressed in $1$ rotation of road roller $=$ curved surface area of road roller
$\therefore$ Curved surface area of the road roller $= 2\pi rh$
$=\pi d h, \ldots[\because d=2 r]$
$=\frac{22}{7} \times 0.9 \times 1.47$
$= 22 \times 0.9 \times 0.2$
$= 3.96$ sq.m.ii. Area of land pressed in $1$ rotation $= 3.96$ sq.m.
$\therefore$ Area of land pressed in $500$ rotations $= 500 \times 3.96$
$= 1980$ sq.m.
$\therefore 1980$ sq.m, land will be pressed in $500$ rotations of the road roller.

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Question 33 Marks

Volume of a hemisphere is $18000 \pi$ cubic cm . Find its diameter.
Given: Volume of hemisphere $=18000 \pi$ cubic cm .
To find: Diameter of the hemisphere

Answer

i. Volume of hemisphere $=\frac{2}{3} \pi r^3$
$
\begin{array}{ll}
\therefore & 18000 \pi=\frac{2}{3} \pi r^3 \\
\therefore & 18000=\frac{2}{3} r^3 \\
\therefore & r^3=\frac{18000 \times 3}{2}
\end{array}
$
$= 9000 \times 3$
$\therefore r^3 = 27000$
$\therefore r =\sqrt[3]{27000} \ldots$ [Taking cube root on both sides] $=30 cm$
ii. Diameter = $2r$
$= 2 \times 30 = 60\ cm$
∴ The diameter of the hemisphere is $60\ cm.$

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Question 43 Marks

Find the surface area of a sphere, if its volume is $38808$ cubic $cm .\left(\pi=\frac{22}{7}\right)$
Given: Volume of sphere $= 38808$ cubic cm.
To find: Surface area of sphere

Answer

$\text { i. } \quad \text { Volume of sphere }=\frac{4}{3} \pi r^3$
$\therefore \quad 38808=\frac{4}{3} \times \frac{22}{7} \times r^3$
$\therefore \quad r^3=\frac{38808 \times 3 \times 7}{4 \times 22}$
$=\frac{9702 \times 3 \times 7}{22}$
$\therefore r^3=441 \times 21=21 \times 21 \times 21$
$\therefore r =21 cm \ldots$... [Taking cube root on both sides]
ii. Surface area of sphere $=4 \pi r ^2$
$=4 \times \frac{22}{7} \times 21$
$=4 \times \frac{22}{7} \times 21 \times 21$
$=4 \times 22 \times 3 \times 21$
$=5544 \text { sq.cm. }$
$\therefore$ The surface area of sphere is $5544\ sq.cm.$

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Question 53 Marks

If the surface area of a sphere is $2826 cm^2$ then find its volume. $(\pi = 3.14)$
Given: Surface area of sphere = $2826 sq.cm.$
To find: Volume of sphere

Answer

i. Surface area of sphere $= 4\pi r^2$
$\therefore 2826 = 4 \times 3.14 \times r^2$
$2826 = 282600 = 900$
$\therefore r^2=\frac{2826}{4 \times 3.14}=\frac{282600}{4 \times 314}=\frac{900}{4}$
$\therefore r^2=225$
$\therefore r=\sqrt{225} \ldots \text { [Taking square root on both sides] }$
$=15 cm $
$ \text { ii. Volume of sphere }=\frac{4}{3} \pi r^3$
$=\frac{4}{3} \times 3.14 \times 15^3$
$=\frac{4}{3} \times 3.14 \times 15 \times 15 \times 15$
$=4 \times 3.14 \times 5 \times 15 \times 15$
$=14130 \text { cubic } cm . $
$\therefore$ The volume of the sphere is $14130$ cubic $cm$.

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Question 63 Marks

There are 25 persons in a tent which is conical in shape. Every person needs an area of 4 sq.m, of the ground inside the tent. If height of the tent is 18 m , find the volume of the tent.
Given: For the tent,
height $(h)=18 m$,
number of people in the tent $=25$,
area required for each person $=4 sq \cdot m$
To find: Volume of the tent

Answer

i. Every person needs an area of 4 sq.m, of the ground inside the tent.
Surface area of the base of the tent = number of people in the tent $\times$ area required for each person
$=25 \times 4$
$=100 \text { sq.mii. Surface area of the base of the tent }=\pi r^2$
$\therefore 100=\pi r^2$
$\therefore \pi r^2=100$
iii. Volume of the tent $=\frac{1}{3} \pi r^2 h=\frac{1}{3} \times 100 \times 18 \ldots \ldots .\left[\because \pi r^2=100\right]$
$=100 \times 6$
$=600$ cubic metre
$\therefore$ The volume of the tent is 600 cubic metre.

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Question 73 Marks

Volume of a cone is $1232 cm ^3$ and its height is $24 cm$. Find the surface area of the cone. ( $\pi$ $=\frac{22}{7}$ )
Given: Height $( h )=24 cm$,
Volume of cone $=1232 cm ^3$
To find: Surface area of the cone

Answer

$\text { i. Volume of cone }=\frac{1}{3} \pi r^2 h$
$\therefore \quad 1232=\frac{1}{3} \times \frac{22}{7} \times r^2 \times 24$
$\therefore \quad r^2=\frac{1232 \times 3 \times 7}{22 \times 24}$
$\qquad=\frac{56 \times 1 \times 7}{1 \times 8}$
$\therefore r^2=49$
$\therefore r=\sqrt{49} \ldots \text { [Taking square root on both sides] }$
$=7 cm $
ii. Now, $r^2=r^2+h^2$
$ \therefore 1^2=7^2+24^2$
$=49+576=625$
$\therefore I=\sqrt{625} \ldots \text { [Taking square root on both sides] }$
$=25 $
iii. Curved surface area of cone $=\pi r$ l
$ =\frac{22}{7} \times 7 \times 25$
$=22 \times 25$
$=550 sq . cm $
$\therefore$ The surface area of the cone is 550 sq.cm.

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Question 83 Marks

Surface area of a cone is $188.4 sq \cdot cm$ and its slant height is 10 cm . Find its perpendicular height ( $\pi=3.14$ ).
Given: Length $( I )=10 cm$, curved surface area of the cone $=188.4 sq . cm$
To find: Perpendicular height $( h )$ of the cone

Answer

i. Curved surface area of the cone $=\pi rl$
$\therefore 188.4 =3.14 \times r \times 10$
$\therefore \quad r =\frac{188.4}{3.14 \times 10}$
$=\frac{188.4}{31.4}$
$ =\frac{1884}{314}$
$ =6 cm$
ii. Now, $l ^2= r ^2+ h ^2$
$\therefore 10^2=6^2+h^2$
$\therefore 100=36+h^2$
$\therefore 100-36=h^2$
$\therefore h^2=64$
$\therefore h=\sqrt{64} \ldots$ [Taking square root on both sides]
$=8 cm$
$\therefore$ The perpendicular height of the cone is 8 cm .

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Question 93 Marks

Curved surface area of a cone is $251.2 cm^2$ and radius of its base is $8\ cm $. Find its slant height and perpendicular height, ( $\pi=3.14$ )
Given: Radius $( r )=8 cm$, curved surface area
of cone $=251.2 cm^2$
To find: Slant height $( l )$ and the perpendicular height $( h )$ of the cone

Answer

i. Curved surface area of cone $=\pi r \mid$
$\therefore 251.2=3.14 \times 8 \times 1$
$\therefore \quad l=\frac{251.2}{3.14 \times 8}$
$=\frac{25120}{314 \times 8}$
$=\frac{3140}{314}$
$\therefore I=10 cm$
ii. Now, $l ^2= r ^2+ h ^2$
$\therefore 10^2=8^2+h^2$
$\therefore 100=64+h^2$
$\therefore 100-64=h^2$
$\therefore h^2=36$
$\therefore h =\sqrt{ } 36 \ldots$ [Taking square root on both sides]
$=6 cm$
$\therefore$ The slant height and the perpendicular height of the cone are $10\ cm$ and $6\ cm$ respectively.

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Question 103 Marks

Find $(i)$ the slant height, $(ii)$ the curved surface area and $(iii)$ total surface area of a cone, if its base radius is $12 \ cm$ and height is $16 \ cm. (\pi= 3.14)$

Answer

$ r=12 \mathrm{~cm}, h=16 \mathrm{~cm}$
$ l^2=r^2+h^2$
$\therefore l^2=(12)^2+(16)^2$
$\therefore l^2=144+256$
$\therefore l^2=400$
$\therefore l =20 \mathrm{~cm}$
$\text { Curved surface area } =\pi r l$
$ =3.14 \times 12 \times 20$
$ =753.6 \mathrm{~cm}^2$
Total surface area of cone
$=\pi r(l+r)$
$=3.14 \times 12(20+12)$
$=3.14 \times 12 \times 32$
$=1205.76 \mathrm{~cm}^2$

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