Chapter 12 Surface Areas and Volumes Exercise-02
EXERCISE 12.2
Unless stated otherwise, take $\pi=\dfrac{22}{7}$.
1. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to $1 \mathrm{~cm}$ and the height of the cone is equal to its radius. Find the volume of the solid in terms of $\pi$.
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Solution
Given that,
Height $(h)$ of conical part $=$ Radius $(r)$ of conical part $=1 cm$
Radius $(r)$ of hemispherical part $=$ Radius of conical part $(r)=1 cm$
Volume of solid $=$ Volume of conical part + Volume of hemispherical part
$=\dfrac{1}{3} \pi r^{2} h+\dfrac{2}{3} \pi r^{3}$
$=\dfrac{1}{3} \pi(1)^{2}(1)+\dfrac{2}{3} \pi(1)^{2}=\dfrac{\pi}{3}+\dfrac{2 \pi}{3}=\pi cm^{3}$
2. Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is $3 \mathrm{~cm}$ and its length is $12 \mathrm{~cm}$. If each cone has a height of $2 \mathrm{~cm}$, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)
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Solution
From the figure, it can be observed that
Height $(h_1)$ of each conical part $=2 cm$
Height $(h_2)$ of cylindrical part $=12-2 \times$ Height of conical part
$=12-2 \times 2=8 cm$
Radius $(r)$ of cylindrical part $=$ Radius of conical part $=\dfrac{3}{2} cm$
Volume of air present in the model $=$ Volume of cylinder $+2 \times$ Volume of cones
$=\pi r^{2} h_2+2 \times \dfrac{1}{3} \pi r^{2} h_1$
$\begin{aligned}=\pi(\dfrac{3}{2})^{2}(8)+2 \times \dfrac{1}{3} \pi(\dfrac{3}{2})^{2}(2) & =\pi \times \dfrac{9}{4} \times 8+\dfrac{2}{3} \pi \times \dfrac{9}{4} \times 2 \\ & =18 \pi+3 \pi=21 \pi=66 cm^{2}\end{aligned}$
3. A gulab jamun, contains sugar syrup up to about $30 %$ of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length $5 \mathrm{~cm}$ and diameter $2.8 \mathrm{~cm}$ (see Fig. 12.15).
Fig. 12.15
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Solution
It can be observed that
Radius $(r)$ of cylindrical part $=$ Radius $(r)$ of hemispherical part $=\dfrac{2.8}{2}=1.4 cm$
Length of each hemispherical part $=$ Radius of hemispherical part $=1.4 cm$
Length $(h)$ of cylindrical part $=5-2 \times$ Length of hemispherical part
$=5-2 \times 1.4=2.2 cm$
Volume of one gulab jamun $=$ Vol. of cylindrical part $+2 \times$ Vol. of hemispherical part $=\pi r^{2} h+2 \times \dfrac{2}{3} \pi r^{3}=\pi r^{2} h+\dfrac{4}{3} \pi r^{3}$
$=\pi \times(1.4)^{2} \times 2.2+\dfrac{4}{3} \pi(1.4)^{3}$
$=\dfrac{22}{7} \times 1.4 \times 1.4 \times 2.2+\dfrac{4}{3} \times \dfrac{22}{7} \times 1.4 \times 1.4 \times 1.4$
$=13.552+11.498=25.05 cm^{3}$
Volume of 45 gulab jamuns $=45 \times 25.05=1,127.25 cm^{3}$
Volume of sugar syrup $=30 %$ of volume
$=\dfrac{30}{100} \times 1,127.25$
$=338.17 cm^{3}$
$\simeq 338 cm^{3}$
4. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are $15 \mathrm{~cm}$ by $10 \mathrm{~cm}$ by $3.5 \mathrm{~cm}$. The radius of each of the depressions is 0.5 $\mathrm{cm}$ and the depth is $1.4 \mathrm{~cm}$. Find the volume of wood in the entire stand (see Fig. 12.16).
Fig. 12.16
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Solution
Depth $(h)$ of each conical depression $=1.4 cm$
Radius ( $r$ ) of each conical depression $=0.5 cm$
Volume of wood $=$ Volume of cuboid $-4 \times$ Volume of cones
$ \begin{aligned} & =lbh-4 \times \dfrac{1}{3} \pi r^{2} h \\ & =15 \times 10 \times 3.5-4 \times \dfrac{1}{3} \times \dfrac{22}{7} \times(\dfrac{1}{2})^{2} \times 1.4 \\ & =525-1.47 \\ & =523.53 cm^{3} \end{aligned} $
5. A vessel is in the form of an inverted cone. Its height is $8 \mathrm{~cm}$ and the radius of its top, which is open, is $5 \mathrm{~cm}$. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius $0.5 \mathrm{~cm}$ are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.
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Solution
Height $(h)$ of conical vessel $=8 cm$
Radius $(r_1)$ of conical vessel $=5 cm$
Radius $(r_2)$ of lead shots $=0.5 cm$
Let $n$ number of lead shots were dropped in the vessel.
Volume of water spilled $=$ Volume of dropped lead shots
$\dfrac{1}{4} \times$ Volume of cone $n \times \dfrac{4}{3} r_2^{3}$
$\dfrac{1}{4} \times \dfrac{1}{3} \pi r_1^{2} h=n \times \dfrac{4}{3} \pi r_2^{3}$
$r_1^{2} h=n \times 16 r_2^{3}$
$5^{2} \times 8=n \times 16 \times(0.5)^{3}$
$n=\dfrac{25 \times 8}{16 \times(\dfrac{1}{2})^{3}}=100$
Hence, the number of lead shots dropped in the vessel is 100.
6. A solid iron pole consists of a cylinder of height $220 \mathrm{~cm}$ and base diameter $24 \mathrm{~cm}$, which is surmounted by another cylinder of height $60 \mathrm{~cm}$ and radius $8 \mathrm{~cm}$. Find the mass of the pole, given that $1 \mathrm{~cm}^{3}$ of iron has approximately $8 \mathrm{~g}$ mass. (Use $\pi=3.14$ )
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Solution
From the figure, it can be observed that
Height $(h_1)$ of larger cylinder $=220 cm$
Radius $(r_1)$ of larger cylinder $=\dfrac{24}{2}=12 cm$
Height $(h_2)$ of smaller cylinder $=60 cm$
Radius $(r_2.$ ) of smaller cylinder $=8 cm$
Total volume of pole $=$ Volume of larger cylinder + Volume of smaller cylinder
$ \begin{aligned} & =\pi r_1^{2} h_1+\pi r_2^{2} h_2 \\ & =\pi(12)^{2} \times 220+\pi(8)^{2} \times 60 \\ & =\pi[144 \times 220+64 \times 60] \\ & =35520 \times 3.14=1,11,532.8 cm^{3} \end{aligned} $
Mass of $1 cm^{3}$ iron $=8 g$
Mass of $111532.8 cm^{3}$ iron $=111532.8 \times 8=892262.4 g=892.262 kg$
7. A solid consisting of a right circular cone of height $120 \mathrm{~cm}$ and radius $60 \mathrm{~cm}$ standing on a hemisphere of radius $60 \mathrm{~cm}$ is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is $60 \mathrm{~cm}$ and its height is $180 \mathrm{~cm}$.
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Solution
Radius $(r)$ of hemispherical part $=$ Radius $(r)$ of conical part $=60 cm$
Height $(h_2)$ of conical part of solid $=120 cm$
Height $(h_1)$ of cylinder $=180 cm$
Radius ( $r$ ) of cylinder $=60 cm$
Volume of water left $=$ Volume of cylinder - Volume of solid
$=$ Volume of cylinder $-($ Volume of cone + Volume of hemisphere $)$
$=\pi r^{2} h_1-(\dfrac{1}{3} \pi r^{2} h_2+\dfrac{2}{3} \pi r^{3})$
$=\pi(60)^{2}(180)-(\dfrac{1}{3} \pi(60)^{2} \times 120+\dfrac{2}{3} \pi(60)^{3})$
$=\pi(60)^{2}[(180)-(40+40)]$
$=\pi(3,600)(100)=3,60,000 \pi cm^{3}=1131428.57 cm^{3}=1.131 m^{3}$
8. A spherical glass vessel has a cylindrical neck $8 \mathrm{~cm}$ long, $2 \mathrm{~cm}$ in diameter; the diameter of the spherical part is $8.5 \mathrm{~cm}$. By measuring the amount of water it holds, a child finds its volume to be $345 \mathrm{~cm}^{3}$. Check whether she is correct, taking the above as the inside measurements, and $\pi=3.14$.
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Solution
Height $(h)$ of cylindrical part $=8 cm$
Radius $(r_2)$ of cylindrical part $=\dfrac{2}{2}=1 cm$
Radius $(r_1)$ spherical part $=\dfrac{8.5}{2}=4.25 cm$
Volume of vessel $=$ Volume of sphere + Volume of cylinder
$=\dfrac{4}{3} \pi r_1^{3}+\pi r_2^{2} h$
$=\dfrac{4}{3} \pi(\dfrac{8.5}{2})^{3}+\pi(1)^{2}(8)$
$=\dfrac{4}{3} \times 3.14 \times 76.765625+8 \times 3.14$
$=321.392+25.12$
$=346.512$
$=346.51 cm^{3}$
Hence, she is wrong.
