Exercise 12.1 (Revised) - Chapter 12 - Surface Areas And Volumes - Ncert Solutions class 10 - Maths

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Chapter 12: Surface Areas and Volumes - NCERT Solutions Class 10 Maths

Unless stated otherwise, take $\pi=\frac{22}{7}$.
Ex 12.1 Question 1.

2 cubes each of volume $64 \mathrm{~cm}^3$ are joined end to end. Find the surface area of the resulting cuboid.

Answer.

Volume of cube $=(\text { Side })^3$
According to question, $(\text { Side })^3=64$
$
\begin{aligned}
& \Rightarrow(\text { Side })^3=4^3 \\
& \Rightarrow \text { Side }=4 \mathrm{~cm}
\end{aligned}
$

For the resulting cuboid, length $(l)=4+4=8 \mathrm{~cm}$, breadth $(b)=4 \mathrm{~cm}$ and height $(h)=4$ $\mathrm{cm}$
Surface area of resulting cuboid $=2(l b+b h+h l)$
$
\begin{aligned}
& =2(8 \times 4+4 \times 4+4 \times 8) \\
& =2(32+16+32) \\
& =2 \times 80=160 \mathrm{~cm}^2
\end{aligned}
$
Ex 12.1 Question 2.

A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is $14 \mathrm{~cm}$ and the total height of the vessel is $13 \mathrm{~cm}$. Find the inner surface area of the vessel.

Answer.

$\because$ Diameter of the hollow hemisphere $=14 \mathrm{~cm}$
$\therefore$ Radius of the hollow hemisphere $=\frac{14}{2}=7 \mathrm{~cm}$

Total height of the vessel $=13 \mathrm{~cm}$
$\therefore$ Height of the hollow cylinder $=13-7=6 \mathrm{~cm}$
$\therefore$ Inner surface area of the vessel
= Inner surface area of the hollow hemisphere + Inner surface area of the hollow cylinder
$=2 \pi(7)^2+2 \pi(7)(6)$
$=98 \pi+84 \pi=182 \pi$
$=182 \times \frac{22}{7}=26 \times 22=572 \mathrm{~cm}^2$
Ex 12.1 Question 3.

A toy is in the form of a cone of radius $3.5 \mathrm{~cm}$ mounted on a hemisphere of same radius. The total height of the toy is $15.5 \mathrm{~cm}$. Find the total surface area of the toy.

Answer.

Radius of the cone $=3.5 \mathrm{~cm}$
$\therefore$ Radius of the hemisphere $=3.5 \mathrm{~cm}$

Total height of the toy $=15.5 \mathrm{~cm}$
$
\begin{aligned}
& \therefore \text { Height of the cone }=15.5-3.5=12 \mathrm{~cm} \\
& \text { Slant height of the cone }=\sqrt{(3.5)^2+(12)^2} \\
& =\sqrt{12.25+144} \\
& =\sqrt{156.25}=12.5 \mathrm{~cm} \\
& \therefore \text { TSA of the toy }=\text { CSA of hemisphere }+ \text { CSA of cone } \\
& =2 \pi r^2+\pi r l \\
& =2 \pi(3.5)^2+\pi(3.5)(12.5) \\
& =24.5 \pi+43.75 \pi=68.25 \pi \\
& =68.25 \times \frac{22}{7}=214.5 \mathrm{~cm}^2
\end{aligned}
$
Ex 12.1 Question 4.

A cubical block of side $7 \mathrm{~cm}$ is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

Answer.

Greatest diameter of the hemisphere $=$ Side of the cubical block $=7 \mathrm{~cm}$
$\therefore$ TSA of the solid = External surface area of the cubical block + CSA of hemisphere

$
\begin{aligned}
& =\left\{6(7)^2-\pi\left(\frac{7}{2}\right)^2\right\}+2 \pi\left(\frac{7}{2}\right)^2 \\
& =>\left(294-\frac{49}{4} \pi\right)+\frac{49}{2} \pi \\
& =294+\frac{49}{4} \pi \\
& =294+\frac{49}{4} \times \frac{22}{7} \\
& =294+\frac{77}{2} \\
& =294+38.5=332.5 \mathrm{~cm}^2
\end{aligned}
$
Ex 12.1 Question 5.

A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter $l$ of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

Answer.

$\because$ Diameter of the hemisphere $=l$, therefore radius of the hemisphere $=\frac{l}{2}$
Also, length of the edge of the cube $=l$
$\therefore$ Surface area of the remaining solid $=$ total surface area of cubical block + curved surface area of hemispherical - area of circular base
$
\begin{aligned}
& =2 \pi\left(\frac{l}{2}\right)^2+6 l^2-\pi\left(\frac{l}{2}\right)^2 \\
& =\pi\left(\frac{l}{2}\right)^2+6 l^2 \\
& =\frac{\pi l^2}{4}+6 l^2
\end{aligned}
$

$
=\frac{1}{4} l^2(\pi+24)
$
Ex 12.1 Question 6.

A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see figure). The length of the entire capsule is $14 \mathrm{~mm}$ and the diameter of the capsule is $5 \mathrm{~mm}$. Find its surface area.

Answer.

Radius of the hemisphere $=\frac{5}{2} \mathrm{~mm}$
Let radius $=r=2.5 \mathrm{~mm}$
Cylindrical height $=$ Total height - Diameter of sphere $=h=14-(2.5+2.5)=9 \mathrm{~mm}$
Surface area of the capsule $=$ CSA of cylinder + curved Surface area of 2 hemispheres

$
\begin{aligned}
& =2 \pi r h+2\left(2 \pi r^2\right) \\
& =2 \pi\left(\frac{5}{2}\right)(9)+2\left\{2 \pi\left(\frac{5}{2}\right)^2\right\} \\
& =45 \pi+25 \pi \\
& =70 \pi=70 \times \frac{22}{7}=220 \mathrm{~mm}^2
\end{aligned}
$
Ex 12.1 Question 7.

A tent is in the shape of a cylinder surmounted buy a conical top. If the height and diameter of the cylindrical part are $2.1 \mathrm{~m}$ and $4 \mathrm{~m}$ respectively and the slant height of the top is $2.8 \mathrm{~m}$, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs. 500 per $_{\mathrm{m}^2}$. (Note that the base of the tent will not be covered with canvas.)

Answer.

Diameter of the cylindrical part $=4 \mathrm{~cm}$

$\therefore$ Radius of the cylindrical part $=2 \mathrm{~cm}$
TSA of the tent = CSA of the cylindrical part + CSA of conical top

$
\begin{aligned}
& =2 \pi(2)(2.1)+\pi(2)(2.8) \\
& =8.4 \pi+5.6 \pi \\
& =14 \pi \\
& =14 \times \frac{22}{7} \\
& =44 \mathrm{~m}^2
\end{aligned}
$
$\therefore$ Cost of the canvas of the tent of $1 \mathrm{~m}^2=$ Rs. 500
cost of canvas of the tent of $44 \mathrm{~m}^2=$
$
=44 \times 500=\text { Rs. } 22000
$
Ex 12.1 Question 8.

From a solid cylinder whose height is $2.4 \mathrm{~cm}$ and diameter $1.4 \mathrm{~cm}$, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest $\mathrm{cm}^2$.

Answer.

Diameter of the solid cylinder $=1.4 \mathrm{~cm}$
$\therefore$ Radius of the solid cylinder $=0.7 \mathrm{~cm}$
$\therefore$ Radius of the base of the conical cavity $=0.7 \mathrm{~cm}$

Height of the solid cylinder $=2.4 \mathrm{~cm}$
$\therefore$ Height of the conical cavity $=2.4 \mathrm{~cm}$
$\therefore$ Slant height of the conical cavity $=\sqrt{(0.7)^2+(2.4)^2}$
$
\begin{aligned}
& =\sqrt{0.49+5.76} \\
& =\sqrt{6.25}=2.5 \mathrm{~cm}
\end{aligned}
$
$\therefore$ TSA of remaining solid = curved surface area of cylinder + area of upper circular part + curved surface area of conical part
$
\begin{aligned}
& =2 \pi(0.7)(2.4)+\pi(0.7)^2+\pi(0.7)(2.5) \\
& =3.36 \pi+0.49 \pi+1.75 \pi \\
& =5.6 \pi \\
& =5.6 \times \frac{22}{7}=17.6 \mathrm{~cm}^2 \\
& =18 \mathrm{~cm}^2 \text { (to the nearest } \mathrm{cm}^2 \text { ) }
\end{aligned}
$

Ex 12.1 Question 9.

A wooden article was made by scooping out a hemisphere from each end of a solid cylinder as shown in figure. If the height of the cylinder is $10 \mathrm{~cm}$ and its base is of radius $3.5 \mathrm{~cm}$, find the total surface area of the article.

Answer.

TSA of the article $=2 \pi r H+2\left(2 \pi r^2\right)=$ curved surface area of cylinder + curved surface area of 2 hemispheres
$
\begin{aligned}
& =2 \pi(3.5)(10)+2\left[2 \pi(3.5)^2\right] \\
& =70 \pi+49 \pi \\
& =119 \pi \\
& =119 \times \frac{22}{7} \\
& =374 \mathrm{~cm}^2
\end{aligned}
$