CBSE Solutions for Class 10 Maths Chapter 13: Surface Areas and Volumes || CBSE Class 10 Maths Chapter 13 Surface Areas and Volumes solutions Ex 13.3

To find: Height of the cylinder (h)
Given:
Radius \( \left(\mathrm{r}_{1}\right) \) of sphere \( =4.2 \mathrm{~cm} \)
Radius \(( \mathrm{r}_{2} )\) of cylinder \( =6 \mathrm{~cm} \)
Let the height of the cylinder be \( h \).
The object formed by recasting the sphere will be the same in volume.
Volume of sphere \( = \) Volume of cylinder
\(=\frac{4}{3} \pi r_{1}^{3}=\pi r_{2}^{2} \mathrm{~h}\)
\(=\frac{4}{3} \times \frac{22}{7} \times 4.2 \times 4.2 \times 4.2=\frac{22}{7} \times 6 \times 6 \times \mathrm{h}\)
\(\mathrm{h}=\frac{4 \times 4.2 \times 4.2 \times 4.2}{3 \times 6 \times 6}\)
\(\mathrm{h}=2.74 \mathrm{~cm}\)
Hence, the height of the cylinder \( =2.74 \mathrm{~cm} \)

Radius \( \left(r_{1}\right) \) of \( 1^{\text {st}} \) sphere \( =6 \mathrm{~cm} \)
Radius \( \left(\mathrm{r}_{2}\right) \) of \( 2^{\text {nd}} \) sphere \( =8 \mathrm{~cm} \)
Radius \( \left(\mathrm{r}_{3}\right) \) of \( 3^{\text {rd}} \) sphere \( =10 \mathrm{~cm} \)
Let the radius of the resulting sphere be \(r \).
The object formed by recasting these spheres will be the same in volume as the sum of the volumes of these spheres.
The volume of 3 spheres \( = \) Volume of resulting sphere
\(\frac{4}{3} \pi r_{1}^{3}+\frac{4}{3} \pi r_{2}^{3}+\frac{4}{3} \pi r_{3}^{3}=\frac{4}{3} \pi r^{3}\)
\(\frac{4}{3} \pi\left[\mathrm{r}_{1}^{3}+\mathrm{r}_{2}^{3}+\mathrm{r}_{3}^{3}\right]=\frac{4}{3} \pi \mathrm{r}^{3}\)
\(\frac{4}{3} \pi\left[6^{3}+83+103\right]=\frac{4}{3} \pi \mathrm{r}^{3}\)
\(\mathrm{r}^{3}=216+512+1000\)
\(\mathrm{r}^{3}=1728\)
\(\mathrm{r}=12 \mathrm{~cm}\)
Hence, the radius of the resulting sphere, \( \mathrm{r}=12 \mathrm{~cm} \).

The diagram for the situation is as follows:

As per the question,
The shape of the well will be cylindrical
Depth (h) of well \( =20 \mathrm{~m} \)
Radius (r) of circular end of well \( =\frac{7}{2} \)
Area of platform \( = \) Length \( \times \) Breadth \( =22 \times 14 \mathrm{~m}^{2} \)
Let height of the platform \( =\mathrm{H} \)
Volume of soil dug from the well would be equal to the volume of soil that is scattered on the platform.
Volume of soil from well \(=\) Volume of soil used to make such platform
\(\pi \mathrm{r}^{2} \mathrm{h}=\text {Area} \times \text {Height }\)
\(\pi \times\left(\frac{7}{2}\right)^{2} \times 20=22 \times 14 \times \mathrm{H}\)
\(\mathrm{H}=\frac{22}{7} \times \frac{49}{4} \times \frac{20}{22 \times 14}\)
\( =2.5 \mathrm{~m} \)
Hence, the height of the platform \( =2.5 \mathrm{~m} \)

The shape of the well will be cylindrical.
Depth \( \left(h_{1}\right) \) of well \( =14 \mathrm{~m} \)
Radius (r) of the circular end of well \( =\frac{ 3 }{ 2 } \mathrm{~m} \)
Width of embankment \( =4 \mathrm{~m} \)
As per the question,
Our embankment will be in a cylindrical shape having outer radius (R)as \( \frac{ 11 }{ 2 } \mathrm{~m} \) and inner radius \( (\mathrm{r})=\frac{ 3 }{ 2 } \mathrm{~m} \)
Let the height of embankment be \( \mathrm{h}_{2} \)
Volume of soil dug from well \(=\) Volume of earth used to form embankment
\(\pi \mathrm{r}^{2} \times \mathrm{h}_{1}=\pi \times\left(\mathrm{R}^{2}-\mathrm{r}^{2}\right) \times \mathrm{h}_{2}\)
\(=\pi \times\left(\frac{3}{2}\right)^{2} \times 14=\pi \times\left[\left(\frac{11}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}\right] \times \mathrm{h}\)
\(=\pi \times\left(\frac{3}{2}\right)^{2} \times 14=\pi \times\left[\frac{121}{4}-\frac{9}{4}\right] \times \mathrm{h}\)
\(=\pi \times \frac{9}{4} \times 14=\pi \times\left[\frac{112}{4}\right] \times \mathrm{h}\)
\(\Rightarrow 31.5=28 \mathrm{~h}\)
\(=\frac{31.5}{28}=\mathrm{h}\)
\(\Rightarrow \mathrm{h}=1.125 \mathrm{~m}\)

Height \( (\mathrm{H}) \) of cylindrical container \( =15 \mathrm{~cm} \)
Radius (\( \mathrm{r}_{1} \)) of circular end of container \( =\frac{12}{2}=6 \mathrm{~cm} \)
Radius (\( \mathrm{r}_{2} \)) of circular end of ice-cream cone \( =\frac{6}{2}=3 \mathrm{~cm} \)
Height (h) of conical part of ice-cream cone \( =12 \mathrm{~cm} \)
Let n cones be filled with ice-cream of the container
Volume of ice-cream in cylinder \( =\mathrm{n} \times \) (Volume of 1 ice-cream cone \(+\) Volume of hemispherical shape on the top)
\(\pi \times\left(r_{1}\right)^{2} \times H=n\left[\left(\frac{1}{3} \pi \times\left(r_{2}\right)^{2} \times h\right)+\left(\frac{2}{3} \pi \times\left(r_{2}\right)^{2}\right)\right]\)
\(\mathrm{n}=\frac{\pi \times\left(r_{1}\right)^{2} \times H}{\pi\left(\frac{1}{3} \times\left(r_{2}\right)^{2} \times h\right)+\left(\frac{2}{3} \times\left(r_{2}\right)^{2}\right)}\)
\(\mathrm{n}=\frac{\left(r_{1}\right)^{2} \times H}{\left(\frac{1}{3} \times\left(r_{2}\right)^{2} \times h\right)+\left(\frac{2}{3} \times\left(r_{2}\right)^{2}\right)}\)
\(=\mathrm{n}=\frac{6 \times 6 \times 15}{\left(\frac{1}{3} \times 9 \times 12\right)+\left(\frac{2}{3} \times 3 \times 3 \times 3\right)}\)
\(=\mathrm{n}=\frac{540}{36+18}=\frac{540}{54}\)
\(=\mathrm{n}=10\)

Coins are cylindrical in shape.
Height \( \left(h_{1}\right) \) of cylindrical coins \( =2 \mathrm{~mm}=0.2 \mathrm{~cm} \)
Radius (r) of circular end of coins \( =\frac{1.75}{2}=0.875 \mathrm{~cm} \)
Let n coins be melted to form the required cuboids
Volume of \( n \) coins \( = \) Volume of cuboids
\(\mathrm{n} \times \pi \times \mathrm{r}^{2} \times \mathrm{h}_{1}=\mathrm{l}\times \mathrm{b} \times \mathrm{h}\)
\(\mathrm{n} \times \pi \times(0.875)^{2} \times 0.2=5.5 \times 10 \times 3.5\)
\(\mathrm{n}=\frac{5.5 * 1 . * 3.5 * 7}{0.875 * 0.875 * 0.2 * 22}\)
\(=400\)
Hence,
The number of coins melted to form the given cuboid is 400.

Given:
Height \( \left(\mathrm{h}_{1}\right) \) of cylindrical bucket \( =32 \mathrm{~cm} \)
Radius \( \left(\mathrm{r}_{1}\right) \) of circular end of bucket \( =18 \mathrm{~cm} \)
Height \( \left(\mathrm{h}_{2}\right) \) of conical heap \( =24 \mathrm{~cm} \)
Let the radius of the circular end of conical heap be \( \mathrm{r}_{2} \)
The volume of sand in the cylindrical bucket will be equal to the volume of sand in the conical heap
Volume of sand in the cylindrical bucket \( = \) Volume of sand in conical heap
\(\pi \mathrm{r}_{1}^{2} \times \mathrm{h}_{1}=\frac{1}{3} \pi \times \mathrm{r}_{2}^{2} \times \mathrm{h}_{2}\)
\(\pi \times(18)^{2} \times 32=\frac{1}{3} \times \frac{22}{7} \times \mathrm{r}_{2}^{2} \times 24\)
\(r_{2}^{2}=\frac{18 \times 18 \times 32 \times 3}{24}\)
\(\mathrm{r}_{2}=\sqrt{18 \times 18 \times 4}\)
\(=36 \mathrm{~cm}\)
If \( \mathrm{h} \) is the height and \(\mathrm{r}\) is the radius of cone, then slant height is given by
\( \mathrm{l}=\sqrt{h^{2}+r^{2}} \)
\( \mathrm{l}=\sqrt{36^{2}+24^{2}} \)
\( \mathrm{l}=12 \sqrt{13} \mathrm{~cm} \)

Area of cross-section \( =6 \times 1.5=9 \mathrm{~m}^{2} \)
Speed of water \( =10 \mathrm{~km} / \mathrm{h} \)
Water flows through canal in \( 60 \mathrm{~min}=10 \mathrm{~km} \)
Water flows through canal in \( 1 \mathrm{~min}=\frac{1}{60} \times 10 \mathrm{~km} \)
Water flows through canal in \( 30 \mathrm{~min}=\frac{30}{60} \times 10 \)
\(=5 \mathrm{~km}=5000 \mathrm{~m}\)
Hence length of the canal is 5000 m.
\(\text{The volume of canal} =\mathrm{L} \times \mathrm{B} \times \mathrm{H}=(5000 \times 6 \times 1.5)=45000 \mathrm{~m}^{3} \)
Let the irrigated area be A.
Now, we know that,
The volume of water irrigating the required area will be equal to the volume of water that flowed in 30 minutes from the canal.
Vol. of water flowing in 30 minutes from canal \( = \) Vol. of water irrigating the reqd. area \( \Rightarrow 45000= \) Area \( \times \) height
Given height \( =8 \mathrm{~cm}=0.08 \mathrm{~m} \Rightarrow 45000= \) Area \( \times 0.08 \)
\( = \) area \( =\frac{45000}{0.08} \)
Area \( =562500 \mathrm{~m}^{2} \)

As we know that \( 1 \mathrm{~m}=100 \mathrm{~cm} \) Therefore, \( 1 \mathrm{~cm}=\frac{1}{100} \mathrm{~m} \)
\(20 \mathrm{~cm}=\frac{20}{200} \mathrm{~m}\)
Radius (r1) of circular end of pipe \( =\frac{20}{200}=0.1 \mathrm{~m} \)
Area of cross-section \( =\pi \times \mathrm{r}_{1}^{2} \)
\( =\pi \times(0.1)^{2} \)
\( =0.01 \pi \mathrm{m}^{2} \)
Speed of water \( =3 \mathrm{~km} / \mathrm{h} \)
\( =\frac{3000}{60} \)
\( =\frac{300}{6} \)
\( =50 \) meter/min
The volume of water that flows in 1 minute from pipe \( =50 \times 0.01 \pi =0.5 \pi \mathrm{~m}^{3} \)
The volume of water that flows in t minutes from pipe \( =\mathrm{t} \times 0.5 \pi \mathrm{~m}^{3} \)
Radius (\( \mathrm{r}_{2} \)) of circular end of cylindrical tank \( =\frac{10}{2}=5 \mathrm{~m} \)
Depth \( \left(\mathrm{h}_{2}\right) \) of cylindrical tank \( =2 \mathrm{~m} \)
Let the tank be filled completely in \( t \) minutes
The volume of water filled in the tank in \( t \) minutes is equal to the volume of water flowed in \( t \) minutes from the pipe
The volume of water that flows in \( t \) minutes from pipe \( = \) Volume of water in the tank
\(\mathrm{t} \times 0.5 \pi=\pi \times\left(\mathrm{r}_{2}\right)^{2} \times \mathrm{h}_{2}\)
\(\mathrm{t} \times 0.5=(5)^{2} \times 2\)
\( \mathrm{t}=100 \)
Hence, the cylindrical tank will be filled in 100 minutes.