NCERT Solutions for Class 10 Maths Chapter 13: Surface Areas and Volumes || CBSE Class 10 Maths Chapter 13 Surface Areas and Volumes solutions Ex 13.2

Given:
Height (h) of conical part \( = \) Radius (r) of conical part \( =1 \mathrm{~cm} \)
Radius (r) of hemispherical part \( = \) Radius of conical part \( (\mathrm{r})=1 \mathrm{~cm} \)
Volume of solid \(=\) Volume of conical part \(+\) Volume of hemispherical part
\(=\frac{1}{3} \pi r^{2} \mathrm{h}+\frac{2}{3} \pi \mathrm{r}^{3}\)
\(=\frac{1}{3} \pi(1)^{2}(1)+\frac{2 \pi}{3}(1)^{3}\)
\(=\frac{\pi}{3}+\frac{2 \pi}{3}\)
\(=\frac{3 \pi}{3}\)
\(=\pi \mathrm{cm}^{3}\)
Hence, Volume of the solid \( =\pi \mathrm{cm}^{3} \)

Height \( \left(\mathrm{h}^{1}\right) \) of each conical part \( =2 \mathrm{~cm} \)
Height \( \left(\mathrm{h}^{2}\right) \) of cylindrical part \( =1^{2}-2 \times \) Height of conical part \( =1^{2}-2 \times 2=8 \mathrm{~cm} \)
Radius \( = \text{Diameter}/ 2\)
Radius (r) of cylindrical part \( =\frac{3}{2} \mathrm{~cm} \)
Radius of conical part \( =\frac{3}{2} \mathrm{~cm} \)
Volume of air present in the model \( = \) Volume of cylinder \( +2 \times \) Volume of cones
volume of cylinder \( =\frac{1}{3} \pi \mathrm{r}^{2} \mathrm{~h} \)
Volume of air present in the model \( =\pi \mathrm{r}^{2} \mathrm{~h}^{2}+2 \times \frac{1}{3} \pi \mathrm{r}^{2} \mathrm{~h}^{1} \)
Volume of air present in the model
\( =\pi\left(\frac{3}{2}\right)^{2}(8) 2 \times \frac{1}{3} \pi\left(\frac{3}{2}\right)^{2}(2) \)
\(=18 \pi+3 \pi\)
\(=21 \pi\)
\(=21 \times \frac{ 22 }{ 7 }\)
\(=3 \times 22\)
\(=66 \mathrm{~cm}^{3}\)
Hence, volume of air contained in the model that Rachel made \( =66 \mathrm{~cm}^{3}\)

Radius (r) of cylindrical part \( =\frac{2.8}{2}=1.4 \mathrm{~cm} \)
Radius \( (\mathrm{R}) \) of hemispherical part \( =\frac{2.8}{2}=1.4 \mathrm{~cm} \)
Length of each hemispherical part \( =1.4 \mathrm{~cm} \)
The radius of hemispherical part \( =1.4 \mathrm{~cm} \)
Length (h) of cylindrical part \( =5-2 \times \) Length of hemispherical part
\(=5-(2 \times 1.4)\)
\(=5-2.8\)
\(=2.2 \mathrm{~cm}\)
The volume of one gulabjamun \( = \) Vol. of cylindrical part \( +2 \times \) Vol. of hemispherical part
\(=\left(\pi \mathrm{r}^{2} \mathrm{~h}\right)+\left(2 \times \frac{2}{3} \pi \mathrm{R}^{3}\right)\)
\(=\left(\pi \mathrm{r}^{2} \mathrm{~h}\right)+\left(\frac{4}{3} \pi \mathrm{R}^{3}\right)\)
\(=\left[\pi \times(1.4)^{2} \times 2.2\right]+\left[\frac{4}{3} \pi(1.4)^{3}\right]\)
\(=\left(\frac{22}{7} \times 1.4 \times 1.4 \times 2.2\right)+\left(\frac{4}{3} \times \frac{22}{7} \times 1.4 \times 1.4 \times 1.4\right)\)
\(=\left(\frac{94.864}{7}\right)+\left(\frac{241.472}{21}\right)\)
\(=13.552+11.498\)
\(=25.05 \mathrm{~cm}^{3}\)
Volume of 45 gulabjamun \( =45 \times 25.05 \)
\(=1,127.25 \mathrm{~cm}^{3}\)
The volume of sugar syrup \( =30 \% \) of the volume
\(=\frac{30}{100} \times 1127.25\)
\(=338.17 \mathrm{~cm}^{3}\)
\(=338 \mathrm{~cm}^{3} \text { (approx.) }\)

Given:
Depth (h) of each conical depression \( =1.4 \mathrm{~cm} \)
Radius (r) of each conical depression \( =0.5 \mathrm{~cm} \)
Volume of wood \( = \) Volume of cuboid \( -4 \times \) Volume of cones
\(= \mathrm{lbh}-4 \times \frac{1}{3} \pi \mathrm{r}^{2} \mathrm{~h}\)
\(=15 \times 10 \times 3.5-4 \times \frac{1}{3} \times \frac{22}{7} \times\left(\frac{1}{2}\right)^{2} \times 1.4\)
\(=525-1.47\)
\( =523.53 \mathrm{~cm}^{3} \)
Hence,
The volume of wood in the entire stand \( =523.53 \mathrm{~cm}^{3} \)

When lead shots are dropped in the Vessel, the same amount of water will come out of the vessel as the Total Volume of Lead Shots.
Height (h) of conical vessel \( =8 \mathrm{~cm} \)
Radius (\( \mathrm{r}_{1} \)) of conical vessel \( =5 \mathrm{~cm} \)
Radius (\( \mathrm{r}_{2} \)) of lead shots \( =0.5 \mathrm{~cm} \)
Volume of water \( = \) Volume of Vessel Let n number of lead shots were dropped in the vessel
Volume of water spilled = Volume of dropped lead shots
Volume of vessel \( =\frac{1}{3} \pi r_{1}^{2} \mathrm{~h} \)
Volume of vessel \( =\frac{1}{3} \pi(5)^{2} 8 \)
Volume of water spilled \( =\frac{1}{4} \times \) volume of vessel
Volume of water spilled \( =\frac{1}{12} \pi \times 25 \times 8 \)
Volume of lead shots \( =\mathrm{n} \times \frac{4}{3} \pi r_{2}^{3} \)
Volume of lead shots \( =\mathrm{n} \times \frac{4}{3} \pi(0.5)^{3} \)
Volume of lead shots \( = \) Volume of water spilled
\( \mathrm{n} \times \frac{4}{3} \pi(0.5)^{3}=\frac{1}{12} \pi \times 25 \times 8 \)
\( \mathrm{n}=\frac{25 \times 8 \times 3}{12 \times 4 \times 0.5 \times 0.5 \times 0.5} \)
\( \mathrm{n}=100 \)
Hence, the number of lead shots dropped in the vessel is 100.