NCERT Solutions For Maths Chapter 12 Exercise 12.2 Class 10 - Surface Areas and Volumes

Class 10 Ex 12.2

1. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.

Ans. From the given figure, we know that

Height (h) of conical part = Radius(r) of conical part \[{\text{ = 1cm}}\]

Radius(r) of hemispherical part = Radius of conical part (r) \[{\text{ = 1cm}}\]

Volume of solid = Volume of conical part + Volume of hemispherical part

\[{\text{ = }}\dfrac{1}{3}\pi {r^2}h + \dfrac{2}{3}\pi {r^3}\]

\[{\text{ = }}\dfrac{1}{3}\pi {\left( 1 \right)^2}\left( 1 \right) + \dfrac{2}{3}\pi {\left( 1 \right)^3} = \dfrac{{2\pi }}{3} + \dfrac{1}{3} = \pi {\text{c}}{{\text{m}}^3}\]

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 aluminum sheet. The diameter of the model is \[{\text{3cm}}\] and its length is\[{\text{12cm}}\]. If each cone has a height of \[{\text{2cm}}\], 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.)

Ans:

From the given figure, we know that 

Height (h1) of each conical part \[{\text{ = 2cm}}\]

Height (h2) of cylindrical part \[{\text{ = 12  -  2}} \times {\text{ Height of conical part}}\]

\[{\text{ = 12  -  2}} \times {\text{2  =  8cm}}\]

Radius (r) of cylindrical part = Radius of conical part \[{\text{ =  }}\dfrac{3}{2}{\text{cm}}\]

Volume of air present in the model = Volume of cylinder \[ + {\text{ }}2\, \times \]Volume of cones

\[ = \pi {r^2}{h_2} + 2 \times \dfrac{1}{3}\pi {r^2}{h_1}\]

\[ = \pi  \times {\left( {\dfrac{3}{2}} \right)^2} \times 8 + 2 \times \dfrac{1}{3}\pi {\left( {\dfrac{3}{2}} \right)^2} \times 2\]

\[ = \pi  \times \dfrac{9}{4} \times 8 + \dfrac{2}{3}\pi  \times \dfrac{9}{4} \times 2\]

\[ = 18\pi  + 3\pi  = 21\pi  = 66{\text{c}}{{\text{m}}^3}\]

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 cm and diameter 2.8 cm (see the given figure).

Ans:

We can observe that,

Radius (r) of cylindrical part = Radius (r) of hemispherical part

\[{\text{ = }}\dfrac{{{\text{2}}{\text{.8}}}}{{\text{2}}}{\text{ = 1}}{\text{.4cm}}\]

Length of each hemispherical part = Radius of hemispherical part \[{\text{ = 1}}{\text{.4cm}}\]

Length (h) of cylindrical part \[{\text{ = 5  - }}\,{\text{2}} \times {\text{Length of hemispherical part }}\]

\[{\text{ = 5  - }}\,{\text{2}} \times {\text{1.4 = 2}}{\text{.2cm}}\]

Volume of one gulab jamun = Vol. of cylindrical part + 2 × Vol. of hemispherical part

\[ = \pi {r^2}h + 2 \times \dfrac{2}{3}\pi {r^3} = \pi {r^2}h + \dfrac{4}{3}\pi {r^3}\]

\[ = \pi {\left( {1.4} \right)^2}\left( {2.2} \right) + \dfrac{4}{3}\pi {\left( {1.4} \right)^3}\]

\[ = \dfrac{{22}}{7} \times 1.4 \times 1.4 \times 2.2 + \dfrac{4}{3} \times \dfrac{{22}}{7} \times 1.4 \times 1.4 \times 1.4\]

\[{\text{ = 13}}{\text{.552 + 11}}{\text{.498 = 25}}{\text{.05c}}{{\text{m}}^{\text{3}}}\]

Volume of \[{\text{45}}\] gulab jamuns \[{\text{ = 45}} \times {\text{25}}{\text{.05 = 1127}}{\text{.25c}}{{\text{m}}^{\text{3}}}\]

Volume of sugar syrup \[ = 30\% \]  of volume

\[ = \dfrac{{30}}{{100}} \times 1127.25\]

\[{\text{ = 338}}{\text{.17c}}{{\text{m}}^{\text{3}}}\]

\[ \cong {\text{338c}}{{\text{m}}^{\text{3}}}\]

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 cuboids are \[{\text{15cm}}\]  by \[{\text{10cm}}\]  by \[{\text{3}}{\text{.5cm}}\]. The radius of each of the depressions is \[{\text{0}}{\text{.5cm}}\] and the depth is \[{\text{1}}{\text{.4cm}}\]. Find the volume of wood in the entire stand (see the following figure).

Ans:

Depth (h) of each conical depression\[{\text{ = 1}}{\text{.4cm}}\]

Radius (r) of each conical depression\[{\text{ = 0}}{\text{.5cm}}\]

Volume of wood = Volume of cuboid \[ - {\text{ }}4 \times \] Volume of cones

\[{\text{ = l * b * h}}\]\[ - 4 \times \dfrac{1}{3}\pi {r^2}h\]

\[ = 15 \times 10 \times 3.5 - 4 \times \dfrac{1}{3} \times \dfrac{{22}}{7} \times {\left( {\dfrac{1}{2}} \right)^2} \times 1.4\]

\[ = 525 - 1.47\]

\[{\text{ = 523}}{\text{.53c}}{{\text{m}}^{\text{3}}}\]

5. A vessel is in the form of an inverted cone. Its height is \[{\text{8cm}}\] and the radius of its top, which is open, is \[{\text{5cm}}\]. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius \[{\text{0}}{\text{.5cm}}\]are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

Ans:

Height (h) of conical vessel\[{\text{ = 8cm}}\]

Radius (r1) of conical vessel\[{\text{ = 5cm}}\]

Radius (r2) of lead shots\[{\text{ = 0}}{\text{.5cm}}\]

Let the number of lead shots dropped in the vessels be n.

Volume of water spilled = Volume of dropped lead shots

\[\dfrac{1}{4} \times \]Volume of cone \[{\text{ = n *}}\dfrac{{\text{4}}}{{\text{3}}}{{\text{r}}_{\text{2}}}^{\text{3}}\]

\[\dfrac{1}{4} \times \dfrac{1}{3}\pi {{\text{r}}_1}^2{\text{h = n * }}\dfrac{{\text{4}}}{{\text{3}}}{{\text{r}}_{\text{2}}}^{\text{3}}\]

\[{{\text{r}}_1}^2{\text{h = n *16}}{{\text{r}}_{\text{2}}}^{\text{3}}\]

\[{5^2} \times {\text{8 = n *16}} \times {\left( {0.5} \right)^{\text{3}}}\]

\[n = \dfrac{{25 \times 8}}{{16 \times {{\left( {\dfrac{1}{2}} \right)}^3}}} = 100\]

Therefore, the number of lead shots dropped in the vessel is \[100\]

6. A solid iron pole consists of a cylinder of height \[{\text{220cm}}\] and base diameter \[{\text{24cm}}\], which is surmounted by another cylinder of height \[{\text{60cm}}\] and radius \[{\text{8cm}}\]. Find the mass of the pole, given that \[{\text{1c}}{{\text{m}}^3}\]of iron has approximately \[{\text{8g}}\] mass.

Ans:

From the above figure, we observe that 

Height (h1) of larger cylinder \[{\text{ = 220cm}}\]

Radius (r1) of larger cylinder \[{\text{ = }}\dfrac{{24}}{2}{\text{ = 12cm}}\]

Height (h2) of smaller cylinder \[{\text{ = 60cm}}\]

Radius (r2) of smaller cylinder \[{\text{ = 8cm}}\]

Total volume of pole = Volume of larger cylinder + volume of smaller cylinder

\[{\text{ =}}\pi{{\text{r}}_{\text{1}}}^{\text{2}}{{\text{h}}_{\text{1}}}{\text{ +}} \pi {{\text{r}}_{\text{2}}}^{\text{2}}{{\text{h}}_{\text{2}}}\]

\[{\text{ = }} \pi {\left( {12} \right)^{\text{2}}}* {\text{220 + }} \pi {\left( 8 \right)^{\text{2}}} * 60\]

${\text{ = }}\pi\left[ {{\text{(144 * 220) + (64 * 60)}}} \right] $

${\text{ = 35520 * 3}}{\text{.14 = 111532}}{\text{.8c}}{{\text{m}}^{\text{3}}} $

Mass of \[{\text{1c}}{{\text{m}}^3}\]iron = \[{\text{8g}}\]

Mass of \[{\text{111532}}{\text{.8c}}{{\text{m}}^{\text{3}}}\]iron \[{\text{ = 111532}}{\text{.8 * 8  =  892262}}{\text{.4g  =  892}}{\text{.262kg}}\]

7. A solid consisting of a right circular cone of height \[{\text{120cm}}\] and radius \[{\text{60cm}}\] standing on a hemisphere of radius \[{\text{60cm}}\] 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 \[{\text{60cm}}\] and its height is \[{\text{180cm}}\].

Ans:

Radius (r) of hemispherical part = Radius (r) of conical part\[{\text{ = 60cm}}\]

Height (h2) of conical part of solid\[{\text{ = 120cm}}\]

Height (h1) of cylinder\[{\text{ = 180cm}}\]

Radius (r) of cylinder\[{\text{ = 60cm}}\]

Volume of water left = Volume of cylinder − Volume of solid 

= Volume of cylinder – (Volume of cone + Volume of hemisphere)

\[{\text{ = }}\pi{{\text{r}}^{\text{2}}}{{\text{h}}_{\text{1}}}{\text{ - }}\left( {\dfrac{{\text{1}}}{{\text{3}}}{\pi}{{\text{r}}^{\text{2}}}{{\text{h}}_{\text{2}}}{\text{ + }}\dfrac{{\text{2}}}{{\text{3}}} \pi {{\text{r}}^{\text{3}}}} \right)\]

\[{\text{ = }}\pi {\left( {60} \right)^{\text{2}}} \times 180{\text{ - }}\left( {\dfrac{{\text{1}}}{{\text{3}}}{\pi }{{\left( {60} \right)}^{\text{2}}} \times {\text{120 + }}\dfrac{{\text{2}}}{{\text{3}}}{\pi }{{\left( {60} \right)}^{\text{3}}}} \right)\]

\[{\text{ = }}\pi{\left( {60} \right)^{\text{2}}}\left[ {\left( {180} \right) - \left( {40 + 40} \right)} \right]\]

\[{\text{ = }}\pi\left( {{\text{3600}}} \right)\left( {{\text{100}}} \right){\text{ = 3,60,000 }}\pi{\text{c}}{{\text{m}}^{\text{3}}}{\text{ = 11311428}}{\text{.57c}}{{\text{m}}^{\text{3}}}{\text{ = 1}}{\text{.131}}{{\text{m}}^{\text{3}}}\]

8. A spherical glass vessel has a cylindrical neck \[{\text{8cm}}\] long, \[{\text{2cm}}\] in diameter; the diameter of the spherical part is \[{\text{8}}{\text{.5cm}}\]. By measuring the amount of water it holds, a child finds its volume to be \[{\text{345c}}{{\text{m}}^3}\]. Check whether she is correct, taking the above as the inside measurements, and \[\pi  = 3.14\].

Ans:

Height (h) of cylindrical part \[{\text{ = 8cm}}\]

Radius (r2) of cylindrical part \[{\text{ = }}\dfrac{2}{2}{\text{ = 1cm}}\]

Radius (r1) spherical part \[{\text{ = }}\dfrac{{8.5}}{2}{\text{ = 4}}{\text{.25cm}}\]

Volume of vessel = Volume of sphere + Volume of cylinder \[{\text{ = }}\dfrac{4}{3}\pi {\left( {\dfrac{{8.5}}{2}} \right)^3}{\text{ + }}\pi {\left( 1 \right)^2}\left( 8 \right)\]

\[{\text{ = }}\dfrac{4}{3} \times 3.14 \times 76.765625{\text{ + 8}} \times {\text{3}}{\text{.14}}\]

\[ = 321.392 + 25.12\]

\[ = 346.512\]

\[{\text{ = 346}}{\text{.51c}}{{\text{m}}^{\text{3}}}\]

Therefore, she is wrong.

Conclusion

Ex 12.2 Class 10 Maths Chapter 12, Surface Areas and Volumes, is all about the volumes of different solid shapes and how they can be combined. Understanding how to compute the volumes of various geometric solids, such as cubes, cuboids, cylinders, cones, and spheres, as well as their combinations, requires completion of this task. For detailed solutions and step-by-step explanations, refer to the NCERT Solutions provided by Vedantu​.

Class 10 Maths Chapter 12: Exercises Breakdown

CBSE Class 10 Maths Chapter 12 Other Study Materials

Chapter-Specific NCERT Solutions for Class 10 Maths

Given below are the chapter-wise NCERT Solutions for Class 10 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.