NCERT Solutions for Class 10 Maths Chapter 13 - Surface Areas and Volumes

9. Frustum of a Cone : If a cone is cut by a plane parallel to the base of the cone, then the portion between the plane and base is called the frustum of a cone.

8. From a solid cylinder whose height is 2.4 cm and diameter 14 cm, a conical cavity of the same height and same diameter is followed out. Find the total surface area of the remaining solid to the nearest cm².

Sol. Given, diameter of cylinder = diameter of conical cavity = 1.4 cm

3. A 20 m deep well with diameter 7 m is dug and the Earth from digging is evently spread out to form a platform 22 m by 14 m. Find the height of the platform.

Sol. Given, the height of deep well which form a cylinder (h) = 20 m

$$\text{Radius of deep well (r}1) =\frac{3}{2}\text{m}$$

Let the width and height of the embarkment form a circular ring are d and h₂.

Here, d = 4 cm (Given)

Now, the volume of deep well (cylinder) formed

$$=\pi r_1^{2}h_1\\=\frac{22}{7}×\frac{3}{2}×\frac{3}{2}×14=99\space m^{3}$$

Here, we observe that the embankment form a hollow cylinder.

∴ Volume of hollow cylinder (embankment)

$$=\pi(r_1+d)^{2}.h_2-\pi r_1^{2}h_2\\=\pi\bigg(\frac{3}{2}+4\bigg)^{2}h_2-\pi\bigg(\frac{3}{2}\bigg)^{2}h_2\\=\pi\bigg[\bigg(\frac{11}{2}\bigg)^{2}h_2-\frac{9}{4}h_2\bigg]$$

According to the question

Volume of deep well (cylinder)

= Volume of embankment (hollow cylinder)

$$\Rarr\space 99=\pi\bigg[\bigg(\frac{11}{2}\bigg)^{2}×h_2-\frac{9}{4}h_2\bigg]\\\Rarr\space\frac{7×99}{22}=\frac{121}{4}×h_2-\frac{9}{4}h_{2}-\frac{112h_2}{4}\\\Rarr\space h_2=\frac{7×99×4}{22×112}\\=\frac{9}{8}=1.125\space\text{m}$$

∴ Height of the embankment is 1.125 cm.

1. A copper wire, 3 mm in diameter, is wound about a cylinder whose length is 12 cm and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cm³.

Sol. Since, the diameter of the wire is 3 mm.

When a wire is one round wound about a cylinder, it covers a 3 mm of length of the cylinder.

Given, length of the cylinder = 12 cm = 120 mm

∴ Number of rounds to cover 120 mm

$$=\frac{120}{3}=40$$

Given, diameter of a cylinder is d= 10 cm

$$\therefore\space\text{Radius, r =}\frac{10}{2}=5\space\text{cm}$$

∴ Length of wire required to complete one round = 2πr = 2π(5) = 10 π cm

∴ Length of the wire in covering the whole surface

= Length of the wire in completing 40 rounds

= 10π × 40 = 400π cm

= 400 × 3.14 = 1256 cm

Now, radius of copper wire =

$$\frac{3}{2}\space\text{mm}=\frac{3}{20}\space\text{cm}$$

∴ Volume of wire =

$$\pi\bigg(\frac{3}{20}\bigg)^{2}(400\pi)$$

= 9π² cm³

∴ Mass of wire = 9π² × density

= 9(3.14)² × 8.88

= 88.74 × 8.88

= 788 g (approx)

5. The shape of a gilli, in the gilli-danda game (see figure) is a combination of

Let the radii of the two spheres are r₁ and r₂ respectively.

Volume of the sphere of radii,

$$r_1=\frac{r}{3}\pi r_1^{3}\\\therefore\space\text{...(i)}$$

and Volume of the sphere of radii,

$$r_2=\frac{4}{3}\pi r_2^{3}\\\text{...(ii)}$$

Given, ratio of volumes

$$\lambda=64:27\Rarr\frac{\frac{4}{3}\pi r_1^{3}}{\frac{4}{3}\pi r_2^{3}}=\frac{64}{27}\\\Rarr\space\frac{r_1^{3}}{r_2^{3}}=\frac{64}{27}\\\Rarr\space\frac{r_1}{r_1}=\frac{4}{3}\space\text{...(iii)}$$

Now,

ratio of the surface area

$$=\frac{4\pi r_1^{2}}{4\pi r_2^{2}}=\frac{r_1^{3}}{r_2^{3}}=\frac{r_1^{2}}{\bigg(\frac{3r_1}{4}\bigg)^{2}}\\\bigg[\text{From eq. (iii),}r_2=\frac{3}{4}r_1\bigg]\\=\frac{r_1^{2}}{\frac{9r_1^{2}}{16}}=\frac{16}{9}\space\text{or 16:9}$$

4. The volume of the frustum of a cone is$$\frac{\textbf{1}}{\textbf{3}}\pi \textbf{h[r}_\textbf{1}^{\textbf{2}}+\textbf{r}_\textbf{2}^{\textbf{2}}-\textbf{r}_\textbf{1}\textbf{r}_\textbf{2}]$$, where, h is vertical height of the frustum and r₁, r₂ are the radii of the ends.

Sol. False.

Since, the volume of the frustum of a cone is

$$\frac{1}{3}\pi h[r_1^{2}+r_2^{2}-r_1r_2],$$

where h is vertical height of the frustum and r₁, r₂ are the radii of the ends.

6. An open metallic bucket is in the shape of frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet. The surface area of the metallic sheet used is equal to the curved surface area of frustum of a cone + area of circular base + curved surface area of cylinder.

Sol. True.

Because the resulting figure is

7. A solid cone of radius r and height h is placed over a solid cylinder having same base radius and height as that of a cone.

The total surface area of the combined solid is

$$\pi\text{rl}\sqrt{r^{2}+h^{2}}+2r+2h].$$

Sol. False.

We know that total curved surface area of a cone of radius r and height h = curved surface area + area of base = πrl + πr².

$$\text{where, l=}\sqrt{h^{2}+r^{2}}$$

and total curved surface area of a cylinder of base radius r and height h = curved surface area + area of both base = 2πrh + 2πr².

Here, when we placed a cone over a cylinder one base is common for both

So, total curved surface area of the combined solid

= πrl + 2πrh + 2πr²

= πr[l + 2h + 2r]

$$=\pi r[\sqrt{r^{2}+h^{2}}+2h+2r]$$