NCERT Solutions for Class 9 Maths Chapter 14 — Exercise Set 14.2#
Cylinder of radius $r$ and height $h$:
$$\text{CSA} = 2\pi rh, \qquad \text{Volume} = \pi r^2 h$$
1. Two cylinders, A and B, are given. The radius of cylinder B is twice that of cylinder A, and the height of cylinder B is half that of cylinder A. Find the ratio of the curved surface area of A to the curved surface area of B. Also find the ratio of the volume of A to the volume of B. #
Solution
Let A have radius $r$ and height $h$. Then B has radius $2r$ and height $\dfrac{h}{2}$.
$$\dfrac{\text{CSA of A}}{\text{CSA of B}} = \dfrac{2\pi rh}{2\pi (2r)\left(\dfrac{h}{2}\right)} = \dfrac{rh}{rh} = 1$$$$\dfrac{\text{Volume of A}}{\text{Volume of B}} = \dfrac{\pi r^2 h}{\pi (2r)^2 \left(\dfrac{h}{2}\right)} = \dfrac{r^2h}{2r^2h} = \dfrac{1}{2}$$∴ CSA ratio 1 : 1; volume ratio 1 : 2.
2. The radii of two cylinders are in the ratio 2:3, and their heights are in the ratio 3:2. Find (a) the ratio of their volumes, and (b) the ratio of their curved surface areas. #
Solution
Let the radii be $2k$, $3k$ and the heights $3m$, $2m$.
(a)
$$\dfrac{V_1}{V_2} = \dfrac{\pi (2k)^2 (3m)}{\pi (3k)^2 (2m)} = \dfrac{12}{18} = \dfrac{2}{3}$$(b)
$$\dfrac{\text{CSA}_1}{\text{CSA}_2} = \dfrac{2\pi (2k)(3m)}{2\pi (3k)(2m)} = \dfrac{6}{6} = 1$$∴ Volumes 2 : 3; curved surface areas 1 : 1.
3. The edge of a cube measures r cm. The largest possible right circular cylinder is cut out of the cube. What do you think is the volume of the cylinder (in cm³)? #
Solution
The largest cylinder stands on one face: its base circle fits inside a square face of side $r$, so its radius is $\dfrac{r}{2}$, and its height is the full edge $r$:
$$V = \pi \left(\dfrac{r}{2}\right)^2 \times r$$$$V = \dfrac{\pi r^3}{4} \text{ cm}^3$$
(This is about 79% of the cube’s volume.)
*4. The radius of the base of a cylinder is increased by 10%. At the same time, the height of the cylinder is decreased by x%. Given that the volume of the cylinder remains unchanged, find the value of x. #
Solution
The new radius is $1.1r$ and the new height is $\left(1 - \dfrac{x}{100}\right)h$. The volume is unchanged, so:
$$\pi (1.1r)^2 \left(1 - \dfrac{x}{100}\right)h = \pi r^2 h$$$$\Rightarrow 1.21\left(1 - \dfrac{x}{100}\right) = 1$$
$$\Rightarrow 1 - \dfrac{x}{100} = \dfrac{100}{121}$$
$$\Rightarrow \dfrac{x}{100} = \dfrac{21}{121}$$
$$\Rightarrow x = \dfrac{2100}{121} \approx 17.36$$
∴ The height is decreased by about 17.36%.
5. A solid metallic cube of side 12 cm is melted and recast into solid cylindrical rods, each having radius 2 cm and height 12 cm. Find: (i) the volume of the cube, (ii) the volume of one cylindrical rod, (iii) the approximate number of complete cylindrical rods that can be formed. (π ≈ 22/7) #
Solution
(i)
$$V_{\text{cube}} = 12^3 = 1728 \text{ cm}^3$$(ii)
$$V_{\text{rod}} = \dfrac{22}{7} \times 2^2 \times 12$$$$V_{\text{rod}} = \dfrac{1056}{7} \approx 150.86 \text{ cm}^3$$
(iii)
$$\text{Number of rods} = \dfrac{1728}{1056/7}$$$$\text{Number of rods} = \dfrac{12096}{1056} \approx 11.45$$
∴ 11 complete rods can be formed (with some metal left over).