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NCERT Solutions · Class 9 Mathematics Math of Space: Surface Area and Volume

52 questions · 52 still being checked

Exercise Set 14.2 1–5 (part 2 of 6)

  1. Exercise 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.

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    Let cylinder A have radius \(\displaystyle r\) and height \(\displaystyle h\). Then B has radius \(\displaystyle 2r\) and height \(\displaystyle \tfrac{h}{2}\).\[\text{CSA}_A = 2\pi r h \] \[\text{CSA}_B = 2\pi (2r)\left(\tfrac{h}{2}\right) = 2\pi r h \] \[\text{CSA}_A : \text{CSA}_B = 1 : 1 \]\[V_A = \pi r^2 h \] \[V_B = \pi (2r)^2 \left(\tfrac{h}{2}\right) = 2\pi r^2 h \] \[V_A : V_B = 1 : 2 \]Answer: curved surface areas \(\displaystyle 1:1\); volumes \(\displaystyle 1:2\).

  2. Exercise 2

    The radii of two cylinders are in the ratio 2\displaystyle 2:3\displaystyle 3, and their heights are in the ratio 3\displaystyle 3:2. Find
    (a)
    the ratio of their volumes, and
    (b)
    the ratio of their curved surface areas.

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    Take radii \(\displaystyle 2k,\ 3k\) and heights \(\displaystyle 3m,\ 2m\).(a) Volume ratio:\[\frac{V_1}{V_2} = \frac{\pi (2k)^2 (3m)}{\pi (3k)^2 (2m)} = \frac{12}{18} = \frac{2}{3} \](b) Curved surface area ratio:\[\frac{S_1}{S_2} = \frac{2\pi (2k)(3m)}{2\pi (3k)(2m)} = \frac{12}{12} = 1 \]Answer: (a) \(\displaystyle 2:3\); (b) \(\displaystyle 1:1\).

  3. Exercise 3

    The edge of a cube measures r cm\displaystyle r \mathrm{~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\displaystyle \mathrm{cm}^3 )?

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    The cylinder must fit inside the cube. Its height is at most \(\displaystyle r\), and its base circle must fit inside a square face of side \(\displaystyle r\). Both are largest when the base circle touches all four sides of the face.NCERT_Solution_Class9_Maths_Ch14_Ex14-2_Q3\[h = r, \qquad 2R = r \ \Rightarrow\ R = \tfrac{r}{2} \] \[V = \pi R^2 h = \pi \left(\tfrac{r}{2}\right)^2 r = \tfrac{\pi r^3}{4} \]Answer: \(\displaystyle \dfrac{\pi r^3}{4}\ \text{cm}^3\).

  4. Exercise 4

    The radius of the base of a cylinder is increased by 10\displaystyle 10%. At the same time, the height of the cylinder is decreased by x%\displaystyle x \%. Given that the volume of the cylinder remains unchanged, find the value of x\displaystyle x.

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    Let the original radius be \(\displaystyle r\) and height \(\displaystyle h\). The new radius is \(\displaystyle 1.1\,r\) and the new height is \(\displaystyle h\left(1 - \tfrac{x}{100}\right)\). The volume is unchanged, so:\[\pi (1.1r)^2 \, h\left(1 - \tfrac{x}{100}\right) = \pi r^2 h \] \[1.21\left(1 - \tfrac{x}{100}\right) = 1 \] \[1 - \tfrac{x}{100} = \tfrac{100}{121} \] \[\tfrac{x}{100} = \tfrac{21}{121} \] \[x = \tfrac{2100}{121} \approx 17.36 \]Answer: \(\displaystyle x = \tfrac{2100}{121} \approx 17.36\).

  5. Exercise 5

    A solid metallic cube of side 12\displaystyle 12 cm is melted and recast into solid cylindrical rods, each having radius 2\displaystyle 2 cm and height 12\displaystyle 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≈227)\displaystyle \left(\pi \approx \frac{22}{7}\right)

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    (i)\[V_{\text{cube}} = 12^3 = 1728\ \text{cm}^3 \](ii)\[V_{\text{rod}} = \pi r^2 h = \tfrac{22}{7} \times 2^2 \times 12 = \tfrac{1056}{7} \approx 150.86\ \text{cm}^3 \](iii) Recasting keeps the volume the same, so the number of rods is:\[n = \frac{1728}{1056/7} = \frac{12096}{1056} \approx 11.45 \]Only complete rods count, so \(\displaystyle n = 11\).Answer: (i) \(\displaystyle 1728\ \text{cm}^3\); (ii) \(\displaystyle \tfrac{1056}{7} \approx 150.86\ \text{cm}^3\); (iii) \(\displaystyle 11\) rods.