SolveIt is under development
SolveItClass 9 · NCERT

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

52 questions · 52 still being checked

Exercise Set 14.4 1–8 (part 4 of 6)

  1. Exercise 1

    A ball bearing has a radius of 0.7\displaystyle 0.7 cm . Find its surface area.

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    Surface area of a sphere is \(\displaystyle 4\pi r^2\). Take \(\displaystyle \pi = \tfrac{22}{7}\).\[A = 4\pi r^2 \] \[A = 4 \times \tfrac{22}{7} \times 0.7 \times 0.7 \] \[A = 6.16 \ \text{cm}^2 \]Answer: \(\displaystyle 6.16\ \text{cm}^2\).

  2. Exercise 2

    Two solid spheres made of the same metal have weights 5920\displaystyle 5920 g and 740\displaystyle 740 g . Determine the radius of the larger sphere, if the diameter of the smaller one is 5\displaystyle 5 cm.

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    Same metal, so equal density: weight is proportional to volume. Let \(\displaystyle R\) and \(\displaystyle r\) be the radii of the larger and smaller spheres.\[\frac{V_R}{V_r} = \frac{5920}{740} = 8 \] \[\frac{\tfrac{4}{3}\pi R^3}{\tfrac{4}{3}\pi r^3} = \left(\frac{R}{r}\right)^3 = 8 \ \Rightarrow\ \frac{R}{r} = 2 \] \[r = \frac{5}{2} = 2.5 \ \text{cm} \] \[R = 2 \times 2.5 = 5 \ \text{cm} \]Answer: radius of the larger sphere \(\displaystyle = 5\) cm.

  3. Exercise 3

    The diameter of the moon is approximately one fourth the diameter of the earth. Given that the moon and earth are both roughly spherical, find the ratio of their surface areas.

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    Diameter of the moon is \(\displaystyle \tfrac14\) that of the earth, so its radius is \(\displaystyle \tfrac14\) of the earth's. Let the earth's radius be \(\displaystyle R\).\[\frac{S_{\text{moon}}}{S_{\text{earth}}} = \frac{4\pi \left(\tfrac{R}{4}\right)^2}{4\pi R^2} = \frac{1}{16} \]Answer: \(\displaystyle 1 : 16\).

  4. Exercise 4

    Find
    (i)
    the curved surface area and
    (ii)
    the total surface area of a hemisphere of radius 21\displaystyle 21 cm.

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    Take \(\displaystyle \pi = \tfrac{22}{7}\), \(\displaystyle r = 21\) cm.(i) Curved surface area:\[2\pi r^2 = 2 \times \tfrac{22}{7} \times 21 \times 21 = 2772 \ \text{cm}^2 \](ii) Total surface area (curved surface plus the circular base):\[3\pi r^2 = 3 \times \tfrac{22}{7} \times 21 \times 21 = 4158 \ \text{cm}^2 \]Answer: (i) \(\displaystyle 2772\ \text{cm}^2\); (ii) \(\displaystyle 4158\ \text{cm}^2\).

  5. Exercise 5

    Metal spheres, each of radius 2\displaystyle 2 cm, are packed into a rectangular box of internal dimensions 16\displaystyle 16 cm × 8\displaystyle 8 cm × 8\displaystyle 8 cm. When 16\displaystyle 16 spheres are packed, the box is filled with preservative liquid. Find the volume of this liquid. Round your answer to the nearest integer.

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    The liquid fills the space the spheres leave in the box. Take \(\displaystyle \pi = \tfrac{22}{7}\).\[V_{\text{box}} = 16 \times 8 \times 8 = 1024 \ \text{cm}^3 \] \[V_{\text{16 spheres}} = 16 \times \tfrac{4}{3}\pi (2)^3 = \tfrac{512}{3} \times \tfrac{22}{7} \approx 536.4 \ \text{cm}^3 \] \[V_{\text{liquid}} = 1024 - 536.4 \approx 487.6 \ \text{cm}^3 \]Answer: \(\displaystyle \approx 488\ \text{cm}^3\).

  6. Exercise 6

    The radius of a sphere is increased by 10\displaystyle 10%. Show that the volume increases by approximately 33.1\displaystyle 33.1%.

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    Let the radius be \(\displaystyle r\), so \(\displaystyle V = \tfrac43 \pi r^3\). The new radius is \(\displaystyle 1.1\,r\).\[V' = \tfrac{4}{3}\pi (1.1\,r)^3 = 1.331 \times \tfrac{4}{3}\pi r^3 = 1.331\,V \] \[\frac{V' - V}{V} \times 100 = (1.331 - 1) \times 100 = 33.1 \]Answer: the volume increases by \(\displaystyle 33.1\%\).

  7. Exercise 7

    The radius of a sphere is increased by x%\displaystyle x \%. The volume of the sphere increases by 72.8\displaystyle 72.8%. Find the value of x\displaystyle x.

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    The new radius is \(\displaystyle \left(1 + \tfrac{x}{100}\right) r\), and the new volume is \(\displaystyle 1.728\,V\).\[\frac{V'}{V} = \left(1 + \frac{x}{100}\right)^3 = 1 + \frac{72.8}{100} = 1.728 \] \[1.728 = (1.2)^3 \ \Rightarrow\ 1 + \frac{x}{100} = 1.2 \] \[x = 20 \]Answer: \(\displaystyle x = 20\).

  8. Exercise 8

    NCERT_Question_Class9_Maths_Ch14_Ex14-4_Q8 The hemispherical dome of a building needs to be painted (see Fig. 14.16\displaystyle 14.16). If the circumference of the base of the dome is 35.2\displaystyle 35.2 m, find the cost of painting it, given the cost of painting is ₹10\displaystyle 10 per 100 cm2\displaystyle 100 \mathrm{~cm}^2.

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    The painted dome in Fig. $\displaystyle 14.16$ is the curved surface of a hemisphere. Take \(\displaystyle \pi = \tfrac{22}{7}\).\[2\pi r = 35.2 \ \Rightarrow\ r = \frac{35.2 \times 7}{2 \times 22} = 5.6 \ \text{m} \] \[\text{CSA} = 2\pi r^2 = 2 \times \tfrac{22}{7} \times 5.6 \times 5.6 = 197.12 \ \text{m}^2 \]Rate: \(\displaystyle \text{₹}\,10\) per \(\displaystyle 100\ \text{cm}^2\) means \(\displaystyle \text{₹}\,10\) per \(\displaystyle 0.01\ \text{m}^2\), i.e. \(\displaystyle \text{₹}\,1000\) per \(\displaystyle \text{m}^2\).\[\text{Cost} = 197.12 \times 1000 = \text{₹}\,1{,}97{,}120 \]Answer: \(\displaystyle \text{₹}\,1{,}97{,}120\).