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NCERT Exemplar · Class 9 Mathematics Surface Areas and Volumes

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EXERCISE 13.2 1–10 (part 2 of 4)

  1. Write True or False and justify your answer in each of the following :

    Exercise 1

    The volume of a sphere is equal to two-third of the volume of a cylinder whose height and diameter are equal to the diameter of the sphere.

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    NCERT’s answer
    True, $\displaystyle \frac{4}{3} \pi r^{3}=\frac{2}{3} \pi r^{2}(2 r)$
    True. Let the sphere have radius \(\displaystyle r\); the cylinder then has radius \(\displaystyle r\) and height \(\displaystyle 2r\). \[V_{\text{cyl}} = \pi r^2(2r) = 2\pi r^3 \] \[\frac{2}{3}V_{\text{cyl}} = \frac{2}{3}(2\pi r^3) = \frac{4}{3}\pi r^3 = V_{\text{sphere}} \]

  2. Exercise 2

    If the radius of a right circular cone is halved and height is doubled, the volume will remain unchanged.

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    NCERT’s answer
    False, since new volume $\displaystyle =\frac{1}{3} \pi\left(\frac{r}{2}\right)^{2} \cdot 2 h=\frac{1}{2}$ (Original volume)
    False. Radius \(\displaystyle r\) and height \(\displaystyle h\) become \(\displaystyle \frac{r}{2}\) and \(\displaystyle 2h\). \[V' = \frac{1}{3}\pi\left(\frac{r}{2}\right)^2(2h) = \frac{1}{3}\pi r^2 h \cdot \frac{1}{2} = \frac{V}{2} \] The volume is halved, not unchanged.

  3. Exercise 3

    In a right circular cone, height, radius and slant height do not always be sides of a right triangle.

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    NCERT’s printed answer for this exercise does not match its own question. This working follows the question as printed.

    False. For a right circular cone, height \(\displaystyle h\), radius \(\displaystyle r\) and slant height \(\displaystyle l\) satisfy \[l^2 = r^2 + h^2 \quad \text{(Pythagoras, on the triangle formed by the axis, the radius and the slant height)} \] so \(\displaystyle h\), \(\displaystyle r\) and \(\displaystyle l\) always are the sides of a right triangle.NCERT prints: True, since r² + h² = l² — but that relation means h, r and l always form a right triangle, the opposite of the stem's own claim.

  4. Exercise 4

    If the radius of a cylinder is doubled and its curved surface area is not changed, the height must be halved.

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    NCERT’s answer
    True, $\displaystyle 2 \pi r h=2 \pi(2 r) \cdot \frac{h}{2}$
    True. Curved surface area \(\displaystyle S = 2\pi r h\). With radius \(\displaystyle 2r\) and height \(\displaystyle h'\), \[2\pi(2r)h' = 2\pi r h \] \[h' = \frac{h}{2} \]

  5. Exercise 5

    The volume of the largest right circular cone that can be fitted in a cube whose edge is 2r\displaystyle 2 r equals to the volume of a hemisphere of radius r\displaystyle r.

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    NCERT’s answer
    True, since volume of cone $\displaystyle =\frac{1}{3} \pi r^{2} \cdot(2 r)=\frac{2}{3} \pi r^{3}=$ volume of hemisphere
    True. The largest cone that fits in a cube of edge \(\displaystyle 2r\) has base radius \(\displaystyle r\) and height \(\displaystyle 2r\). \[V_{\text{cone}} = \frac{1}{3}\pi r^2(2r) = \frac{2}{3}\pi r^3 \] \[V_{\text{hemisphere}} = \frac{2}{3}\pi r^3 \] The two volumes are equal.

  6. Exercise 6

    A cylinder and a right circular cone are having the same base and same height. The volume of the cylinder is three times the volume of the cone.

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    NCERT’s answer
    True, since $\displaystyle \mathrm{V}_{1}=$ volume of cylinder $\displaystyle =\pi r^{2} h$ since $\displaystyle \mathrm{V}_{2}=$ volume of cone $\displaystyle =\frac{1}{3} \pi r^{2} h$ Therefore, $\displaystyle \mathrm{V}_{1}=3 \mathrm{~V}_{2}$
    True. Same base radius \(\displaystyle r\) and height \(\displaystyle h\) give \[V_{\text{cyl}} = \pi r^2 h, \qquad V_{\text{cone}} = \frac{1}{3}\pi r^2 h \] \[V_{\text{cyl}} = 3\,V_{\text{cone}} \]

  7. Exercise 7

    A cone, a hemisphere and a cylinder stand on equal bases and have the same height. The ratio of their volumes is 1\displaystyle 1 : 2\displaystyle 2 : 3.

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    NCERT’s answer
    True, $\displaystyle \mathrm{V}_{1}=\frac{1}{3} \pi r^{2} r, \mathrm{~V}_{2}=\frac{2}{3} \pi r^{3}, \mathrm{~V}_{3}=\pi r^{2} r$
    True. Common radius \(\displaystyle r\); the shared height equals \(\displaystyle r\), since a hemisphere's height is its radius. \[V_{\text{cone}} = \frac{1}{3}\pi r^3, \qquad V_{\text{hemi}} = \frac{2}{3}\pi r^3, \qquad V_{\text{cyl}} = \pi r^3 \] \[V_{\text{cone}} : V_{\text{hemi}} : V_{\text{cyl}} = 1 : 2 : 3 \]

  8. Exercise 8

    If the length of the diagonal of a cube is 63 cm\displaystyle 6 \sqrt{3} \mathrm{~cm}, then the length of the edge of the cube is 3\displaystyle 3 cm.

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    NCERT’s answer
    False, $\displaystyle \sqrt{3} a=6 \sqrt{3}=a=6$ Therefore, edge = $\displaystyle 6$ cm
    False. The diagonal of a cube of edge \(\displaystyle a\) is \(\displaystyle a\sqrt{3}\). \[a\sqrt{3} = 6\sqrt{3} \] \[a = 6 \text{ cm} \] The edge is \(\displaystyle 6\) cm, not \(\displaystyle 3\) cm.

  9. Exercise 9

    If a sphere is inscribed in a cube, then the ratio of the volume of the cube to the volume of the sphere will be 6:π\displaystyle 6: \pi.

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    NCERT’s answer
    True, $\displaystyle \mathrm{V}_{1}$ (volume of cube) $\displaystyle =a^{3}$ Radius of sphere $\displaystyle =\frac{a}{2} . \mathrm{V}_{2}($ Volume of sphere $\displaystyle )=\frac{4}{3} \pi \frac{a^{3}}{8}$ $\displaystyle \mathrm{V}_{1}: \mathrm{V}_{2}=6: \pi$
    True. The sphere touches every face of the cube, so its diameter equals the edge.NCERT_Solution_Class9_Maths_Exemplar_Ch13_Ex13-2_Q9\[a = 2r \] \[V_{\text{cube}} = a^{3} = (2r)^{3} = 8r^{3} \] \[V_{\text{sphere}} = \frac{4}{3}\pi r^{3} \] \[\frac{V_{\text{cube}}}{V_{\text{sphere}}} = \frac{8r^{3}}{\dfrac{4}{3}\pi r^{3}} = \frac{6}{\pi} \]

  10. Exercise 10

    If the radius of a cylinder is doubled and height is halved, the volume will be doubled.

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    NCERT’s answer
    True, new volume $\displaystyle =\pi(2 r)^{2} \cdot\left(\frac{h}{2}\right)=2\left[\pi r^{2} h\right]$. Therefore, volume is doubled.
    True. Volume scales as \(\displaystyle r^{2}h \); doubling the radius and halving the height multiplies it by exactly \(\displaystyle 2 \).\[V = \pi r^{2} h \] \[V' = \pi (2r)^{2}\left(\dfrac{h}{2}\right) = \pi \cdot 4r^{2} \cdot \dfrac{h}{2} = 2\pi r^{2} h \] \[V' = 2V \]