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MATHS M.C.Q

Surface Area And Volume of Sphere · Rajasthan Board (RBSE) STD 9 (Rajasthan - English Medium). 15 questions.

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Question 11 Mark
If a solid sphere of radius 10cm is moulded into 8 spherical solid balls of equal radius, then the surface area of each ball (in sq.cm) is:
  1. $100\pi$
  2. $75\pi$
  3. $60\pi$
  4. $50\pi$
Answer
  1. $100\pi$
Solution:
Volume of solid sphere $=\frac{4}{3}\pi(10)^3=\frac{4000\pi}{3}\text{cm}^3$
Vomule 8 solid sphere of radius (say) $\text{r}=8\times\frac{4}{3}\pi\text{r}^3=\frac{32\pi\text{r}^3}{3}\text{cm}^3$
Now, $\frac{32\pi\text{r}^3}{3}=\frac{4000\pi}{3}$
$\Rightarrow\text{r}=\Big(\frac{1000}{8}\Big)^\frac{1}{3}=\frac{10}{2}=5\text{cm}$
Surface Area of each small ball $=4\pi\text{r}^2=4\pi(5)^2=100\pi\text{ cm}^2 $
Hence, correct option is (a).
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Question 21 Mark
A cone, a hemisphere and a cylinder stand on equal bases and have the same height.
The ratio of their volumes is:
  1. 1 : 2 : 3
  2. 2 : 1 : 3
  3. 2 : 3 : 1
  4. 3 : 2 : 1
Answer
  1. 1 : 2 : 3
Solution:

If all of these have equal bases, then their radii are equal.
Their heights are same. (given)
$\text{r}=\text{h}_1=\text{h}_2$
$\text{V}_\text{cone}=\frac{1}{3}\pi\text{r}^2\text{h}_1=\frac{1}{3}\pi\text{r}^2(\text{r})=\frac{1}{3}\pi\text{r}^3$
$\text{V}_\text{hemisphere}=\frac{2}{3}\pi\text{r}^3$
$\text{V}_\text{cylinder}=\pi\text{r}^2\text{h}_2=\pi\text{r}^2(\text{r})=\pi\text{r}^3$
$\text{V}_\text{cone}:\text{V}_\text{hemisphere}:\text{V}_\text{cylinder}=\frac{1}{2}:\frac{2}{3}:1=1:2:3$
Hence, correct option is (a).
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Question 31 Mark
The largest sphere is cut off from a cube of side 6cm. The volume of the sphere will be:
  1. $27\pi\ \text{cm}^2$
  2. $36\pi\ \text{cm}^3$
  3. $108\pi\ \text{cm}^3$
  4. $12\pi\ \text{cm}^3$
Answer
  1. $36\pi\ \text{cm}^2$
Solution:
The largest sphere that can be cut from a cube of side 6cm will have its diameter = side of cube.
i. e. 2r = 6cm ⇒ r = 3cm
Volume of that sphere $=\frac{4}{3}\pi\text{r}^3=\frac{4}{3}\pi\times3\times3\times3=36\pi\text{ cm}^3$
Hence, correct option is (b)
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Question 41 Mark
A sphere is placed inside a right circular cylinder so as to touch the top, base and lateral surface of the cylinder. If the radius of the sphere is r, then the volume of the cylinder is
  1. $4\pi\text{r}^3$
  2. $\frac{8}{3}\pi\text{r}^3$
  3. $2\pi\text{r}^3$
  4. $8\pi\text{r}^3$
Answer
  1. $2\pi\text{r}^3$
Solution:

Radius of sphere = r
Sphere touches cylinder at Top, Base and Lateral Surface.
Then,
2r = height of cylinder = h
r = Radius of cylinder
Volume of cylinder $=\pi\text{r}^2\text{h}$
$=\pi\text{r}^2(2\text{r})$
$=2\pi\text{r}^3$
Hence, correct option is (c).
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Question 51 Mark
The total surface area of a hemisphere of radius r is:
  1. $\pi\text{r}^2$
  2. $2\pi\text{r}^2$
  3. $3\pi\text{r}^2$
  4. $4\pi\text{r}^2$
Answer
  1. $3\pi\text{r}^2$
Solution:
A hemisphere has two surfaces: one top surface and other curved surface.
T.S.A $=2\pi\text{r}^2+(\pi\text{r}^2)$ $\{$Area of Top-face = $\pi\text{r}^2\}$
$=3\pi\text{r}^2$
Hence, correct option is (c).
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Question 61 Mark
If the ratio of volumes of two spheres is 1 : 8, then the ratio of their surface areas is
  1. 1 : 2
  2. 1 : 4
  3. 1 : 8
  4. 1 : 16
Answer
  1. 1 : 4
Solution:
Volume of sphere $=\frac{4}{3}\pi\text{r}^3=\text{v}$
$\frac{\text{V}_1}{\text{V}_1}=\frac{\frac{4}{3}\pi\text{r}^3}{\frac{4}{3}\pi\text{r}^3_2}=\frac{\text{r}^3_1}{\text{r}^3_2}=\frac{1}{8}$
$\Rightarrow\frac{\text{r}_1}{\text{r}_2}=\frac{1}{2}$
now, Surface Area of Sphere $=4\pi\text{r}^2=\text{S}$
$\frac{\text{S}_1}{\text{S}_2}=\frac{4\pi\text{r}^2_1}{4\pi\text{r}^2_2}=\Big(\frac{\text{r}_1}{\text{r}_2}\Big)^2=\frac{1}{4}=1:4$
Hence, correct option is (b).
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Question 71 Mark
In a sphere the number of faces is:
  1. 1
  2. 2
  3. 3
  4. 4
Answer
  1. 1
Solution:
Sphere has only one surface i.e. curved surface, so number of faces = 1
Hence, correct option is (a).
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Question 81 Mark
The ratio between the volume of a sphere and volume of a circumscribing right circular cylinder is:
  1. 2 : 1
  2. 1 : 1
  3. 2 : 3
  4. 1 : 2
Answer
  1. 2 : 3
Solution:
Volume of sphere of radius radius r $=\frac{4}{3}\pi\text{r}^3=\text{v}_1\ ...(1)$
If a cylinder is circumscibibing the sphre, then 
diameter of cylinder = diameter of sphere
height of cylinder = Radius of sphere
Height of cylinder = 2r
Volume of cylinder = $\text{V}_2=\pi\text{r}^2\text{h}$
$=\pi\text{r}^2(2\text{r})$
$\Rightarrow \text{V}_2=2\pi\text{r}^3\ ....(2)$
dividing equation (1) and (2)
$\frac{\text{V}_1}{\text{V}_2}=\frac{\frac{4}{3}\pi\text{r}^3}{2\pi\text{r}^3}$
$\Rightarrow\frac{\text{V}_1}{\text{V}_2}=\frac{2}{3}$
hence, correct option is (c).
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MCQ 91 Mark
If a sphere is inscribed in a cube, then the ratio of the volume of the sphere to the volume of the cube is:
  • A
    $\pi:2$
  • B
    $\pi:3$
  • C
    $\pi:4$
  • ✓
    $\pi:6$
Answer
Correct option: D.
$\pi:6$

Edge of cube $= a$
$\Rightarrow$ Volume of cube $= a^3$
If Sphere is inscribed inside cube then $a =2\text{r}$
$\Rightarrow\text{r}=\frac{\text{a}}{2}$
Volume of sphere $=\frac{4}{3}\pi\text{r}^3=\frac{4}{3}\pi\Big(\frac{\text{a}}{2}\Big)^3=\frac{\pi}{6}\text{a}^3$
Ratio of volume of sphere to volume of cube $=\frac{\frac{\pi}{6}\text{a}^3}{\text{a}^3}=\frac{\pi}{6}$
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Question 101 Mark
A cylindrical rod whose height is 8 times of its radius is melted and recast into spherical balls of same radius. The number of balls will be
  1. 4
  2. 3
  3. 6
  4. 8
Answer
  1. 6
Solution:
Volume of cylindrical rod $=\pi\text{r}^\text{h}$
$=\pi\text{r}^2(8\text{r})$ [h = 8r (given)]
$=8\pi\text{r}^3$
Now, if spherical balls have same radius, then the volume of one ball $=\frac{4}{3}\pi\text{r}^3$
$\therefore$ No. of balls $=\frac{\text{Volume of Cylindrical Rod}}{\text{Volume of one Rod}}=\frac{8\pi\text{r}^3}{\frac{4}{3}\pi\text{r}^3}=6$
Hence, correct option is (c).
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MCQ 111 Mark
If the surface area of a sphere is $144\pi\text{ m}^2,$ then its volume $($in $m^3)$ is:
  • ✓
    $288\pi$
  • B
    $316\pi$
  • C
    $300\pi$
  • D
    $188\pi$
Answer
Correct option: A.
$288\pi$
Surface Area of Sphere
$\Rightarrow 4\pi\text{r}^2=144\pi$
$\Rightarrow\text{r}^2=36$
$\Rightarrow\text{r}=6$
Volume of Sphere $=\frac{4}{3}\pi\text{r}^3$
$=\frac{4}{3}\pi(6)^3$
$=288\pi\text{ m}^3$
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Question 121 Mark
A cone and a hemisphere have equal bases and equal volume the ratio of their heights is:
  1. $1:2$
  2. $2:1$
  3. $4:1$
  4. $\sqrt{2}:1$
Answer
  1. 2  : 1
Solution:
In the given problem, we are given a cone and a hemisphere which have equal bases equal volumes. We need to find the ratio of their heights.
So,
Let the radius of the cone and hemisphere be x cm.
Also, height of the hemisphere is equal to the radius of the hemisphere.
Now, let the height of the cone = hcm
So, the ratio of the height of cone to the height of the hemisphere $=\frac{\text{h}}{\text{x}}$
Here Volume of the hemisphere = volume of the cone
$\Big(\frac{2}{3}\Big)\pi\text{r}^3_\text{h}=\Big(\frac{1}{3}\Big)\pi\text{r}^2_\text{c}\text{h}$
$\Big(\frac{2}{3}\Big)\pi(\text{x})^3=\Big(\frac{1}{3}\Big)\pi(\text{x})^2\text{h}$
$\Big(\frac{2}{3}\Big)(\text{x})=\Big(\frac{1}{3}\Big)\text{h}$
$2\text{x}=\text{l}\text{h}$
$\frac{\text{h}}{\text{x}}=\frac{2}{1}$
Therefore, the ratio of the heights of the cone and the hemisphere is 2 : 1. 
So, the correct option is (b).
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Question 131 Mark
If a solid sphere of radius r is melted and cast into the shape of a solid cone of height r, then the radius of the base of the cone is
  1. 2r
  2. 3r
  3. r
  4. 4r
Answer
  1. 2r
Solution:
Volume of sphere $=\frac{4}{3}\pi\text{r}^3$
sphere costed into a cone of height r.
Let the radius of cone = R
$\therefore$ Volume of cone $=\frac{1}{3}\pi\text{R}^2(\text{r})$
Volume of cone = volume of sphere
$\Rightarrow\frac{1}{3}\pi\text{R}^2\text{r}=\frac{4}{3}\pi\text{r}^3$
$\Rightarrow\text{R}^2=4\text{r}^2$
$\Rightarrow\text{R}=2\text{r}$
Hence, correct option is (a).
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Question 141 Mark
The ratio of the total surface area of a sphere and a hemisphere of same radius is:
  1. 2 :1
  2. 3 : 2
  3. 4 : 1
  4. 4 : 3
Answer
  1. 4 : 3
Solution:
Total surface area of sphere $=4\pi\text{r}^2$
Total surface area of hemisphere $3\pi\text{r}^2$
$\therefore$ Required ratio $=\frac{4\pi\text{r}^2}{3\pi\text{r}^2}=\frac{4}{3}=4:3$
Hence, correct option is (d).
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Question 151 Mark
A sphere and a cube are of the same height. The ratio of their volumes is
  1. 3 : 4
  2. 21 : 11
  3. 4 : 3
  4. 11 : 21
Answer
  1. 11 : 21
 
Solution:
Height of sphere = diameter = 2r
Height of cube = Side of cube = Height of sphere = 2r
Volume of sphere $=\frac{4}{3}\pi\text{r}^3$
Volume of cube $(2\text{r})^3=8\text{r}^3$
Ratio of their volumes $=\frac{\frac{4}{3}\pi\text{r}^3}{8\text{r}^3}=\frac{\pi}{6}=\frac{22^{11}}{7\times6_3}=\frac{11}{21}=11:21$
Hence, correct option is (d).
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