RS AGGARWAL CLASS 9 CHAPTER 15 VOLUMES AND SURFACE AREA OF SOLIDS MCQ

 MULTIPLE CHOICE QUESTIONS

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Question 1:

The length, breadth and height of a cuboid are 15 cm, 12 cm and 4.5 cm respectively. Its volume is
(a) 243 cm³
(b) 405 cm³
(c) 810 cm³
(d) 603 cm³

Answer 1:

(c) 810 cm³

 $\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{cuboid}=l\times b\times h=15\times 12\times 4.5{\mathrm{cm}}^3=810{\mathrm{cm}}^3$$$\begin{gathered} \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{cuboid}=l\times b\times h \\ = 15\times 12\times 4.5 {\mathrm{cm}}^3 \\ =810 {\mathrm{cm}}^3 \end{gathered}$$

Question 2:

A cuboid is 12 cm long, 9 cm broad and 8 cm high. Its total surface area is
(a) 864 cm²
(b) 552 cm²
(c) 432 cm²
(d) 276 cm²

Answer 2:

(b) 552 cm²

$\mathrm{Total}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{cuboid}=2\left(lb+bh+lh\right){\mathrm{cm}}^2=2(12\times 9+8\times 9+12\times 8){\mathrm{cm}}^2=2\left(108+72+96\right){\mathrm{cm}}^2=552{\mathrm{cm}}^2$$$\begin{gathered} \mathrm{Total} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cuboid} =2\left(lb+bh+lh\right) {\mathrm{cm}}^2 \\ =2(12\times 9+8\times 9+12\times 8) {\mathrm{cm}}^2 \\ =2\left(108+72+96\right) {\mathrm{cm}}^2 \\ =552 {\mathrm{cm}}^2 \end{gathered}$$

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Question 3:

The length, breadth and height of a cuboid are 15 m, 6 m and 5 dm respectively. The lateral surface area of the cuboid is
(a) 45 m²
(b) 21 m²
(c) 201 m²
(d) 90 m²

Answer 3:

Length of the cuboid, l = 15 m

Breadth of the cuboid, b = 6 m

Height of the cuboid, h = 5 dm = 0.5 m           (1 m = 10 dm)

∴ Lateral surface area of the cuboid = 2h(l + b) = 2 × 0.5 × (15 + 6) = 2 × 0.5 × 21 = 21 cm²

Hence, the correct answer is option (b).

Question 4:

A beam 9 m long, 40 cm wide and 20 cm high is made up of iron which weighs 50 kg per cubic metre. The weight of the beam is
(a) 27 kg
(b) 48 kg
(c) 36 kg
(d) 56 kg

Answer 4:

(c) 36 kg
Length = 9 m
Breadth = 40 cm = 0.4 m
Height = 20 cm = 0.2 m
∴ Weight of the beam = volume of the beam $\times$$\times$weight of iron per cubic metre
                              $=9\times 0.4\times 0.2\times 50=36\mathrm{kg}$$$\begin{gathered} =9\times 0.4\times 0.2\times 50 \\ =36 \mathrm{kg} \end{gathered}$$

Question 5:

The length of the longest rod that can be placed in a room of dimensions (10 m × 10 m × 5 m) is
(a) 15 m
(b) 16 m
(c) $10\sqrt{5}m$$10\sqrt{5} \mathrm{m}$
(d) 12 m

Answer 5:

(a) 15 m

Length of longest rod = diagonal of the room
                                 = diagonal of a cuboid
                                 $=\sqrt{l^2+b^2+h^2}=\sqrt{100+100+25}m=\sqrt{225}m=15m$$$\begin{gathered} =\sqrt{l^2+b^2+h^2} \\ =\sqrt{100+100+25} \mathrm{m} \\ =\sqrt{225} \mathrm{m} \\ = 15 \mathrm{m} \end{gathered}$$

Question 6:

What is the maximum length of a pencil that can be placed in a rectangular box of dimensions (8 cm × 6 cm × 5 cm)?
(a) 8 cm
(b) 9.5 cm
(c) 19 cm
(d) 11.2 cm

Answer 6:

(d) 11.2 cm
Maximum length of the pencil =  diagonal of the box
                                             $=\sqrt{l^2+b^2+h^2}=\sqrt{8^2+6^2+5^2}\mathrm{cm}=\sqrt{64+36+25}\mathrm{cm}=\sqrt{125}\mathrm{cm}=11.2\mathrm{cm}$$$\begin{gathered} =\sqrt{l^2+b^2+h^2} \\ =\sqrt{8^2+6^2+5^2 }\mathrm{cm} \\ =\sqrt{64+36+25} \mathrm{cm} \\ =\sqrt{125} \mathrm{cm} \\ =11.2 \mathrm{cm} \end{gathered}$$

Question 7:

The number of planks of dimensions (4 m × 5 m × 2 m) that can be stored in a pit which is 40 m long, 12 m wide and 16 m deep, is
(a) 190
(b) 192
(c) 184
(d) 180

Answer 7:

(b)  192

Number of planks =  $\frac{\mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{pit}}{\mathrm{volume} \mathrm{of} 1 \mathrm{plank}}$
                             $$\begin{gathered} =\frac{40\times 12\times 16}{4\times 5\times 2} \\ =\frac{7680}{40} \\ =192 \end{gathered}$$
        

Question 8:

How many planks of dimensions (5 m × 25 cm × 10 cm) can be stored in a pit which is 20 m long, 6 m wide and 50 cm deep?
(a) 480
(b) 450
(c) 320
(d) 360

Answer 8:

(a)  480

Length of the pit = 20 m
Breadth of the pit = 6 m
Height of the pit = 50 cm = 0.5 m

Length of the plank = 5m
Breadth of the plank = 25 cm = 0.25 m
Height of the plank = 10 cm = 0.1 m

∴ Number of planks = $\frac{\mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{pit}}{\mathrm{volume} \mathrm{of} 1 \mathrm{plank}}$

                                  $$\begin{gathered} =\frac{20\times 6\times 0.5}{5\times 0.25\times 0.1} \\ =\frac{60\times 1000}{125} \\ =480 \end{gathered}$$

Question 9:

How many bricks will be required to construct a wall 8 m long, 6 m high and 22.5 cm thick if each brick measures (25 cm × 11.25 cm × 6 cm)?
(a) 4800
(b) 5600
(c) 6400
(d) 5200

Answer 9:

(c)  6400

Length of the wall = 8 m = 800 cm
Breadth of the wall = 6 m = 600 cm
Height of the wall = 22.5 cm

Length of the brick = 25 cm
Breadth of the brick = 11.25 cm
Height of the brick = 6 cm

∴ Number of bricks required = $\frac{\mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{wall}}{\mathrm{volume} \mathrm{of} 1 \mathrm{brick}}$

                                             $$\begin{gathered} =\frac{800\times 600\times 22.5}{25\times 11.25\times 6} \\ =\frac{10800000}{1687.5} \\ =6400 \end{gathered}$$

Question 10:

How many persons can be accommodated in a dining hall of dimensions (20 m × 15 m × 4.5 m), assuming that each person requires 5 m³ of air?
(a) 250
(b) 270
(c) 320
(d) 300

Answer 10:

(b) 270

Number of persons  = $\frac{\mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{hall}}{\mathrm{volume} \mathrm{of} \mathrm{air} \mathrm{required} \mathrm{by} 1 \mathrm{person}}$

                               $$\begin{gathered} =\frac{20\times 15\times 4.5}{5} \\ =20\times 3\times 4.5 \\ =270 \end{gathered}$$

∴ 270 persons can be accommodated.

Question 11:

A river 1.5 m deep and 30 m wide is flowing at the rate of 3 km per hour. The volume of water that runs into the sea per minute is
(a) 2000 m³
(b) 2250 m³
(c) 2500 m³
(d) 2750 m³

Answer 11:

(b) 2250 m³

Length of the river = 1.5 m
Breadth of the river = 30 m
Depth of the river = 3 km = 3000 m

  $$\begin{gathered} \mathrm{Now}, \mathrm{volume} \mathrm{of} \mathrm{water} \mathrm{that} \mathrm{runs} \mathrm{into} \mathrm{the} \mathrm{sea} =1.5\times 30\times 3000 {\mathrm{m}}^3 \\ =135000 {\mathrm{m}}^3 \end{gathered}$$

∴ Volume of water that runs into the sea per minute = $\frac{135000}{60}=2250 {\mathrm{m}}^3$

Question 12:

The lateral surface area of a cube is 256 m². The volume of the cube is
(a) 64 m³
(b) 216 m³
(c) 256 m³
(d) 512 m³

Answer 12:

(d) 512  m³

Suppose that a m be the edge of the cube.
We have:
$$\begin{gathered} 4a^2=256 \\ \Rightarrow a^2=\frac{256}{4}= 64 \\ \Rightarrow a =8 \mathrm{m} \end{gathered}$$

∴ Volume of the cube = $a^3 {\mathrm{m}}^3 =8^3 {\mathrm{m}}^3 = 512 {\mathrm{m}}^3$

Question 13:

The total surface area of a cube is 96 cm². The volume of the cube is
(a) 8 cm³
(b) 27 cm³
(c) 64 cm³
(d) 512 cm³

Answer 13:

(c)  64 cm³

Let a cm be the edge of the cube.
We have:
$$\begin{gathered} 6a^2=96 \\ \Rightarrow a^2=16 \\ \Rightarrow a=4 \mathrm{cm} \end{gathered}$$

∴ Volume of the cube = $a^3 {\mathrm{cm}}^3 =4^3 {\mathrm{cm}}^3 = 64 {\mathrm{cm}}^3$

Question 14:

The volume of a cube is 512 cm³. Its total surface area is
(a) 256 cm²
(b) 384 cm²
(c) 512 cm²
(d) 64 cm²

Answer 14:

(b) 384 cm²

Suppose that a cm is the edge of the cube.
We have:
$$\begin{gathered} a^3 = 512 \\ \Rightarrow a =\sqrt[3]{512} = 8 \mathrm{cm} \end{gathered}$$

$$\begin{gathered} \therefore \mathrm{Total} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{cube} =6a^2 {\mathrm{cm}}^2 \\ =6\times 8\times 8 {\mathrm{cm}}^2= 384 {\mathrm{cm}}^2 \end{gathered}$$

Question 15:

The length of the longest rod that can fit in a cubical vessel of side 10 cm, is
(a) 10 cm
(b) 20 cm
(c) $10\sqrt{2} \mathrm{cm}$
(d) $10\sqrt{3} \mathrm{cm}$

Answer 15:

(d)  $10\sqrt{3}$ cm103√ cm
Length of the longest rod = body diagonal of the vessel
                                 $$\begin{gathered} =\sqrt{3}a \\ =\sqrt{3}\times 10 \\ =10\sqrt{3} \mathrm{cm} \end{gathered}$$

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Question 16:

If the length of diagonal of a cube is $8\sqrt{3} \mathrm{cm}$, then its surface area is
(a) 192 cm²
(b) 384 cm²
(c) 512 cm²
(d) 768 cm²

Answer 16:

(b)  384 cm²
We have:
$$\begin{gathered} \sqrt{3}a=8\sqrt{3} \\ \Rightarrow a =8 \mathrm{cm} \end{gathered}$$

∴ Surface area of the cube = $6a^2=6\times 8\times 8= 384 {\mathrm{cm}}^2$

Question 17:

If each edge of a cube is increased by 50%, then the percentage increase in its surface area is
(a) 50%
(b) 75%
(c) 100%
(d) 12%

Answer 17:

Let a be the edge of the cube.
Then the surface area is $6a^2=S \left(\mathrm{say}\right)$

Now, increased edge = $\left(a+\frac{50}{100}a\right)$ = $\frac{150}{100}a = \frac{3}{2}a$

$$\begin{gathered} \mathrm{Then}, \mathrm{new} \mathrm{surface} \mathrm{area}=6{\left(\frac{3}{2}a\right)}^2 =6\times \frac{9}{4}{a}^2 =\frac{9}{4}S \\ \mathrm{In}\mathrm{crease} \mathrm{in} \mathrm{surface} \mathrm{area} =\frac{9}{4}S - S = \frac{5}{4}S \\ \therefore \mathrm{Percentage} \mathrm{increase} \mathrm{in} \mathrm{surface} \mathrm{area} =\frac{\frac{5}{4}S}{S} = \frac{5}{4}\times 100\% = 125\% \end{gathered}$$

Question 18:

Three cubes of metal with edges 3 cm, 4 cm and 5 cm respectively are melted to form a single cube. The lateral surface area of the new cube formed is
(a) 72 cm²
(b) 144 cm²
(c) 128 cm²
(d) 256 cm²

Answer 18:

(b)  144  cm²
Volume of the new cube formed = total volume of the three cubes
Suppose that a cm is the edge of the new cube, then

$$\begin{gathered} a^3=3^3+4^3+5^3 \\ =27+64+125 \\ =216 \\ = 6 \mathrm{cm} \end{gathered}$$

∴ Lateral surface area of the new cube = $4a^2 = 4\times 6\times 6 = 144 {\mathrm{cm}}^2$

Question 19:

In a shower, 5 cm of rain falls. What is the volume of water that falls on 2 hectares of ground?
(a) 500 m³
(b) 750 m³
(c) 800 m³
(d) 1000 m³

Answer 19:

(d) 1000 m³

 $$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{land}=2 \mathrm{sq} \mathrm{hec}=2000 \mathrm{sq} \mathrm{m} \\ \mathrm{Amount} \mathrm{of} \mathrm{rainfall}=5 \mathrm{cm}=0.05 \mathrm{m} \\ \therefore \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{water}=\mathrm{area} \mathrm{of} \mathrm{the} \mathrm{land} \times \mathrm{amount} \mathrm{of} \mathrm{rainfall} \\ =2000\times 0.05 \\ =1000 {\mathrm{m}}^3 \end{gathered}$$

Question 20:

Two cubes have their volumes in the ratio 1 : 27. The ratio of their surface areas is
(a) 1 : 3
(b) 1 : 8
(c) 1 : 9
(d) 1 : 18

Answer 20:

(c) 1 : 9
Suppose that the edges of the cubes are a and b.
We have:
$$\begin{gathered} \frac{a^3}{b^3}=\frac{1}{27} \\ \Rightarrow {\left(\frac{a}{b}\right)}^3=\frac{1}{27} \\ \Rightarrow \frac{a}{b}=\frac{1}{3} \end{gathered}$$

∴ Ratio of the surface areas = $\frac{6a^2}{6b^2}={\left(\frac{a}{b}\right)}^2={\left(\frac{1}{3}\right)}^2 =\frac{1}{9}$

Question 21:

If each side of a cube is doubled, then its volume
(a) is doubled
(b) becomes 4 times
(c) becomes 6 times
(d) becomes 8 times

Answer 21:

(d) becomes 8 times

Suppose that the side of the cube is a.
When it is doubled, it becomes 2a.
New volume of the cube = ${\left(2a\right)}^3=8a^3$

Hence, the volume becomes 8 times the original volume.

Question 22:

The diameter of the base of a cylinder is 6 cm and its height is 14 cm. The volume of the cylinder is
(a) 198 cm³
(b) 396 cm³
(c) 495 cm³
(d) 297 cm³

Answer 22:

(b)  396 cm³

$$\begin{gathered} \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{cylinder}=\mathrm{\pi }{r}^{\mathit{2}}h \\ =\frac{22}{7}\times 3^2\times 14 \\ =22\times 9\times 2 \\ = 396 {\mathrm{cm}}^3 \end{gathered}$$

Question 23:

If the diameter of a cylinder is 28 cm and its height is 20 cm, then its curved surface area is
(a) 880 cm²
(b) 1760 cm²
(c) 3520 cm²
(d) 2640 cm²

Answer 23:

(b)  1760 cm²

$$\begin{gathered} \mathrm{Curved} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cylinder}=2\mathrm{\pi rh} \\ =2\times \frac{22}{7}\times 14\times 20 \\ =44\times 40 \\ =1760 {\mathrm{cm}}^2 \end{gathered}$$

Question 24:

If the curved surface area of a cylinder is 1760 cm² and its base radius is 14 cm, then its height is
(a) 10 cm
(b) 15 cm
(c) 20 cm
(d) 40 cm

Answer 24:

(c) 20 cm

Curved surface area = 1760 cm²
Suppose that h cm is the height of the cylinder.
Then we have:
$$\begin{gathered} 2\mathrm{\pi r}h = 1760 \\ \Rightarrow 2\times \frac{22}{7}\times 14\times h\mathit{ }= 1760 \\ \Rightarrow h=\frac{1760\times 7}{44\times 14}=20 \mathrm{cm} \end{gathered}$$

Question 25:

The height of a cylinder is 14 cm and its curved surface area is 264 cm². The volume of the cylinder is
(a) 308 cm³
(b) 396 cm³
(c) 1232 cm³
(d) 1848 cm³

Answer 25:

(b)  396 cm³

Curved surface area = 264 cm².
Let r cm be the radius of the cylinder.
Then we have:

$$\begin{gathered} 2\mathrm{\pi }rh =264 \\ \Rightarrow 2\times \frac{22}{7}\times r\times 14 =264 \\ \Rightarrow r=\frac{264\times 7}{44\times 14}=3 \mathrm{cm} \end{gathered}$$

 $$\begin{gathered} \therefore \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{cylinder}=\frac{22}{7}\times 3^2\times 14 \\ =22\times 9\times 2 \\ = 396 {\mathrm{cm}}^3 \end{gathered}$$

Question 26:

The curved surface area of a cylindrical pillar is 264 m² and its volume is 924 m³. The height of the pillar is
(a) 4 m
(b) 5 m
(c) 6 m
(d) 7 m

Answer 26:

(c)  6 m

Curved surface area = 264 m² 
Volume = 924 m³
Let r m be the radius and h m be the height of the cylinder.
Then we have:
$$\begin{gathered} 2\mathrm{\pi }rh = 264 \mathrm{and} \mathrm{\pi }{r}^{\mathit{2}}h =924 \\ \Rightarrow rh\mathit{ }=\frac{264}{2\mathrm{\pi }} \\ \Rightarrow h\mathit{ }=\frac{264}{2r\times \mathrm{\pi }} \\ \mathrm{Now}, \mathrm{\pi }{r}^{\mathit{2}}h =\mathrm{\pi }\times r^{\mathit{2}}\times \frac{264}{2r\times \mathrm{\pi }}=924 \\ \Rightarrow r=\frac{924\times 2}{264} \\ \Rightarrow r=7 \mathrm{m} \\ \therefore h\mathit{ }=\frac{264\times 7}{2\times 7\times 22}=6 \mathrm{m} \end{gathered}$$

Question 27:

The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. The ratio of their curved surface area is
(a) 2 : 5
(b) 8 : 7
(c) 10 : 9
(d) 16 : 9

Answer 27:

(c)  10 : 9

Suppose that the radii of the cylinders are 2r and 3r and their respective heights are 5h and 3h.
Then, ratio of the curved surface areas = $\frac{2\mathrm{\pi }\left(2r\right)\left(5h\right)}{2\mathrm{\pi }\left(3r\right)\left(3h\right)}$=10 : 9

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Question 28:

The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. The ratio of their volumes is
(a) 27 : 20
(b) 20 : 27
(c) 4 : 9
(d) 9 : 4

Answer 28:

(b)  20 : 27

Suppose that the radii of the cylinders are 2r and 3r and their respective heights are 5h and 3h..
$$\begin{gathered} \therefore \mathrm{Ratio} \mathrm{of} \mathrm{their} \mathrm{volumes}=\frac{\mathrm{\pi }{\left(2r\right)}^2\times 5h}{\mathrm{\pi }{\left(3r\right)}^2\times 3h} \\ =\frac{4\times 5}{9\times 3}=\frac{20}{27} \end{gathered}$$

Question 29:

The ratio between the radius of the base and the height of a cylinder is 2 : 3. If its volume is 1617 cm³, then its total surface area is
(a) 308 cm²
(b) 462 cm²
(c) 540 cm²
(d) 770 cm²

Answer 29:

(d) 770 cm²
We have:
  r: h = 2 : 3
 $$\begin{gathered} \Rightarrow \frac{r}{h}=\frac{2}{3} \\ \Rightarrow h= \frac{3}{2}r \end{gathered}$$
Now, volume = 1617 cm³
$$\begin{gathered} \Rightarrow \mathrm{\pi }{r}^{\mathit{2}}h\mathit{ }= 1617 \\ \Rightarrow \frac{22}{7}\times r^2\times \frac{3}{2}r\mathit{ }=1617 \\ \Rightarrow r^3=\frac{1617\times 14}{66}=343 \\ \Rightarrow r= 7 \mathrm{cm} \end{gathered}$$

∴ h = 10.5 cm

$$\begin{gathered} \mathrm{Hence}, \mathrm{total} \mathrm{surface} \mathrm{area}=2\mathrm{\pi }rh+2\mathrm{\pi }{r}^2 \\ =\frac{22}{7}(2\times 7\times 10.5 +2\times 7^2) \\ =\frac{22}{7}\left(147+2\times 49\right) \\ =\frac{22}{7}\times 245 \\ =770 {\mathrm{cm}}^2 \end{gathered}$$

Question 30:

Two circular cylinders of equal volume have their heights in the ratio 1 : 2. The ratio of their radii is
(a) $1 : \sqrt{2}$
(b) $\sqrt{2} : 1$
(c) 1 : 2
(d) 1 : 4

Answer 30:

(b) $\sqrt{2} : 1$

Suppose that the heights of two cylinders are h and 2h whose radii are r and R, respectively.
Since the volumes of the cylinders are equal, we have:

$$\begin{gathered} \mathrm{\pi }{r}^{\mathit{2}}h=\mathrm{\pi }{R}^2\times 2h \\ \Rightarrow \frac{r^{\mathit{2}}}{R^{\mathit{2}}}=\frac{2}{1} \\ \Rightarrow \left(\frac{r}{R}\right)=\sqrt{\frac{2}{1}} \\ \Rightarrow r\mathit{:}R\mathit{ }= \sqrt{2} :1 \end{gathered}$$

Question 31:

The ratio between the curved surface area and the total surface area of a right circular cylinder is 1 : 2. If the total surface area is 616 cm², then the volume of the cylinder is
(a) 1078 cm³
(b) 1232 cm³
(c) 1848 cm³
(d) 924 cm³

Answer 31:

(a) 1078 cm³

We have:

$$\begin{gathered} \frac{2\mathrm{\pi }rh}{2\mathrm{\pi }rh+2\mathrm{\pi }{r}^2}=\frac{1}{2} \\ \Rightarrow 4\mathrm{\pi }rh = 2\mathrm{\pi }rh +2\mathrm{\pi }{r}^2 \\ \Rightarrow 2\mathrm{\pi }rh =2\mathrm{\pi }{r}^2 \\ \Rightarrow \frac{r}{h}=\frac{1}{1} \end{gathered}$$

$$\begin{gathered} \mathrm{Also}, 2\mathrm{\pi }rh\mathit{ }+2\mathrm{\pi }{r}^2=616 \\ \Rightarrow 2\mathrm{\pi }{r}^2+2\mathrm{\pi }{r}^2=616 \\ \Rightarrow 4\mathrm{\pi }{r}^2=616 \\ \Rightarrow \mathrm{\pi }{r}^2=154 \\ \Rightarrow r^2=\frac{154}{\mathrm{\pi }}{\mathrm{cm}}^2 \end{gathered}$$

$$\begin{gathered} \therefore \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{cylinder}=\mathrm{\pi }{r}^{\mathit{2}}h \\ =\mathrm{\pi }\times \frac{154}{\mathrm{\pi }}\times r \\ =154\times \sqrt{\frac{154\times 7}{22}} \\ =154\times 7 \\ =1078 {\mathrm{cm}}^3 \end{gathered}$$

Question 32:

In a cylinder, if the radius is halved and the height is doubled, then the volume will be
(a) the same
(b) doubled
(c) halved
(d) four times

Answer 32:

(c) halved

Suppose that the new radius is $\frac{1}{2}$r and the height is 2h.
$$\begin{gathered} \therefore \mathrm{Volume} =\mathrm{\pi }\times {\left(\frac{1}{2}r\right)}^2\times 2h \\ =\mathrm{\pi }\times \frac{r^{\mathit{2}}}{4}\times 2h \\ =\frac{1}{2}\mathrm{\pi }{r}^{\mathit{2}}h \end{gathered}$$

Question 33:

The number of coins 1.5 cm in diameter and 0.2 cm thick to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm is
(a) 540
(b) 450
(c) 380
(d) 472

Answer 33:

(b) 450

Number of coins = $\frac{\mathrm{volume} \mathrm{of} \mathrm{cylinder} \mathrm{to} \mathrm{be} \mathrm{formed}}{\mathrm{volume} \mathrm{of} \mathrm{a} \mathrm{coin}}$
 
                          $$\begin{gathered} =\frac{\mathrm{\pi }\times 2.25\times 2.25\times 10}{\mathrm{\pi }\times 0.75\times 0.75\times 0.2} \\ =\frac{225\times 225\times 5}{75\times 75} \\ =450 \end{gathered}$$

Question 34:

The radius of a wire is decreased to one-third. If volume remains the same, the length will become
(a) 2 times
(b) 3 times
(c) 6 times
(d) 9 times

Answer 34:

(d) 9 times

Let the new radius be $\frac{1}{3}$r.
 Suppose that the new height is H.
The volume remains the same.
$$\begin{gathered} \mathrm{i}.\mathrm{e}., \mathrm{\pi }{r}^{\mathit{2}}h =\mathrm{\pi }\times {\left(\frac{1}{3}r\right)}^2\times H \\ \Rightarrow h = \frac{1}{9}H \\ \Rightarrow H =9h\mathit{ } \\ \therefore \mathrm{The} \mathrm{new} \mathrm{height} \mathrm{becomes} \mathrm{nine} \mathrm{times} \mathrm{the} \mathrm{original} \mathrm{height}. \end{gathered}$$

Question 35:

The diameter of a roller, 1 m long, is 84 cm. If it takes 500 complete revolutions to level a playground, the area of the playground is
(a) 1440 m²
(b) 1320 m²
(c) 1260 m²
(d) 1550 m²

Answer 35:

(b)  1320  m²

Area covered by the roller in 1 revolution = 2$\mathrm{\pi }$rh
 $$\begin{gathered} =2\times \frac{22}{7}\times 42\times 100 \\ =44\times 600 \\ =26400 {\mathrm{cm}}^2 \\ \therefore \mathrm{Area} \mathrm{covered} \mathrm{in} 50 \mathrm{revolutions} =26400 \times 500 {\mathrm{cm}}^2 \\ =1320 {\mathrm{m}}^2 \end{gathered}$$

Question 36:

2.2 dm³ of lead is to be drawn into a cylindrical wire 0.50 cm in diameter. The length of the wire is
(a) 110 m
(b) 112 m
(c) 98 m
(d) 124 m

Answer 36:

(b) 112 m

Volume of the lead = volume of the cylindrical wire
Suppose that h m is the length of the wire.
As 2.2 dm³ of lead is to be drawn into a cylindrical wire of diameter 0.50 cm, we have:
$$\begin{gathered} \mathrm{\pi }{\left(0.25\right)}^2\times h =2.2\times 10\times 10\times 10 \\ \Rightarrow h\mathit{ }=\frac{2200\times 7}{22\times 0.0625} \\ =\frac{700}{0.0625} \\ =11200 \mathrm{cm} \\ = 112 \mathrm{m} \end{gathered}$$

Question 37:

The lateral surface area of a cylinder is
(a) πr²h
(b) πrh
(c) 2πrh
(d) 2πr²

Answer 37:

(c) 2πrh
The lateral surface area of a cylinder is equal to its curved surface area.
∴ Lateral surface area of a cylinder = $2\mathrm{\pi }rh\mathit{ }\mathit{ }$

Question 38:

The height of a cone is 24 cm and the diameter of its base is 14 cm. The curved surface area of the cone is
(a) 528 cm²
(b) 550 cm²
(c) 616 cm²
(d) 704 cm²

Answer 38:

(b) 550 cm²

   $$\begin{gathered} l =\sqrt{r^2+h^2} \\ =\sqrt{7^2+{24}^2} \\ =\sqrt{49+576} \\ =\sqrt{625} \\ =25 \mathrm{cm} \\ \therefore \mathrm{Curved} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cone} =\mathrm{\pi }rl \\ = \frac{22}{7}\times 7\times 25 \\ =550 {\mathrm{cm}}^2 \end{gathered}$$

Question 39:

The volume of a right circular cone of height 12 cm and base radius 6 cm, is
(a) (12π) cm³
(b) (36π) cm³
(c) (72π) cm³
(d) (144π) cm³

Answer 39:

(d) (144π) cm³

Volume of the cone = $\frac{1}{3}\mathrm{\pi }{r}^{\mathit{2}}h\mathit{ }$
                       $$\begin{gathered} =\frac{1}{3}\mathrm{\pi }\times 6^2\times 12 \\ =\mathrm{\pi }\times 36\times 4 \\ =144\mathrm{\pi } {\mathrm{cm}}^3 \end{gathered}$$

PAGE NO-608

Question 40:

How much cloth 2.5 m wide will be required to make a conical tent having base radius 7 m and height 24 m?
(a) 120 m
(b) 180 m
(c) 220 m
(d) 550 m

Answer 40:

(c)  220 m
Let the length of the required cloth be L m and its breadth be B m.
Here, B = 2.5 m
Radius of the conical tent =7 m
Height of the tent = 24 m
Area of cloth required = curved surface area of the conical tent
$$\begin{gathered} \Rightarrow L\times B=\mathrm{\pi rl} \\ \Rightarrow \mathrm{L}\times 2.5=\frac{22}{7}\times 7\times \sqrt{7^2+{24}^2} \\ \Rightarrow 2.5L=22\times \sqrt{49 + 576} \\ \therefore L=\frac{22\times 25}{2.5}=220 \mathrm{m} \end{gathered}$$

Question 41:

The volume of a cone is 1570 cm³ and its height is 15 cm. What is the radius of the cone?
(a) 10 cm
(b) 9 cm
(c) 12 cm
(d) 8.5 cm

Answer 41:

(a) 10 cm
Let r cm be the radius of the cone.
Volume = 1570  cm³

$$\begin{gathered} \mathrm{Then} \frac{1}{3}\times 3.14\times r^2\times 15 = 1570 \\ \Rightarrow r^2 =\frac{1570}{3.14\times 5}=100 \\ \Rightarrow r = 10 \mathrm{cm} \end{gathered}$$

Question 42:

The height of a cone is 21 cm and its slant height is 28 cm. The volume of the cone is
(a)  7356 cm³
(b) 7546 cm³
(c) 7506 cm³
(d) 7564 cm³

Answer 42:

(b)  7546 cm³

$$\begin{gathered} \mathrm{Radius} \mathrm{of} \mathrm{the} \mathrm{cone}, r=\sqrt{l^2-h^2} \\ =\sqrt{{28}^2-{21}^2} \\ =\sqrt{784-441} \\ =\sqrt{343} \mathrm{cm} \end{gathered}$$

$$\begin{gathered} \therefore \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{cone} =\frac{1}{3}\mathrm{\pi }{r}^{\mathit{2}}h \\ =\frac{1}{3}\times \frac{22}{7}\times 343\times 21 \\ =22\times 343 \\ = 7546 {\mathrm{cm}}^3 \end{gathered}$$

Question 43:

The volume of a right circular cone of height 24 cm is 1232 cm³. Its curved surface area is
(a)  1254 cm²
(b) 704 cm²
(c) 550 cm²
(d) 462 cm²

Answer 43:

(c) 550 cm²
Let r cm be the radius of the cone.
Volume of the right circular cone  = 1232 cm³
Then we have:
$$\begin{gathered} \frac{1}{3}\times \frac{22}{7}\times r^2\times 24=1232 \\ \Rightarrow r^2=\frac{1232\times 21}{22\times 24} \\ \Rightarrow r^2=49 \\ \Rightarrow r = 7 \mathrm{cm} \end{gathered}$$

$$\begin{gathered} \therefore \mathrm{Curved} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cone} =\mathrm{\pi }rl \\ =\frac{22}{7}\times 7\times \sqrt{7^2+{24}^2} \\ =22\times \sqrt{49+576} \\ =22\times 25 \\ =550 {\mathrm{cm}}^2 \end{gathered}$$

Question 44:

If the volumes of two cones be in the ratio 1 : 4 and the radii of their bases be in the ratio 4 : 5, then the ratio of their heights is
(a) 1 : 5
(b) 5 : 4
(c) 25 : 16
(d) 25 : 64

Answer 44:

(d) 25 : 64

Suppose that the radii of the cones are 4r and 5r and their heights are h and H, respectively .
It is given that the ratio of the volumes of the two cones is 1:4
Then we have:
$$\begin{gathered} \frac{{\displaystyle \frac{1}{3}}\mathrm{\pi }{\left(4r\right)}^{2}h}{{\displaystyle \frac{1}{3}}\mathrm{\pi }{\left(5r\right)}^{2}H}=\frac{1}{4} \\ \Rightarrow \frac{16r^{\mathit{2}}h}{25r^{2}H}=\frac{1}{4} \\ \Rightarrow \frac{h}{H}=\frac{25}{64} \\ \therefore \mathit{ }h : H = 25 : 64 \end{gathered}$$

Question 45:

If the height of a cone is doubled, then its volume is increased by
(a) 100%
(b) 200%
(c) 300%
(d) 400%

Answer 45:

(a) 100 %

Suppose that height of the cone becomes 2h and let its radius be r.
Then new volume of the cone = $\frac{1}{3}\mathrm{\pi }{r}^{\mathit{2}}\mathit{\left(}2h\right) = \frac{2}{3}\mathrm{\pi }{r}^{\mathit{2}}h\mathit{ }= 2\times \mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{cone}$
Increase in volume = $\frac{2}{3}\mathrm{\pi }{r}^{\mathit{2}}h-\frac{1}{3}\mathrm{\pi }{r}^{\mathit{2}}h=\frac{1}{3}\mathrm{\pi }{r}^{\mathit{2}}h$
$\therefore \mathrm{Percentage} \mathrm{increase}=\frac{\mathrm{increase} \mathrm{in} \mathrm{volume} }{\mathrm{initial} \mathrm{volume} }\times 100 \%$=$\frac{{\displaystyle \frac{1}{3}}\mathrm{\pi }{r}^{\mathit{2}}h}{{\displaystyle \frac{1}{3}}\mathrm{\pi }{r}^{\mathit{2}}h}\times 100\%$=100%

Hence, the volume increases by 100%.

Question 46:

The curved surface area of one cone is twice that of the other while the slant height of the latter is twice that of the former. The ratio of their radii is
(a) 2 : 1
(b) 4 : 1
(c) 8 : 1
(d) 1 : 1

Answer 46:

(b) 4 : 1
If the slant height of the first cone is l, then the slant height of the second cone will be 2l.
Let the radii of the first and second cones be r and R, respectively.
Then we have:

$$\begin{gathered} \mathrm{\pi }rl=2\times (\mathrm{\pi }R\times 2l) \\ \Rightarrow r=4R \\ \Rightarrow \frac{r}{R}=\frac{4}{1} \end{gathered}$$

Question 47:

The ratio of the volumes of a right circular cylinder and a right circular cone of the same base and the same height will be
(a) 1 : 3
(b) 3 : 1
(c) 4 : 3
(d) 3 : 4

Answer 47:

(b)  3 : 1

It is given that the right circular cylinder and the right circular cone have the same base and height.
Suppose that the respective radii of bases and heights are equal to r and h.
Then ratio of their volumes = $\frac{\mathrm{\pi }{r}^{\mathit{2}}h}{{\displaystyle \frac{1}{3}}\mathrm{\pi }{r}^{\mathit{2}}h}=\frac{3}{1}$

Question 48:

A right circular cylinder and a right circular cone have the same radius and the same volume. The ratio of the height of the cylinder to that of the cone is
(a) 3 : 5
(b) 2 : 5
(c) 3 : 1
(d) 1 : 3

Answer 48:

(d)  1 : 3
It is given that the right circular cylinder and the right circular cone have the same radius and volume.
Suppose that the radii of their bases are equal to r and the respective heights of the cylinder and the cone are h and H.
As the volumes of the cylinder and the cone are the same, we have:
$$\begin{gathered} \mathrm{\pi }{r}^{\mathit{2}}h=\frac{1}{3}\mathrm{\pi }{r}^{\mathit{2}}H \\ \Rightarrow \frac{\mathit{h}}{\mathit{H}}=\frac{1}{3} \end{gathered}$$

Question 49:

The radii of the bases of a cylinder and a cone are in the ratio 3 : 4 and their heights are in the ratio 2 : 3. Then, their volumes are in the ratio
(a) 9 : 8
(b) 8 : 9
(c) 3 : 4
(d) 4 : 3

Answer 49:

(a)  9 : 8

Suppose that the respective radii of the cylinder and the cone are 3r and 4r and their respective heights are 2h and 3h.

$\therefore \mathrm{Ratio} \mathrm{of} \mathrm{the} \mathrm{volumes}=\frac{\mathrm{\pi }{\left(3r\right)}^2\times 2h}{{\displaystyle \frac{1}{3}}\mathrm{\pi }{\left(4r\right)}^2\times 3h}=\frac{3\times 9\times 2}{16\times 3}=\frac{9}{8}$

Question 50:

If the height and the radius of a cone are doubled, the volume of the cone becomes
(a) 3 times
(b) 4 times
(c) 6 times
(d) 8 times

Answer 50:

(d) 8 times

Let the new height and radius be 2h and 2r, respectively.

$$\begin{gathered} \mathrm{New} \mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{cone}=\frac{1}{3}\mathrm{\pi }{\left(2r\right)}^2\times 2h \\ =\frac{8}{3}\mathrm{\pi }{r}^{\mathit{2}}h\mathit{ }=8\times \mathrm{initial} \mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{cone} \end{gathered}$$

Question 51:

A solid metallic cylinder of base radius 3 cm and height 5 cm is melted to make n solid cones of height 1 cm and base radius 1 mm. The value of n is
(a) 450
(b) 1350
(c) 4500
(d) 13500

Answer 51:

(d) 13500
Radius of the cylinder = 3 cm
Height of the cylinder = 5 cm
Radius of the cone = 1 mm = 0.1 cm
Height of the cone = 1 cm

∴ Number of cones, n = $\frac{\mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{cylinder}}{\mathrm{volume} \mathrm{of} 1 \mathrm{cone}}$
                             $$\begin{gathered} =\frac{\mathrm{\pi }\times 3^2\times 5}{{\displaystyle \frac{1}{3}}\mathrm{\pi }\times 0.1^2\times 1} \\ =\frac{3\times 9\times 5}{0.01} \\ =13500 \end{gathered}$$ 

PAGE NO-609

Question 52:

A conical tent is to accommodate 11 persons such that each person occupies 4 m² of space on the ground. They have 220 m³ of air to breathe. The height of the cone is
(a) 14 m
(b) 15 m
(c) 16 m
(d) 20 m

Answer 52:

(b) 15 m

Suppose that the height of the cone is h m.
Area of the ground = $11\times 4 = 44 m^2$

$$\begin{gathered} \therefore \mathrm{\pi }{r}^2=44 \\ \Rightarrow r^2=\frac{44\times 7}{22}=14 \\ \mathrm{Also}, \frac{1}{3}\mathrm{\pi }{r}^{\mathit{2}}h =220 \\ \Rightarrow \frac{1}{3}\times \frac{22}{7}\times 14h = 220 \\ \Rightarrow h = \frac{220\times 21}{22\times 14}=15 \mathrm{m} \end{gathered}$$
Hence, the height of the cone is 15 m.

Question 53:

The volume of a sphere of radius 2r is
(a) $\frac{32{\mathrm{\pi r}}^3}{3}$
(b) $\frac{16{\mathrm{\pi r}}^3}{3}$
(c) $\frac{8{\mathrm{\pi r}}^3}{3}$
(d) $\frac{64{\mathrm{\pi r}}^3}{3}$

Answer 53:

(a) $\frac{32{\mathrm{\pi r}}^3}{3}$

$\mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{sphere} \mathrm{of} \mathrm{radius} 2r=\frac{4}{3}\mathrm{\pi }{\left(2r\right)}^3 =\frac{32}{3}\mathrm{\pi }{r}^3$

Question 54:

The volume of a sphere of radius 10.5 cm is
(a) 9702 cm³
(b) 4851 cm³
(c) 19404 cm³
(d) 14553 cm³

Answer 54:

(b)  4851 cm³
Volume of the sphere = $\frac{4}{3}\mathrm{\pi }{r}^3$
                                 $$\begin{gathered} =\frac{4}{3}\times \frac{22}{7}\times 10.5\times 10.5\times 10.5 \\ =88\times 0.5\times 10.5\times 10.5 = 4851 {\mathrm{cm}}^3 \end{gathered}$$

Question 55:

The surface area of a sphere of radius 21 cm is
(a) 2772 cm²
(b) 1386 cm²
(c) 4158 cm²
(d) 5544 cm²

Answer 55:

(d)  5544 cm²

Surface area of sphere = $4\mathrm{\pi }{r}^2$
                                     = $$\begin{gathered} 4\times \frac{22}{7}\times 21\times 21 \\ =5544 {\mathrm{cm}}^2 \end{gathered}$$

Question 56:

The surface area of a sphere is 1386 cm². Its volume is
(a) 1617 cm³
(b) 3234 cm³
(c) 4851 cm³
(d) 9702 cm³

Answer 56:

(c)  4851 cm³

Surface area = 1386 cm²
Let r cm be the radius of the sphere.
Then we have:
$$\begin{gathered} 4\mathrm{\pi }{r}^2=1386 \\ \Rightarrow r^2=\frac{1386\times 7}{4\times 22}=110.25 \\ \Rightarrow r = 10.5 \mathrm{cm} \\ \therefore \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{sphere}=\frac{4}{3}\mathrm{\pi }{r}^3 \\ =\frac{4}{3}\times \frac{22}{7}\times 10.5\times 10.5\times 10.5 \\ =4851 {\mathrm{cm}}^3 \end{gathered}$$

Question 57:

If the surface area of a sphere is (144π) m², then its volume is
(a) (288π) m³
(b) (188π) m³
(c) (300π) m³
(d) (316π) m³

Answer 57:

(a) (288π) m³

Surface area = (144π) m²
Ler r m be the radius of the sphere.
Then we have:

$$\begin{gathered} 4\mathrm{\pi }{r}^2=144\mathrm{\pi } \\ \Rightarrow r^2=\frac{144}{4}=36 \\ \Rightarrow r =6 \mathrm{m} \\ \therefore \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{sphere} = \frac{4}{3}\mathrm{\pi }{r}^3 \\ =\frac{4}{3}\mathrm{\pi }\times 6\times 6\times 6 \\ =288\mathrm{\pi } {\mathrm{m}}^3 \end{gathered}$$

Question 58:

The volume of a sphere is 38808 cm³. Its curved surface area is
(a) 5544 cm²
(b) 8316 cm²
(c) 4158 cm²
(d) 1386 cm²

Answer 58:

(a) 5544 cm²
Let r cm be the radius of the sphere.
Then we have:

$$\begin{gathered} \frac{4}{3}\mathrm{\pi }{r}^3=38808 \\ \Rightarrow r^3=\frac{38808\times 3\times 7}{4\times 22}=9261 \\ \Rightarrow r=21 \mathrm{cm} \\ \therefore \mathrm{Curved} \mathrm{surface} \mathrm{area}=4\mathrm{\pi }{r}^2 = 4\times \frac{22}{7}\times 21\times 21 \\ = 5544 {\mathrm{cm}}^3 \end{gathered}$$

Question 59:

If the ratio of the volumes of two spheres is 1 : 8, then the ratio of their surface area is
(a) 1 : 2
(b) 1 : 4
(c) 1 : 8
(d) 1 : 16

Answer 59:

(b) 1 : 4
Suppose that r and R are the radii of the spheres.
Then we have:

$$\begin{gathered} \frac{{\displaystyle \frac{4}{3}}\mathrm{\pi }{r}^3}{{\displaystyle \frac{4}{3}}\mathrm{\pi }{R}^3}=\frac{1}{8} \\ \Rightarrow {\left(\frac{r}{R}\right)}^3=\frac{1}{8} \\ \Rightarrow \frac{r}{R}=\frac{1}{2} \end{gathered}$$
∴ Ratio of surface area of spheres =$\frac{4\mathrm{\pi }{r}^2}{4\mathrm{\pi }{R}^2}={\left(\frac{r}{R}\right)}^2={\left(\frac{1}{2}\right)}^2=\frac{1}{4}$

Question 60:

A solid metal ball of radius 8 cm is melted and cast into smaller balls, each of radius 2 cm. The number of such balls is
(a) 8
(b) 16
(c) 32
(d) 64

Answer 60:

(d) 64

Number of balls = $\frac{\mathrm{volume} \mathrm{of} \mathrm{solid} \mathrm{metal} \mathrm{ball}}{\mathrm{volumeof} 1 \mathrm{small} \mathrm{ball}}$
                        = $\frac{{\displaystyle \frac{4}{3}}\mathrm{\pi }\times 8^3}{{\displaystyle \frac{4}{3}}\mathrm{\pi }\times 2^3}=\frac{512}{8}=64$

Question 61:

A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere. The radius of the sphere is
(a) 4.2 cm
(b) 2.1 cm
(c) 2.4 cm
(d) 1.6 cm

Answer 61:

(b) 2.1 cm

Let r cm be the radius of the sphere.
Volume of the cone = volume of the sphere
$$\begin{gathered} \Rightarrow \frac{1}{3}\mathrm{\pi }\times 2.1\times 2.1\times 8.4 =\frac{4}{3}\mathrm{\pi }{r}^3 \\ \Rightarrow r^3 =2.1\times 2.1\times 2.1 \\ \Rightarrow r = 2.1 \mathrm{cm} \end{gathered}$$

Question 62:

A solid lead ball of radius 6 cm is melted and then drawn into a wire of diameter 0.2 cm. The length of wire is
(a) 272 m
(b) 288 m
(c) 292 m
(d) 296 m

Answer 62:

(b)  288 m

Let h m be the length of wire.
Volume of the lead ball = volume of the wire
$$\begin{gathered} \Rightarrow \frac{4}{3}\mathrm{\pi }\times 6^3=\mathrm{\pi }\times {\left(0.1\right)}^{2}h \\ \Rightarrow h\mathit{ }= \frac{4\times 216}{3\times 0.01}=28800 \mathrm{cm} =288 \mathrm{m} \end{gathered}$$

Question 63:

A metallic sphere of radius 10.5 cm is melted and then recast into small cones, each of radius 3.5 cm and height 3 cm. Then number of such cones will be
(a) 21
(b) 63
(c) 126
(d) 130

Answer 63:

(c)  126

Number of cones =$\frac{\mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{sphere}}{\mathrm{volume} \mathrm{of} 1 \mathrm{cone}}$
                          $$\begin{gathered} =\frac{{\displaystyle \frac{4}{3}}\mathrm{\pi }\times 10.5\times 10.5\times 10.5}{{\displaystyle \frac{1}{3}}\mathrm{\pi }\times 3.5\times 3.5\times 3} \\ =4\times 3\times 3\times 3.5 =126 \end{gathered}$$

Question 64:

How many lead shots, each 0.3 cm in diameter, can be made from a cuboid of dimensions 9 cm × 11 cm× 12 cm?
(a) 7200
(b) 8400
(c) 72000
(d) 84000

Answer 64:

(d) 84000

$$\begin{gathered} \mathrm{Number} \mathrm{of} \mathrm{lead} \mathrm{shots} =\frac{\mathrm{volume} \mathrm{of} \mathrm{cuboid}}{\mathrm{volume} \mathrm{of} 1 \mathrm{lead} \mathrm{shot}} \\ =\frac{9\times 11\times 12}{{\displaystyle \frac{4}{3}}\times {\displaystyle \frac{22}{7}}\times 0.15\times 0.15\times 0.15} \\ =\frac{9\times 11\times 3\times 3\times 7}{22\times 0.15\times 0.15\times 0.15} \\ =84000 \end{gathered}$$

PAGE NO-610

Question 65:

The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 2 mm. The length of the wire is
(a) 12 m
(b) 18 m
(c) 36 m
(d) 66 m

Answer 65:

(c)  36 m
Let h m be the length of the wire.
Volume of the sphere = volume of the wire
$$\begin{gathered} \Rightarrow \frac{4}{3}\mathrm{\pi }\times 3^3 =\mathrm{\pi }\times {\left(0.1\right)}^2\times h \\ \Rightarrow h= \frac{4\times 9}{0.001} \\ =3600 \mathrm{cm} \\ =36 \mathrm{m} \end{gathered}$$

Question 66:

A sphere of diameter 12.6 cm is melted and cast into a right circular cone of height 25.2 cm. The radius of the base of the cone is
(a) 6.3 cm
(b) 2.1 cm
(c) 6 cm
(d) 4 cm

Answer 66:

(a)  6.3 cm
Let r cm be the radius of base of the cone.
Volume of the sphere = volume of the cone
$$\begin{gathered} \Rightarrow \frac{4}{3}\mathrm{\pi }\times {\left(6.3\right)}^3=\frac{1}{3}\mathrm{\pi }\times r^2\times 25.2 \\ \Rightarrow r^2=\frac{4\times 6.3\times 6.3\times 6.3}{25.2}=39.69 \\ \Rightarrow r = \sqrt{39.69}= 6.3 \mathrm{cm} \end{gathered}$$

Question 67:

A spherical ball of radius 3 cm is melted and recast into three spherical balls. The radii of two of these balls are 1.5 cm and 2 cm. The radius of the third ball is
(a) 1 cm
(b) 1.5 cm
(c) 2.5 cm
(d) 0.5 cm

Answer 67:

(c) 2.5 cm

Let r cm be the radius of the third ball.
Volume of the original ball = volume of the three balls
$$\begin{gathered} \frac{4}{3}\mathrm{\pi }\times 3^3 =\frac{4}{3}\mathrm{\pi }\times 1.5^3+\frac{4}{3}\mathrm{\pi }\times 2^3+\frac{4}{3}\mathrm{\pi }{r}^3 \\ \Rightarrow 27 =3.375+8 +{\mathrm{r}}^3 \\ \Rightarrow r^3 = 27- 11.375 = 15.625 \\ \Rightarrow r = 2.5 \mathrm{cm} \end{gathered}$$

Question 68:

The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. The ratio of the surface areas of the balloons in two cases is
(a) 1 : 4
(b) 1 : 3
(c) 2 : 3
(d) 1 : 2

Answer 68:

(a)  1 : 4
Ratio of the surface areas of balloon = $\frac{2\mathrm{\pi }\times 6^2}{2\mathrm{\pi }\times {12}^2}=\frac{36}{144}=\frac{1}{4}$

Question 69:

The volumes of the two spheres are in the ratio 64 : 27 and the sum of their radii is 7 cm. The difference of their total surface areas is
(a) 38 cm²
(b) 58 cm²
(c) 78 cm²
(d) 88 cm²

Answer 69:

(d) 88 cm²

Suppose that the radii of the spheres are r cm and (7 − r) cm. Then we have:
$$\begin{gathered} \frac{{\displaystyle \frac{4}{3}}\mathrm{\pi }{\left(7-\mathrm{r}\right)}^3}{{\displaystyle \frac{4}{3}}{\mathrm{\pi r}}^3}=\frac{64}{27} \\ \Rightarrow \frac{(7-r)}{r}=\frac{4}{3} \\ \Rightarrow 21-3r=4r \\ \Rightarrow 21=7r \\ \Rightarrow r=3 \mathrm{cm} \end{gathered}$$

Now, the radii of the two spheres are 3 cm and 4 cm.

$$\begin{gathered} \therefore \mathrm{Required} \mathrm{difference}=4\times \frac{22}{7}\times 4^2 -4\times \frac{22}{7}\times 3^2 \\ =4\times \frac{22}{7}\times \left(16-9\right) \\ = 4\times \frac{22}{7}\times 7 \\ =88 {\mathrm{cm}}^2 \end{gathered}$$

Question 70:

A hemispherical bowl of radius 9 cm contains a liquid. This liquid is to be filled into cylindrical small bottles of diameter 3 cm and height 4 cm. How many bottles will be needed to empty thee bowl?
(a) 27
(b) 35
(c) 54
(d) 63

Answer 70:

(c) 54

Number of bottles  = $\frac{\mathrm{volume} \mathrm{of} \mathrm{bowl}}{\mathrm{volume} \mathrm{of} 1 \mathrm{bottle}}$
                         
                           $$\begin{gathered} =\frac{{\displaystyle \frac{2}{3}}\times \mathrm{\pi }\times 9^3}{\mathrm{\pi }\times 1.5^2\times 4} \\ =\frac{2\times 729}{2.25\times 12}=54 \end{gathered}$$

Question 71:

A cone and a hemisphere have equal bases and equal volumes. The ratio of their heights is
(a) 1 : 2
(b) 2 : 1
(c) 4 : 1
(d) $\sqrt{2} : 1$

Answer 71:

(b) 2 : 1

Let the radii of the cone and the hemisphere be r and their respective heights be h and H.
$$\begin{gathered} \mathrm{Then} \frac{1}{3}\mathrm{\pi }\times r^{\mathit{2}}\mathit{\times }h = \frac{2}{3}\mathrm{\pi }\times r^{\mathit{2}}\mathit{\times }H \\ \Rightarrow \frac{h}{H}=\frac{2}{1} \end{gathered}$$

Question 72:

A cone, a hemisphere and a cylinder stand on equal bases and have the same height. The ratio of their volumes is
(a) 1 : 2 : 3
(b) 2 : 1 : 3
(c) 2 : 3 : 1
(d) 3 : 2 : 1

Answer 72:

(a) 1 : 2 : 3
The cone, hemisphere and the cylinder stand on equal bases and have the same height.
We know that radius and height of a hemisphere are the same.
Hence, the height of the cone and the cylinder will be the common radius.
i.e., r = h
Ratio of the volumes of the cone, hemisphere and the cylinder:

$$\begin{gathered} \frac{{\displaystyle \frac{{\displaystyle \frac{1}{3}}\mathrm{\pi }{r}^{\mathit{2}}h}{{\displaystyle \frac{2}{3}}\mathrm{\pi }{r}^3}}}{\mathrm{\pi }{r}^{2}h} \\ =\frac{{\displaystyle \frac{{\displaystyle \frac{1}{3}}\mathrm{\pi }{r}^{\mathit{3}}}{{\displaystyle \frac{2}{3}}\mathrm{\pi }{r}^3}}}{\mathrm{\pi }{r}^3} \\ =\frac{{\displaystyle \frac{1}{2}}}{3} = 1 : 2 : 3 \end{gathered}$$

Question 73:

If the volume and the surface area of a sphere are numerically the same, the nits radius is
(a) 1 unit
(b) 2 units
(c) 3 units
(d) 4 units

Answer 73:

(c) 3 units

We have:

$$\begin{gathered} \frac{4}{3}\mathrm{\pi }{r}^3= 4\mathrm{\pi }{r}^2 \\ \Rightarrow \frac{1}{3}r\mathit{ }=1 \\ \Rightarrow r\mathit{ }= 3 \mathrm{units} \end{gathered}$$