M.C.Q - Page 2 - MATHS STD 9 Questions - Vidyadip

Questions · Page 2 of 2

M.C.Q

Question 511 Mark

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:

3 : 5
2 : 5
3 : 1
1 : 3

Answer

1 : 3

Solution:
Let the height of a circular cylinder and a right circular cone be h cm and H cm respectively.
Since a right circular and a right circular cone have the same radius and the same volume,
$\Rightarrow\pi\text{r}^2\text{h}=\frac{1}{3}\pi\text{r}^2\text{H}$
$\Rightarrow\text{h}=\frac{1}{3}\text{H}$
$\Rightarrow\frac{\text{h}}{\text{H}}=\frac{1}{3}$
⇒ Ratio of the height is 1 : 3.

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MCQ 521 Mark

A cuboid is $12\ cm$ long, $9\ cm$ broad and $8\ cm$ high. Its total surface area is:

A
$864\ cm^2$
✓
$552\ cm^2$
C
$432\ cm^2$
D
$276\ cm^2$

Answer

Correct option: B.

$552\ cm^2$

Total surface area $= 2(lb \times bh \times lh)$
$= 2 [(12 \times 9) + (9 \times 8) + (12 \times 8)]$
$= 2[108 + 72 + 96]$
$= 2 \times 276$
$= 552\ cm^2$

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Question 531 Mark

The length of the longest rod that can fit in a cubical vessel of side 10cm, is:

$10\text{cm}$
$ 20\text{cm}$
$ 10\sqrt{2}\text{cm}$
$ 10\sqrt{3}\text{cm}$

Answer

$ 10\sqrt{3}\text{cm}$

Solution:
The length of the longest rod = length of the diagonal
$=\sqrt{3}\text{a}$
$=\sqrt{3}\times10$
$=10\sqrt{3}\text{cm}$

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Question 541 Mark

How much cloth 2.5m wide will be required to make a conical tent having base radius 7m and height 24m?

120m
180m
220m
550m

Answer

220m

Solution:
The amount of cloth required to make a tent is equal to the curved surface area of a cone.
$\text{l}=\sqrt{\text{r}^2+\text{h}^2}$
$\text{l}=\sqrt{7^2+24^2}$
$\text{l}=\sqrt{49+576}$
$\text{l}=25\text{m}$
Curved surface area of the cone $=\pi\text{rl}$
$=\frac{22}{7}\times7\times25$
$=550\text{m}^2$
Length of the cloth $=\frac{\text{area}}{\text{width}}$
$=\frac{550}{2.5}=220\text{m}$

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Question 551 Mark

A cone is 8.4cm high and the base is 2.1cm. It is melted and recast into a sphere. The radius of the sphere is:

4.2cm
2.1cm
2.4cm
1.6cm

Answer

2.1cm

Solution:
Let the radius of the sphere be r cm.
Since the cone is melted and recast into a sphere,
Volume of the sphere = volume of the cone
$\Rightarrow\frac{4}{3}\pi\times\text{r}^3=\frac{1}{3}\pi\times(2.1)^2\times8.4$
$\Rightarrow\text{r}^3=2.1\times2.1\times2.1$
$\Rightarrow\text{r}=2.1\text{cm}$

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Question 561 Mark

A sphere of diameter 12.6cm is melted and cast into a right circular cone of height 25.2cm. The radius of the base of the cone is:

6.3cm
2.1cm
6cm
4cm

Answer

6.3cm

Solution:
Let the radius of the base be r cm.
Since the sphere is melted and cast into a cone,
Volume of the sphere = volume of the cone
$\Rightarrow\frac{4}{3}\pi(6.3)^3=\frac{1}{3}\pi\text{r}^2(25.2)$
$\Rightarrow4(6.3)^3=\text{r}^2(25.2)$
$\Rightarrow\frac{4\times(6.3)^3}{(25.2)}=\text{r}^2$
$\Rightarrow\text{r}=6.3\text{cm}$

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MCQ 571 Mark

Two circular cylinders of equal volume have their heights in the ratio $1 : 2.$ The ratio of their radii is:

A
$1:\sqrt2$
✓
$\sqrt2:1$
C
$1:2$
D
$1:4$

Answer

Correct option: B.

$\sqrt2:1$

Let their heights be $x \ cm$ and $2x \ cm$ respectively and let their radii be $R_1$ and $R_2$ respectively.
$\Rightarrow\pi\text{R}_1^2\text{h}=\pi\text{R}_2^2\text{h}$
$\Rightarrow\pi\times\text{R}_1^2\times\text{x}=\pi\times\text{R}_2^2\times2\text{x}$
$\Rightarrow\text{R}_1^2=\text{R}_2^2\times2$
$\Rightarrow\Big(\frac{\text{R}_1}{\text{R}_2}\Big)^2=2$
$\Rightarrow\frac{\text{R}_1}{\text{R}_2}=\frac{\sqrt2}{1}$
$\Rightarrow$ Ratio of their radii is $\sqrt2:1$

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Question 581 Mark

The volume of a right circular cone of height 12cm and base radius 6cm, is:

$(12\pi)\text{cm}^3$
$(36\pi)\text{cm}^3$
$(72\pi)\text{cm}^3$
$(144\pi)\text{cm}^3$

Answer

$(144\pi)\text{cm}^3$

Solution:
The height of the cone is 12cm and the radius of its base is 6cm.
Volume of the cone $=\frac{1}{3}\pi\text{r}^2\text{h}$
$=\frac{1}{3}\times\pi\times6\times6\times12$
$=144\pi\text{cm}^3$

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Question 591 Mark

If the height and the radius of a cone are doubled, the volumes of the cone becomes:

3 times.
4 times.
6 times.
8 times.

Answer

8 times.

Solution:
The volume of a cone of height h and radius r $=\frac{1}{3}\pi\text{r}^2\text{h}=\text{v}$
Since the height and the radius of cone are doubled,
New height = 2h and new radius = 2r
⇒ New volume $=\frac{1}{3}\pi(2\text{r})^2\times2\text{h}$
$=\frac{1}{3}\pi\times4\text{r}^2\times2\text{h}$
$=8\times\Big(\frac{1}{3}\pi\text{r}^2\text{h}\Big)$
$=8\text{v}$

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MCQ 601 Mark

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
$72\ cm^2$
✓
$144\ cm^2$
C
$128\ cm^2$
D
$256\ cm^2$

Answer

Correct option: B.

$144\ cm^2$

Let the side of the cube be $x.$
Volume of new cube formed $= 3^3 + 4^3 + 5^3$
$\Rightarrow x^3 = 27 + 64 + 125$
$\Rightarrow x^3 = 216\ cm^3$
$\Rightarrow x = 6\ cm$
Lateral surface area of the new cube $= 4x^2 = 4 \times (6)^2 = 144\ cm^2$

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Question 611 Mark

Two cubes have their volumes in the ratio 1 : 27. The ratio of their surface areas is:

1 : 3
1 : 8
1 : 9
1 : 18

Answer

1 : 9

Solution:
$\frac{\text{Volume}\ \text{of}\ \text{cube}\ 1}{\text{Volume}\ \text{of}\ \text{cube}\ 2}=\frac{\text{a}^3}{\text{b}^3}$
$=\frac{1}{27}$
$\Rightarrow\Big(\frac{\text{a}}{\text{b}}\Big)^3=\Big(\frac{1}{3}\Big)^3$
$\Rightarrow\frac{\text{a}}{\text{b}}=\frac{1}{3}$
$\frac{\text{a}^2}{\text{b}^2}=\Big(\frac{\text{a}}{\text{b}}\Big)^2$
$=\Big(\frac{1}{3}\Big)^2$
$=\frac{1}{9}$
Surface area $=\frac{6\text{a}^2}{6\text{b}^2}$
$=\frac{6\times1}{6\times9}$
$=\frac{1}{9}$
$\therefore$ Ratio of their surface areas = 1 : 9.

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MCQ 621 Mark

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 areas is:

A
$2 : 5$
B
$8 : 7$
✓
$10 : 9$
D
$16 : 9$

Answer

Correct option: C.

$10 : 9$

Let $r_1$ and $r_2$ be the radii and $h_1$ and $h_2$ be the height of two cylinders.
$\Rightarrow\frac{\text{r}_1}{\text{r}_2}=\frac{2}{3}$ and $\frac{\text{h}_1}{\text{h}_2}=\frac{5}{3}$
Ratio of the surface area $=\frac{2\pi\text{r}_1\text{h}_1}{2\pi\text{r}_2\text{h}_2}$
$=\frac{\text{r}_1}{\text{r}_2}\times\frac{\text{h}_1}{\text{h}_2} $
$=\frac{2}{3}\times\frac{5}{3}$
$=\frac{10}{9}$
$=10:9$

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Question 631 Mark

If the volume and the surface area of a sphere are numerically the same then its radius is:

1 unit
2 unit
3 unit
4 unit

Answer

3 unit

Solution:
Volume of the sphere and the surface area of the sphere are numerically the same,
$\Rightarrow\frac{4}{3}\pi\text{r}^3=4\pi\text{r}^2$
$\Rightarrow\text{r}=3\ \text{units}$

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Question 641 Mark

The length of the longest rod that can be placed in a room of dimension (10m × 10m × 5m) is:

15m
16m
$10\sqrt{5}\text{m}$
12m

Answer

15m

Solution:
Length of the longest rod = length of the diagonal
$=\sqrt{\text{l}^2+\text{b}^2+\text{h}^2}$
$=\sqrt{10^2+10^2+5^2}$
$=\sqrt{100+100+25}$
$=\sqrt{225}$
$=15\text{m}$

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MCQ 651 Mark

The volume of a sphere is $38808\ cm^3.$ Its curved surface area is:

✓
$5544\ cm^2$
B
$8316\ cm^2$
C
$4158\ cm^2$
D
$1386\ cm^2$

Answer

Correct option: A.

$5544\ cm^2$

Volume of sphere $= 38808\ cm^3$
$\Rightarrow\frac{4}{3}\pi\text{r}^3=38808$
$\Rightarrow\frac{4}{3}\times\frac{22}{7}\times\text{r}^3=38808$
$\Rightarrow\text{r}^3=\frac{38808\times7\times3}{88}$
$\Rightarrow\text{r}^3=441\times21$
$\Rightarrow\text{r}^3=(21)^3$
$\Rightarrow\text{r}=21\ cm$
Curved surface area of a sphere $=4\pi\text{r}^2$
$=4\times\frac{22}{7}\times21\times21$
$=4\times22\times3\times21$
$=5544\ cm^2$

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Question 661 Mark

If the height of a cone is doubled then its volume is increased by:

100%
200%
300%
400%

Answer

100%

Solution:
Let the original height of the cone be h and the radius be r.
Volume of the cone $=\frac{1}{3}\pi\text{r}^2\text{h}$
New height is 2h and the radius is the same
So, the new volume of the cone $=\frac{1}{3}\pi\text{r}^2(2\text{h})=\frac{2}{3}\pi\text{r}^2\text{h}$
Increase in the volume $=\frac{2}{3}\pi\text{r}^2\text{h}-\frac{1}{3}\pi\text{r}^2\text{h}=\frac{1}{3}\pi\text{r}^2\text{h}$
Increase $\%=\frac{\text{Increase}}{\text{Original}\ \text{volume}}\times100$
⇒ Increase $=\frac{\frac{1}{3}\pi\text{r}^2\text{h}}{\frac{1}{3}\pi\text{r}^2\text{h}}\times100$
⇒ Increase $\%=100$

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Question 671 Mark

The volume of a sphere of radius 2r is:

$\frac{32\pi\text{r}^3}{3}$
$\frac{16\pi\text{r}^3}{3}$
$\frac{8\pi\text{r}^3}{3}$
$\frac{64\pi\text{r}^3}{3}$

Answer

$\frac{32\pi\text{r}^3}{3}$

Solution:
Volume of sphere $=\frac{4}{3}\pi\text{r}^3$
$=\frac{4}{3}\pi\times(2\text{r})^3$
$=\frac{4}{3}\pi\times8\text{r}^3$
$=\frac{32\pi\text{r}^3}{3}$

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Question 681 Mark

The lateral surface area of a cylinder is:

$\pi\text{r}^2\text{h}$
$\pi\text{rh}$
$2\pi\text{rh}$
$2\pi\text{r}^2$

Answer

$2\pi\text{rh}$

Solution:
The lateral surface area of a cylinder is given to be $2\pi\text{rh}.$

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Question 691 Mark

A spherical ball of radius 3cm is melted and recast into three spherical balls. The radii of two of these balls are 1.5cm and 2cm. The radius of third ball is:

1cm
1.5cm
2.5cm
0.5cm

Answer

2.5cm

Solution:
Let the radius of the third ball be r cm.
Volume of the spherical ball = sum of the volume of the three balls
$\Rightarrow\frac{4}{3}\pi(3)^3=\frac{4}{3}\pi(1.5)^3+\frac{4}{3}\pi(2)^3+\frac{4}{3}\pi\text{r}^3$
$\Rightarrow(3)^3=(1.5)^3+(2)^3+\text{r}^3$
$\Rightarrow\text{r}^3=15.625$
$\Rightarrow\text{r}=2.5\text{cm}$

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Question 701 Mark

If the volume 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:

1 : 5
5 : 4
25 : 16
25 : 64

Answer

25 : 64

Solution:
Let the radii of the cones be 4x cm and 5x cm respectively.
Let their be h cm and H cm respectively.
⇒ Ratio of the volume of the cones $=\frac{\frac{1}{3}\pi\times(4\text{x}^2\times\text{h})}{\frac{1}{3}\pi\times(5\text{x})^2\times\text{H}}$
$\Rightarrow\frac{\text{V}}{4\text{v}}=\frac{\frac{1}{3}\pi\times(4\text{x})^2\times\text{h}}{\frac{1}{3}\pi\times(5\text{x})^2\times\text{H}}$
$\Rightarrow\frac{1}{4}=\frac{16\text{h}}{25\text{H}}$
$\Rightarrow\frac{\text{h}}{\text{H}}=\frac{1}{4}\times\frac{25}{16}$
$\Rightarrow\frac{\text{h}}{\text{H}}=\frac{25}{64}$
$\Rightarrow\text{h}:\text{H}=25:64$

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Question 711 Mark

The diameter of a sphere is 6cm. It is melted and drawn into a wire of diameter 2mm. The length of the wire is:

12m
18m
36m
66m

Answer

36m

Solution:
Given that the diameter of the sphere is 6cm.
So, the radius of the sphere is 3cm.
Diameter of the wire = 2mm = 1mm = 0.1cm
Since the sphere is drawn into a wire,
Volume of the sphere = volume of the wire
$\frac{4}{3}\pi(3)^3=\pi(0.1)^2\text{h}$
$\Rightarrow36\pi=\pi(0.1)^2\text{h}$
$\Rightarrow\text{h}=3600\text{cm}=36\text{m}$

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Question 721 Mark

The radius of a wire is decreased to one third. If volume remains the same, the length will become:

2 times.
3 times.
6 times.
9 times.

Answer

9 times.

Solution:
Let the radius of the wire be r and the height be h.
So, the new radius $=\frac{1}{3}\text{r}$
Let the new height be H.
Note that the height of the wire is the same as the length of the wire.
Given that the volume remains the same.
So, $\pi\text{r}^2\text{h}=\pi\Big(\frac{1}{3}\text{r}\Big)^2\text{H}$
$\Rightarrow\text{r}^2\text{h}=\frac{1}{9}\text{r}^2\text{H}$
$\Rightarrow\text{h}=\frac{1}{9}\text{H}$
$\Rightarrow\text{H}=9\text{h}$
Thus, the length will become 9 times the original length.

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Question 731 Mark

What is the maximum length of a pencil that can be placed in a rectangular box of dimension (8cm × 6cm × 5cm)? $\big(\text{Given}\sqrt{5}=2.24\big).$

8cm
9.5cm
19cm
11.2cm

Answer

11.2cm

Solution:
Given: $\sqrt{5}=2.24$
Required length $=\sqrt{\text{l}^2+\text{b}^2+\text{h}^2}$
$=\sqrt{8^2+6^2+5^2}$
$=\sqrt{64+36+25}$
$=\sqrt{125}$
$=5\sqrt{5}$
$=2\times2.24$
$=11.2\text{cm}$

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M.C.Q - Page 2 - MATHS STD 9 Questions - Vidyadip