Questions · Page 2 of 2

M.C.Q

MCQ 511 Mark
The curved surface area of a cylindrical pillar is $264 \mathrm{~m}^2$ and its volume is $924 \mathrm{~m}^3$. The height of the pillar is:
  • A
    $4m$
  • B
    $5m$
  • $6m$
  • D
    $7m$
Answer
Correct option: C.
$6m$

Given that,
Curved surface area $= 264 \mathrm{~m}^2$
Volume $= 924 \mathrm{~m}^3$
Now,
Curved surface area $=2\pi\text{rh}$
$\Rightarrow264=2\times\frac{22}{7}\times\text{r}\times\text{h}$
$\Rightarrow264=\frac{44}{7}\times\text{r}\times\text{h}$
$\Rightarrow\text{r}=\frac{264\times7}{44\times\text{h}}$
$\Rightarrow\text{r}=\frac{42}{\text{h}}$
Volume $=\pi\text{r}^2\text{h}$
$\Rightarrow924=\frac{22}{7}\times\frac{42}{\text{h}}\times\frac{42}{\text{h}}\times\text{h}$
$\Rightarrow924=22\times\frac{6}{\text{h}}\times42$
$\Rightarrow\frac{924}{22\times42\times6}=\frac{1}{\text{h}}$
$\Rightarrow\frac{924}{5544}=\frac{1}{\text{h}}$
$\Rightarrow\text{h}=\frac{5544}{924}$
$\Rightarrow\text{h}=6\text{m}$

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MCQ 521 Mark
The length, breadth and height of a cuboid are $15m, 6m$ and $5dm$ respectively. The lateral area of the cuboid is:
  • A
    $45 \mathrm{~m}^2$
  • $21 \mathrm{~m}^2$
  • C
    $201 \mathrm{~m}^2$
  • D
    $90 \mathrm{~m}^2$
Answer
Correct option: B.
$21 \mathrm{~m}^2$

Lateral surface area of the cuboid $= 2(l + b) × h$
$=2(15+6)\times\frac{5}{10}$ $...\Big(1\text{dm}=\frac{1}{10}\text{m}\Rightarrow5\text{dm}=\frac{5}{10}\text{m}\Big)$
$=2(21)\times\frac{1}{2}$
$=21\text{m}^2$

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MCQ 531 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:
  • A
    $3 : 5$
  • B
    $2 : 5$
  • C
    $3 : 1$
  • $1 : 3$
Answer
Correct option: D.
$1 : 3$
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 541 Mark
A cuboid is $12 \ cm$ long, $9 \ cm$ broad and $8 \ cm$ high. Its total surface area is:
  • A
    $864 \mathrm{~cm}^2$
  • $552 \mathrm{~cm}^2$
  • C
    $432 \mathrm{~cm}^2$
  • D
    $276 \mathrm{~cm}^2$
Answer
Correct option: B.
$552 \mathrm{~cm}^2$
Total surface area $=2(\mathrm{lb} \times \mathrm{bh} \times \mathrm{lh})$
$=2[(12 \times 9)+(9 \times 8)+(12 \times 8)]$
$=2[108+72+96]$
$=2 \times 276$
$=552 \mathrm{~cm}^2$
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MCQ 551 Mark
The length of the longest rod that can fit in a cubical vessel of side $10\ cm,$ is:
  • A
    $10\text{cm}$
  • B
    $ 20\text{cm}$
  • C
    $ 10\sqrt{2}\text{cm}$
  • $ 10\sqrt{3}\text{cm}$
Answer
Correct option: D.
$ 10\sqrt{3}\text{cm}$
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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MCQ 561 Mark
How much cloth $2.5m$ wide will be required to make a conical tent having base radius $7m$ and height $24m?$
  • A
    $120m$
  • B
    $180m$
  • $220m$
  • D
    $550m$
Answer
Correct option: C.
$220m$
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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MCQ 571 Mark
A cone is $8.4\ cm$ high and the base is $2.1\ cm$. It is melted and recast into a sphere. The radius of the sphere is:
  • A
    $4.2\ cm$
  • $2.1\ cm$
  • C
    $2.4\ cm$
  • D
    $1.6\ cm$
Answer
Correct option: B.
$2.1\ cm$
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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MCQ 581 Mark
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:
  • $6.3\ cm$
  • B
    $2.1\ cm$
  • C
    $6\ cm$
  • D
    $4\ cm$
Answer
Correct option: A.
$6.3\ cm$
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 591 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 $\text{R}_1$ and $\text{R}_2$ respectively.
$\Rightarrow\pi\text{R}_1^2\text{h}=\pi\text{R}_2^2\text{h}$
$\Rightarrow\pi\times\text{R}_1^2\times\text{x}=\pi\times\text{R}_2^2\times2\text{x}$
$\Rightarrow\text{R}_1^2=\text{R}_2^2\times2$
$\Rightarrow\Big(\frac{\text{R}_1}{\text{R}_2}\Big)^2=2$
$\Rightarrow\frac{\text{R}_1}{\text{R}_2}=\frac{\sqrt2}{1}$
$⇒$ Ratio of their radii is $\sqrt2:1$

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MCQ 601 Mark
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.
  • $9$ times.
Answer
Correct option: D.
$9$ times.
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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MCQ 611 Mark
What is the maximum length of a pencil that can be placed in a rectangular box of dimension $(8\ cm × 6\ cm × 5\ cm)?$ $\big(\text{Given}\sqrt{5}=2.24\big).$
  • A
    $8\ cm$
  • B
    $9.5\ cm$
  • C
    $19\ cm$
  • $11.2\ cm$
Answer
Correct option: D.
$11.2\ cm$
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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MCQ 621 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:
  • A
    $1 : 5$
  • B
    $5 : 4$
  • C
    $25 : 16$
  • $25 : 64$
Answer
Correct option: D.
$25 : 64$
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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MCQ 631 Mark
The volume of a right circular cone of height $12\ cm$ and base radius $6\ cm,$ is:
  • A
    $(12\pi)\text{cm}^3$
  • B
    $(36\pi)\text{cm}^3$
  • C
    $(72\pi)\text{cm}^3$
  • $(144\pi)\text{cm}^3$
Answer
Correct option: D.
$(144\pi)\text{cm}^3$
The height of the cone is $12\ cm$ and the radius of its base is $6\ cm$
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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MCQ 641 Mark
If the height and the radius of a cone are doubled, the volumes of the cone becomes:
  • A
    $3$ times.
  • B
    $4$ times.
  • C
    $6$ times.
  • $8$ times.
Answer
Correct option: D.
$8$ times.
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 651 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
    $72 \mathrm{~cm}^2$
  • $144 \mathrm{~cm}^2$
  • C
    $128 \mathrm{~cm}^2$
  • D
    $256 \mathrm{~cm}^2$
Answer
Correct option: B.
$144 \mathrm{~cm}^2$

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

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MCQ 661 Mark
Two cubes have their volumes in the ratio $1 : 27.$ The ratio of their surface areas is:
  • A
    $1 : 3$
  • B
    $1 : 8$
  • $1 : 9$
  • D
    $1 : 18$
Answer
Correct option: C.
$1 : 9$
$\frac{\text{Volume}\ \text{of}\ \text{cube}\ 1}{\text{Volume}\ \text{of}\ \text{cube}\ 2}=\frac{\text{a}^3}{\text{b}^3}$
$=\frac{1}{27}$
$\Rightarrow\Big(\frac{\text{a}}{\text{b}}\Big)^3=\Big(\frac{1}{3}\Big)^3$
$\Rightarrow\frac{\text{a}}{\text{b}}=\frac{1}{3}$
$\frac{\text{a}^2}{\text{b}^2}=\Big(\frac{\text{a}}{\text{b}}\Big)^2$
$=\Big(\frac{1}{3}\Big)^2$
$=\frac{1}{9}$
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 671 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.
$\Rightarrow\frac{\text{r}_1}{\text{r}_2}=\frac{2}{3}\ \text{and}\ \frac{\text{h}_1}{\text{h}_2}=\frac{5}{3}$
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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MCQ 681 Mark
If the volume and the surface area of a sphere are numerically the same then its radius is:
  • A
    $1$ unit
  • B
    $2$ unit
  • $3$ unit
  • D
    $4$ unit
Answer
Correct option: C.
$3$ unit
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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MCQ 691 Mark
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 third ball is:
  • A
    $1\ cm$
  • B
    $1.5\ cm$
  • $2.5\ cm$
  • D
    $0.5\ cm$
Answer
Correct option: C.
$2.5\ cm$
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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MCQ 701 Mark
The volume of a sphere is $38808 \mathrm{\sim cm}^3$. Its curved surface area is:
  • $5544 \mathrm{~cm}^2$
  • B
    $8316 \mathrm{~cm}^2$
  • C
    $4158 \mathrm{~cm}^2$
  • D
    $1386 \mathrm{~cm}^2$
Answer
Correct option: A.
$5544 \mathrm{~cm}^2$
Volume of sphere = $38808 \mathrm{~cm}^3$
$\Rightarrow\frac{4}{3}\pi\text{r}^3=38808$
$\Rightarrow\frac{4}{3}\times\frac{22}{7}\times\text{r}^3=38808$
$\Rightarrow\text{r}^3=\frac{38808\times7\times3}{88}$
$\Rightarrow\text{r}^3=441\times21$
$\Rightarrow\text{r}^3=(21)^3$
$\Rightarrow\text{r}=21\text{ cm}$
Curved surface area of a sphere $=4\pi\text{r}^2$
$=4\times\frac{22}{7}\times21\times21$
$=4\times22\times3\times21$
$=5544\text{ cm}^2$
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MCQ 711 Mark
If the height of a cone is doubled then its volume is increased by:
  • $100\%$
  • B
    $200\%$
  • C
    $300\%$
  • D
    $400\%$
Answer
Correct option: A.
$100\%$
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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MCQ 721 Mark
The volume of a sphere of radius $2r$ is:
  • $\frac{32\pi\text{r}^3}{3}$
  • B
    $\frac{16\pi\text{r}^3}{3}$
  • C
    $\frac{8\pi\text{r}^3}{3}$
  • D
    $\frac{64\pi\text{r}^3}{3}$
Answer
Correct option: A.
$\frac{32\pi\text{r}^3}{3}$
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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MCQ 731 Mark
The lateral surface area of a cylinder is:
  • A
    $\pi\text{r}^2\text{h}$
  • B
    $\pi\text{rh}$
  • $2\pi\text{rh}$
  • D
    $2\pi\text{r}^2$
Answer
Correct option: C.
$2\pi\text{rh}$
The lateral surface area of a cylinder is given to be $2\pi\text{rh}.$
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M.C.Q - Page 2 - Maths STD 9 Questions - Vidyadip