M.C.Q (1 Marks)

MCQ 11 Mark

In mean equation $\overline{ X }=a+\frac{\Sigma f i d i}{\Sigma f_i} d i$ _______ .

✓
$x_i-a$
B
$a-x_i$
C
$f_i-a$
D
$a-f_i$

Answer

Correct option: A.

$x_i-a$

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MCQ 21 Mark

The volume of a cube is $2744 \ cm ^3$. Its surface area is

A
$196 \ cm ^2$
✓
$1176 \ cm ^2$
C
$784 \ cm ^2$
D
$588 \ cm ^2$

Answer

Correct option: B.

$1176 \ cm ^2$

We have, $V=a^3=2744=2^3 \times 7^3$
$\Rightarrow a=(2 \times 7)=14 \ cm .$
$\therefore \text { S.A. }=6 \times a^2=6 \times 14^2=1176 \ cm ^2$

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MCQ 31 Mark

The ratio of the total surface area to the lateral surface area of a cylinder with base radius $80 cm$ and height $20 cm$ is

A
$1: 2$
B
$2: 1$
C
$3: 1$
✓
$5: 1$

Answer

Correct option: D.

$5: 1$

(d) : $\frac{\text { Total surface area }}{\text { Lateral surface area }}=\frac{2 \pi r(h+r)}{2 \pi r h}$
$
=\frac{h+r}{h}=\frac{(20+80)}{20}=\frac{100}{20}=\frac{5}{1}
$
$\therefore \quad$ Required ratio is $5: 1$.

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MCQ 41 Mark

The height of a cylinder is $14 \ cm$ and its curved surface area is $264 \ cm ^2$. The volume of the cylinder is

A
$296 \ cm ^3$
✓
$396 \ cm ^3$
C
$369 \ cm ^3$
D
$503 \ cm ^3$

Answer

Correct option: B.

$396 \ cm ^3$

$\text{C.S.A.} =2 \pi r h=2 \times \frac{22}{7} \times r \times 14=264\ ($given$)$
$\Rightarrow r=\frac{264}{88}=3 \ cm$
$\therefore$ Volume $=\pi r^2 h=\frac{22}{7} \times 3 \times 3 \times 14=396 \ cm ^3$

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MCQ 51 Mark

The ratio of the volumes of two spheres is $8: 27$. The ratio between their surface areas is

A
$2: 3$
B
$4: 27$
C
$8: 9$
✓
$4: 9$

Answer

Correct option: D.

$4: 9$

(d) : $\frac{\text { Volume of sphere with radius } r}{\text { Volume of sphere with radius } R}$
$
=\frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3}=\frac{8}{27} \Rightarrow \frac{r^3}{R^3}=\frac{8}{27} \Rightarrow\left(\frac{r}{R}\right)^3=\left(\frac{2}{3}\right)^3 \Rightarrow \frac{r}{R}=\frac{2}{3}
$
Ratio between their surface areas
$
=\frac{4 \pi r^2}{4 \pi R^2}=\frac{r^2}{R^2}=\left(\frac{r}{R}\right)^2=\left(\frac{2}{3}\right)^2=\frac{4}{9}
$

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MCQ 61 Mark

The radii of the base of a cylinder and a cone are in the ratio $3: 4$ and their heights are in the ratio $2: 3$, then ratio of their volumes is

✓
$9: 8$
B
$9: 4$
C
$3: 1$
D
$27: 64$

Answer

Correct option: A.

$9: 8$

(a) : Let the radii of the cylinder and cone be $3 r$ and $4 r$ respectively and their heights be $2 h$ and $3 h$ respectively. Then,
$
\frac{\text { Volume of cylinder }}{\text { Volume of cone }}=\frac{\pi(3 r)^2 \times 2 h}{\frac{1}{3} \pi(4 r)^2 3 h}=\frac{54}{48}=\frac{9}{8}
$

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MCQ 71 Mark

The base radius of the cylinder is $1 \frac{2}{3}$ times its height. The cost of painting its $\text{C.S.A.}$ at $2$ paise $cm ^2$ is $₹\ 92.40$ . The volume of the cylinder is

✓
$80850 \ cm ^3$
B
$88850 \ cm ^3$
C
$80508 \ cm ^3$
D
None of these

Answer

Correct option: A.

$80850 \ cm ^3$

$r=\frac{5}{3} h, \text{C.S.A.} =2 \pi r h=2 \times \frac{22}{7} \times \frac{5}{3} h^2$
$\Rightarrow \frac{220}{21} h^2=\frac{9240}{2} $
$\Rightarrow h=21 \ cm , r=35 \ cm$
$\therefore V=\pi r^2 h=\frac{22}{7} \times 35 \times 35 \times 21=80850 \ cm ^3$

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MCQ 81 Mark

A circus tent is cylindrical up to a height of $4 m$ and conical above it. If its diameter is $105 m$ and its slant height is $40 m$, the total area of canvas required to built the tent is

✓
$7920 m ^2$
B
$7820 m ^2$
C
$9720 m ^2$
D
$2645 m ^2$

Answer

Correct option: A.

$7920 m ^2$

(a) : Area of canvas $=(2 \pi r h+\pi r l)$
$
=\left(2 \times \frac{22}{7} \times \frac{105}{2} \times 4+\frac{22}{7} \times \frac{105}{2} \times 40\right) m ^2=7920 m ^2
$

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MCQ 91 Mark

A glass cylinder with diameter $20 \ cm$ has water to a height of $9 \ cm$. A metal cube of $8 \ cm$ edge is immersed in it completely. The height by which water will rise in the cylinder is $($Take $\pi=3.14 )$

✓
$1.6 \ cm$
B
$2.5 \ cm$
C
$1 \ cm$
D
$2.6 \ cm$

Answer

Correct option: A.

$1.6 \ cm$

Volume of the water displaced $=$ Volume of the cube of edge $8 \ cm$
$\Rightarrow \pi r^2 h=8^3 $
$\Rightarrow 3.14 \times 10^2 \times h=8 \times 8 \times 8$
$\Rightarrow h=\frac{8 \times 8 \times 8}{3.14 \times 100} \ cm =1.6 \ cm$

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MCQ 101 Mark

The volume of the greatest sphere that can be cut off from a cylindrical log of wood of base radius $1 cm$ and height $5 cm$ is

✓
$\frac{4}{3} \pi cm ^3$
B
$\frac{10}{3} \pi cm ^3$
C
$5 \pi cm ^3$
D
$\frac{20}{3} \pi cm ^3$

Answer

Correct option: A.

$\frac{4}{3} \pi cm ^3$

(a) : The radius of the greatest sphere that can be cut off from the cylinder $=1 cm$$\therefore \quad$
Volume of the sphere $=\frac{4}{3} \pi r^3=\frac{4}{3} \pi(1)^3=\frac{4}{3} \pi cm ^3$

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MCQ 111 Mark

The length of the diagonal of a cube is $6 \sqrt{3}$ $cm$. Its total surface area is

A
$144 cm ^2$
✓
$216 cm ^2$
C
$180 cm ^2$
D
$108 cm ^2$

Answer

Correct option: B.

$216 cm ^2$

(b) : Let $a$ be the side of the cube.Given, diagonal of cube $=6 \sqrt{3} cm$
$
\therefore \sqrt{3} a=6 \sqrt{3} \Rightarrow a=6 cm
$
$\therefore$ Total surface area $=6 a^2=(6 \times 6 \times 6) cm ^2=216 cm ^2$

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MCQ 121 Mark

A cube whose edge is $20 cm$ long, has circles on each of its faces painted black. What is the total area of the unpainted surface of the cube, if the circles are of the largest possible areas?

A
$90.72 cm ^2$
B
$256.72 cm ^2$
C
$330.3 cm ^2$
✓
$514.28 cm ^2$

Answer

Correct option: D.

$514.28 cm ^2$

(d) : Diameter of circle $=20 cm$
$\therefore \quad$ Area of a circle $=100 \pi cm ^2$
$\therefore \quad$ Area of 6 circles $=6 \times 100 \pi=600 \pi cm ^2$
Surface area of cube $=6 \times(20)^2=2400 cm ^2$
Area of unpainted surface $=(2400-600 \pi) cm ^2$
$
=2400-600 \times \frac{22}{7}=514.28 cm ^2
$

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MCQ 131 Mark

The total surface area of a cube is $864 cm ^2$. Its volume is

A
$3456 cm ^3$
B
$432 cm ^3$
✓
$1728 cm ^3$
D
$3460 cm ^3$

Answer

Correct option: C.

$1728 cm ^3$

(c) : Let $a$ be the side of the cube.Given, total surface area of cube $=864 cm ^2$
$
\therefore \quad 6 a^2=864 \Rightarrow a^2=144 \Rightarrow a=12 cm
$
$\therefore$ Volume of the cube $=(12 \times 12 \times 12) cm ^3=1728 cm ^3$

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MCQ 141 Mark

Two cubes each of volume $27 \ cm ^3$ are joined end to end to form a solid. Find the surface area of the resulting cuboid.

✓
$90 \ cm ^2$
B
$80 \ cm ^2$
C
$70 \ cm ^2$
D
$60 \ cm ^2$

Answer

Correct option: A.

$90 \ cm ^2$

Let the edge of each cube be $a \ cm$.

$\therefore \quad$ Volume of each cube $=a^3 \ cm ^3$
i.e., $a^3=27=(3)^3$
$\Rightarrow a=3$
$\therefore \quad$ Surface area of the cuboid
$=2(l b+b h+h l)$
$=2(6 \times 3+3 \times 3+3 \times 6)$
$=2 \times 45=90 \ cm ^2$

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MCQ 151 Mark

If the total surface area of a solid hemisphere is $462 cm ^2$, then find its volume. $\left[\right.$ Take $\left.\pi=\frac{22}{7}\right]$

A
$518 cm ^3$
B
$726 cm ^3$
C
$816 cm ^3$
✓
$718.67 cm ^3$

Answer

Correct option: D.

$718.67 cm ^3$

(d) : Let radius of hemisphere $=r cm$Total surface area of hemisphere $=462 cm ^2$
$
\Rightarrow 3 \pi r^2=462 \Rightarrow r^2=\frac{462 \times 7}{3 \times 22}=49 \Rightarrow r=7
$
$\therefore \quad$ Volume of hemisphere
$
=\frac{2}{3} \pi r^3=\frac{2}{3} \times \frac{22}{7} \times 7 \times 7 \times 7=718.67 cm ^3
$

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MCQ 161 Mark

The radius of spherical balloon increases from $8 \ cm$ to $12 \ cm$. The ratio of the surface areas of the balloon in two cases is

A
$2: 3$
B
$3: 2$
C
$8: 27$
✓
$4: 9$

Answer

Correct option: D.

$4: 9$

Let $r_1$ and $r_2$ be the radius of smaller spherical balloon and bigger spherical balloon respectively.
$\therefore r_1=8 \ cm \text { and } r_2=12 \ cm$
Hence, required ratio
$=\frac{\text { Surface area of smaller spehrical sphere }}{\text { Surface area of bigger spherical sphere }}=\frac{4 \pi r_1^2}{4 \pi r_2^2}=\left(\frac{r_1}{r_2}\right)^2$
$=\left(\frac{8}{12}\right)^2=\left(\frac{2}{3}\right)^2=\frac{4}{9}=4: 9$

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MCQ 171 Mark

If two cubes, each of edge $4 cm$ are joined end to end, then the surface area of the resulting cuboid is

A
$100 cm ^2$
✓
$160 cm ^2$
C
$200 cm ^2$
D
$80 cm ^2$

Answer

Correct option: B.

$160 cm ^2$

(b) : Two cubes of edge $4 cm$ each are joined end to end to form a cuboid.
$\therefore$ For resulting cuboid, length $(l)=4+4=8 cm$, breadth $(b)=4 cm$ and height $(h)=4 cm$
$\therefore \quad$ Surface area of cuboid
$
=2(l b+b h+h l)=2(8 \times 4+4 \times 4+4 \times 8)=160 cm ^2
$

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MCQ 181 Mark

The length of the longest pole that can be kept in a room of dimensions $12 m \times 9 m \times 8 m$ is

A
$29 m$
B
$21 m$
C
$19 m$
✓
$17 m$

Answer

Correct option: D.

$17 m$

(d) : Given, length $(l)=12 m$, breadth $(b)=9 m$ and height $(h)=8 m$
Length of the longest pole
$=$ Length of diagonal of the room
$
=\sqrt{l^2+b^2+h^2}=\sqrt{(12)^2+(9)^2+(8)^2}=\sqrt{289}=17 m
$

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MCQ 191 Mark

The ratio of the radii of two spheres is $4: 5$. Then the ratio of their total surface area is

✓
$16: 25$
B
$2: \sqrt{3}$
C
$5: 4$
D
$2: 3$

Answer

Correct option: A.

$16: 25$

(a) : Let $r_1$ and $r_2$ be radii of two spheres.
$
\therefore \quad \frac{r_1}{r_2}=\frac{4}{5}
$
(Given)
Ratio of total surface area of two spheres
$
\frac{4 \pi r_1^2}{4 \pi r_2^2}=\left(\frac{r_1}{r_2}\right)^2=\left(\frac{4}{5}\right)^2=\frac{16}{25}
$

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MCQ 201 Mark

The curved surface area of a right circular cone of height $15 cm$ and base diameter $16 cm$ is

A
$60 \pi cm ^2$
B
$68 \pi cm ^2$
C
$120 \pi cm ^2$
✓
$136 \pi cm ^2$

Answer

Correct option: D.

$136 \pi cm ^2$

(d) : Let $r$ be the radius, $h$ be the height and $l$ be the slant height of right circular cone.
Now, $h=15 cm$, Diameter $=16 cm$
[Given]
$
\therefore r=\frac{16}{2}=8 cm
$Also, $l=\sqrt{r^2+h^2}=\sqrt{8^2+15^2}=\sqrt{289}=17 cm$
$\therefore \quad$ Curved surface area of the cone $=\pi r l$
$
=\pi \times 8 \times 17=136 \pi cm ^2
$

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MCQ 211 Mark

A right triangle with sides $3 cm , 4 cm$ and $5 cm$ is rotated about the side of $3 cm$ to form a cone. The volume of the cone so formed is

A
$12 \pi cm ^3$
B
$15 \pi cm ^3$
✓
$16 \pi cm ^3$
D
$20 \pi cm ^3$

Answer

Correct option: C.

$16 \pi cm ^3$

(c) : Here, radius $(r)=4 cm$ and height $(h)=3 cm$
$\therefore \quad$ Volume of the cone $=\frac{1}{3} \pi r^2 h$
$
=\frac{\pi}{3} \times 4 \times 4 \times 3=16 \pi cm ^3
$

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MCQ 221 Mark

The volume of the largest right circular cone that can be cut out from a cube of edge $4.2 cm$ is

A
$9.7 cm ^3$
B
$72.6 cm ^3$
C
$58.2 cm ^3$
✓
$19.4 cm ^3$

Answer

Correct option: D.

$19.4 cm ^3$

(d) : Diameter of cone $=$ length of cube $=4.2 cm$
$\therefore$ Radius of cone, $r=\frac{4.2}{2}=2.1 cm$
Height of cone $=$ Height of cube $=4.2 cm$
$\therefore \quad$ Volume of the cone $=\frac{1}{3} \pi r^2 h$
$
=\frac{1}{3} \times \frac{22}{7} \times \frac{21}{10} \times \frac{21}{10} \times \frac{42}{10}=19.4 cm ^3 \text { (approx). }
$

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MCQ 231 Mark

The volume of the largest right circular cone that can be carved out of a solid hemisphere of radius $r$ can be calculated as:
Radius of largest cone $=$ Radius of hemisphere $=$ $r$ (Step 1)
Height of largest cone $=$ Radius of hemisphere $=r$ (Step 2)
$\therefore \quad$ Volume of cone $=\frac{1}{3} \pi r^2 h=\frac{1}{3} \pi r^2 r=\frac{1}{3} \pi r^3$
(Step 3)In which step is there an error involving?

A
Step-1
B
Step-2
C
Step-3
✓
There is no error.

Answer

Correct option: D.

There is no error.

(d) : Radius of largest cone $=$ Radius of hemisphere $=r$ (Step 1)
Height of largest cone $=$ Radius of hemisphere $=r$ (Step 2)
$\therefore \quad$ Volume of cone $=\frac{1}{3} \pi r^2 h=\frac{1}{3} \pi r^2 r=\frac{1}{3} \pi r^{\prime}$ (Step 3)

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MCQ 241 Mark

If the length, breadth and height of a cuboid are made 2,4 and 6 times respectively, then by what percent the volume increases?

A
$47 \%$
B
$470 \%$
✓
$4700 \%$
D
none of these

Answer

Correct option: C.

$4700 \%$

(c) : We know that if length, breadth and height of a cuboid are made by $x, y$ and $z$ times respectively then volume be increased by
$
(x y z-1) \times 100 \% \text {. }
$
In our problem, $x=2, y=4, z=6$
$\therefore \%$ increase in volume $=(2 \times 4 \times 6-1) \times 100 \%=4700 \%$

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MCQ 251 Mark

The radius of a sphere $($in $cm)$ whose volume is $12 \pi \ cm ^3$ is

✓
$3^{2 / 3}$
B
$3^{3 / 2}$
C
$3^{1 / 3}$
D
$3^{1 / 2}$

Answer

Correct option: A.

$3^{2 / 3}$

Let radius of the sphere be $r \ cm$.
According to question,
$\frac{4}{3} \pi r^3=12 \pi$
$\Rightarrow r^3=\frac{3 \times 12}{4}=9=3^2 $
$\Rightarrow r=\left(3^2\right)^{1 / 3}=(3)^{2 / 3} \ cm$

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MCQ 261 Mark

The total surface area of solid opened at the top in the given figure is

✓
$2 \pi r(h+r)$
B
$\pi r(2 h+r)$
C
$2 \pi h(h+r)$
D
None of these

Answer

Correct option: A.

$2 \pi r(h+r)$

(a) : Total surface area of solid = Curved surface area of cylinder + Curved surface area of hemisphere $=2 \pi r h+2 \pi r^2=2 \pi r(h+r)$

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MCQ 271 Mark

If curved surface area of cylinder is equal to its volume. The radius of cylinder will be

✓
2
B
3
C
4
D
5

Answer

Correct option: A.

(a) : Let $h$ and $r$ be the height and radius of the base of the right circular cylinder respectively, then
$2 \pi r h=\pi r^2 h$ i.e., $r=2$

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MCQ 281 Mark

A solid is hemispherical at the bottom and conical (of same radius) above it. If the surface areas of the two parts are equal, then what will be the ratio of its radius and the slant height of the conical part?

A
$2: 1$
✓
$1: 2$
C
$3: 2$
D
$2: 3$

Answer

Correct option: B.

$1: 2$

(b) : Let $r$ be the radius of hemisphere and conical part. Also, let $l$ be the slant height of conical part.
Given that, surface area of hemisphere
$=$ surface area of conical part
$\Rightarrow 2 \pi r^2=\pi r l \Rightarrow 2 r=l$
$\Rightarrow \frac{r}{l}=\frac{1}{2}$ i.e., $r: l=1: 2$

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MCQ 291 Mark

The total surface area of a cube is $32 \frac{2}{3} m ^2$. The volume of the cube (in $m ^3$ ) is

✓
$12 \frac{19}{27}$
B
$19 \frac{12}{27}$
C
$12 \frac{7}{27}$
D
None of these

Answer

Correct option: A.

$12 \frac{19}{27}$

(a) : Let the side of the cube be $x m$.
$\therefore$ Total surface area $=6 x^2 m ^2$
$\Rightarrow \quad 6 x^2=32 \frac{2}{3} \Rightarrow 6 x^2=\frac{98}{3}$
$\Rightarrow x^2=\frac{98}{3} \times \frac{1}{6}=\frac{49}{9} \Rightarrow x=\frac{7}{3} m$.
$\therefore \quad$ Volume of the cube $=x^3 m ^3$
$
=\left(\frac{7}{3}\right)^3 m ^3=\frac{343}{27} m ^3=12 \frac{19}{27} m ^3 \text {. }
$

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MCQ 301 Mark

A cone and a cylinder have the same radii but the height of the cone is 3 times that of the cylinder. The ratio of their volumes is/are

A
$1: 2$
B
$2: 1$
✓
$1: 1$
D
$2: 3$

Answer

Correct option: C.

$1: 1$

(c) : Let the radius and the height of the cylinder are $r$ and $h$ respectively.
So, radius of the cone is $r$ and height of the cone is $3 h$.
$\therefore \quad$ Volume of the cylinder $=\pi r^2 h$
Volume of the cone $=\frac{1}{3} \pi r^2 3 h=\pi r^2 h$
So, required ratio $=\frac{\pi r^2 h}{\pi r^2 h}=1: 1$

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MCQ 311 Mark

The volume of a hemisphere is $2425 \frac{1}{2} cm ^3$. Its curved surface area is $\left[\right.$ Use $\left.\pi=\frac{22}{7}\right]$

A
$793 cm ^2$
B
$963 cm ^2$
✓
$693 cm ^2$
D
None of these

Answer

Correct option: C.

$693 cm ^2$

(c) : Given, volume of hemisphere $=\frac{4851}{2} cm ^3$
Let the radius of the hemisphere be ' $r$ ' $cm$.
$\therefore \quad$ Volume of hemisphere $=\frac{2}{3} \pi r^3$
$\Rightarrow \frac{2}{3} \pi r^3=\frac{4851}{2}$
$\Rightarrow r^3=\frac{4851 \times 3 \times 7}{2 \times 2 \times 22} \Rightarrow r=\frac{21}{2} cm$
Now, curved surface area of hemisphere $=2 \pi r^2$
$
=2 \times \frac{22}{7} \times \frac{21 \times 21}{2 \times 2}=693 cm ^2
$

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MCQ 321 Mark

Find the volume of the largest right circular cone that can be cut out from a cube of edge 6.4 $cm$.

A
$49.65 cm ^3$
B
$62.6 cm ^3$
C
$48.2 cm ^3$
✓
$68.65 cm ^3$

Answer

Correct option: D.

$68.65 cm ^3$

(d) : Diameter of cone $=$ length of cube $=6.4 cm$
$\therefore$ Radius of cone, $r=\frac{6.4}{2}=3.2 cm$
Height of cone $=$ Height of cube $=6.4 cm$
$\therefore \quad$ Volume of the cone $=\frac{1}{3} \pi r^2 h$
$=\frac{1}{3} \times \frac{22}{7} \times \frac{32}{10} \times \frac{32}{10} \times \frac{64}{10}=68.65 cm ^3$ (approx.)

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MCQ 331 Mark

A toy is in the form of a right circular cylinder surmounted by a right circular cone as shown in the figure. If $A D=D E=D C=1 cm$, then the volume of the toy $\left(\right.$ in $cm ^3$ ) is

✓
$\frac{4}{3} \pi$
B
$\frac{3}{4} \pi$
C
$3 \pi$
D
$4 \pi$

Answer

Correct option: A.

$\frac{4}{3} \pi$

(a) : Let $r$ be the radius and $h$ be the height of cylindrical part and conical part.
$
\therefore r=h=1 cm
$.
Volume of the toy
$=$ Volume of cylindrical part + Volume of conical part
$
=\pi r^2 h+\frac{1}{3} \pi r^2 h=\pi(1)^2(1)+\frac{1}{3} \pi(1)^2(1)=\frac{4 \pi}{3} cm ^3
$

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MCQ 341 Mark

The internal and external radii of a hemispherical bowl are $r_1$ and $r_2$ respectively. Then, the curved surface area of the bowl is

✓
$2 \pi\left(r_1^2+r_2^2\right)$
B
$\left(r_1^2+r_2{ }^2\right)$
C
$\pi\left(r_1^2+r_2^2\right)$
D
None of these

Answer

Correct option: A.

$2 \pi\left(r_1^2+r_2^2\right)$

(a) : The curved surface area of the hemispherical bowl
$
=2 \pi r_1^2+2 \pi r_2^2=2 \pi\left(r_1^2+r_2^2\right) \text {. }
$

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MCQ 351 Mark

A hemisphere and a cone have equal bases. If their heights are also equal, then the ratio of their curved surface areas is

A
$\sqrt{2}: 3$
B
$1: \sqrt{2}$
✓
$\sqrt{2}: 1$
D
$1: 1$

Answer

Correct option: C.

$\sqrt{2}: 1$

(c) : Let $r$ be the radii of bases of both hemisphere and cone.
$\because \quad$ Height of hemisphere, $h=r$
$\Rightarrow$ Height of cone, $h=r$
$\therefore$ Curved surface area of hemisphere $=2 \pi r^2$
Also, curved surface area of cone $=\pi r l=\pi r \sqrt{r^2+h^2}$
$
=\pi r \sqrt{r^2+r^2}[\because h=r]=\sqrt{2} \pi r^2
$
$\therefore \quad$ Required ratio $=2 \pi r^2: \sqrt{2} \pi r^2=\sqrt{2}: 1$

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MCQ 361 Mark

If the radius and slant height of a cone are in ratio $4: 7$ and its curved surface area is $792 cm ^2$, then its radius will be

A
$10 cm$
B
$11 cm$
✓
$12 cm$
D
$13 cm$

Answer

Correct option: C.

$12 cm$

(c) : Let the radius $(r)$ and the slant height $(l)$ of cone be $4 x$ and $7 x$ respectively.
$\therefore \quad$ Curved surface area of cone $=\pi r l=\pi(4 x)(7 x)$
$\Rightarrow 792=28 \times \frac{22}{7} \times x^2$
[Given]
$\Rightarrow x^2=\frac{792 \times 7}{28 \times 22}=9 \Rightarrow x=3$
$\therefore$ Radius of cone $=4 x=4 \times 3=12 cm$

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MCQ 371 Mark

Two cones have their heights in the ratio $1: 3$ and radii in the ratio $3: 1$. The ratio of their volumes will be

✓
$3: 1$
B
$1: 3$
C
$2: 3$
D
$1: 4$

Answer

Correct option: A.

$3: 1$

(a) : Let height of one cone be $h$ and height of another cone be $3 h$. Radius of one cone is $3 r$ and radius of another cone is $r$.
$\therefore \quad$ Ratio of their volumes $=\frac{\frac{1}{3} \pi(3 r)^2 \times h}{\frac{1}{3} \pi r^2 \times(3 h)}$ $=\frac{9 r^2 h}{3 r^2 h}=\frac{9}{3}=3: 1$

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MCQ 381 Mark

A sphere of maximum volume is cut-out from a solid hemisphere of radius $r$. What is the ratio of the volume of the hemisphere to that of the cut-out sphere?

✓
$4: 1$
B
$2: 3$
C
$1: 1$
D
$1: 4$

Answer

Correct option: A.

$4: 1$

(a) : Here, radius of hemisphere $=r$
$\therefore \quad$ Volume of hemisphere $=\frac{2}{3} \pi r^3$
$\because \quad$ Diameter of sphere of maximum volume that is cutout from it is $r$.
$\Rightarrow$ Radius of sphere $=\frac{r}{2}$
Volume of sphere $=\frac{4}{3} \pi\left(\frac{r}{2}\right)^3=\frac{\pi}{6} r^3$
$\therefore \quad$ Required ratio $=\frac{\frac{2}{3} \pi r^3}{\frac{\pi}{6} r^3}=4: 1$

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MCQ 391 Mark

The volume of the sphere with radius $1.5 cm$ is............