M.C.Q (1 Marks)

MCQ 511 Mark

What is the total surface area of a solid hemisphere of diameter 'd'?

A
$3 \pi d^2$
B
$2 \pi d^2$
C
$\frac{1}{2} \pi d^2$
✓
$\frac{3}{4} \pi d^2$

Answer

Correct option: D.

$\frac{3}{4} \pi d^2$

(d)
Total surface area of a solid hemisphere of radius r is $S=3 \pi r^2=3 \pi\left(\frac{d}{2}\right)^2=\frac{3}{4} \pi d^2$

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MCQ 521 Mark

Water in a river which is 3 m deep and 40 m wide is flowing at the rate of 2km/hr. How much water will fall into the sea in 2 minutes?

A
$800 m^3$
B
$4000 m^3$
✓
$8000 m^3$
D
$2000 m^3$

Answer

Correct option: C.

$8000 m^3$

(c)
Length of water column formed in two minutes in a river flowing at the rate of $2 km / hr =\left(\frac{2000}{60} \times 2\right) m =\frac{200}{3} m$
∴ Volume of water that falls into the sea in 2 minutes
= Volume of a cuboid of dimensions 3 m x 40 m x $\frac{200}{3} m=\left(3 \times 40 \times \frac{200}{3}\right) m ^3=8000 m^3$

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MCQ 531 Mark

The volume of a right circular cone whose area of the base is $156 cm^2$ and the vertical height is 8 cm, is

A
$2496 cm^3$
B
$1248 cm^3$
C
$1664 cm^3$
✓
$416 cm^3$

Answer

Correct option: D.

$416 cm^3$

(d)
Volume of cone = 1/3 (Area of base x Height) $=\frac{1}{3} \times 156 \times 8 cm^3=416 cm^3$

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MCQ 541 Mark

The sum of the length, breadth and height of a cuboid is 6sqrt(3) cm and the length of its diagonal is $2 \sqrt{3} cm$.The total surface area of the cuboid is

A
$48 cm^2$
B
$72 cm^2$
✓
$96 cm^2$
D
$108 cm^2$

Answer

Correct option: C.

$96 cm^2$

(c)
Let the length, breadth and height of the cuboid be a, b and c centimeters long. It is given that
$a+b+c=6 \sqrt{3}$ and $\sqrt{a^2+b^2+c^2}=2 \sqrt{3}$
$\Rightarrow \quad(a+b+c)^2=108$ and $a^2+b^2+c^2=12$
$\therefore \quad(a+b+c)^2=a^2+b^2+c^2+2(a b+b c+c a)$
$\Rightarrow \quad 108=12+$ Surface area of the cuboid $\Rightarrow$ Surface area of the cuboid $=96 cm^2$

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MCQ 551 Mark

Three solid cubes have a face diagonal 4sqrt(2) cm each. Three other solid cubes have a face diagonal of $8 \sqrt{2} cm$ each.All the cubes are melted together to form a cube. The length of an edge of the cube formed is

✓
12 cm
B
24 cm
C
18 cm
D
26 cm

Answer

Correct option: A.

12 cm

(a)
Let the lengths of edges of two sets of three cones each be a cm and b cm respectively. Then,
$\sqrt{2} a=4 \sqrt{2}$ and $\sqrt{2} b=8 \sqrt{2} \Rightarrow a=4$ and $b=8$
Let the length of an edge of new cube formed be c cm. Then,
$c^3=3 \times 4^3+3 \times 8^3=192+1536=1728=12^3 \Rightarrow c=12 cm$

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MCQ 561 Mark

If three cubes of same metal whose edges are 6 cm, 8 cm and 10 cm melted and formed into a single cube, then the length of the diagonal of the larger cube formed is

A
$4 \sqrt{3} cm$
B
$15 \sqrt{3} cm$
✓
$12 \sqrt{3} cm$
D
$1\sqrt{3} cm$

Answer

Correct option: C.

$12 \sqrt{3} cm$

(c)
Let the length of an edge of the larger cube be a cm. Then,
$a^3=6^3+8^3+10^3=216+512+1000=1728=12^3 \Rightarrow a=12$
Length of the diagonal of the larger cube $=\sqrt{3} a=12 \sqrt{3} cm$

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MCQ 571 Mark

The edge of a cube whose volume is equal to that of a cuboid of dimensions 8 cm x 4 cm x 2 cm is

A
6 cm
✓
4 cm
C
2 cm
D
6 cm

Answer

Correct option: B.

4 cm

(b)
Let the length of an edge of the cube be a cm. Then,
$a^3=8 \times 4 \times 2 \Rightarrow a^3=2^3 \times 2^2 \times 2 \Rightarrow a^3=2^6 \Rightarrow a=2^2=4 cm$

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MCQ 581 Mark

The volume of the largest right circular cone that can be carved out from a solid cube of edge 2 cm is

A
$\frac{4 \pi}{3} cm^3$
B
$\frac{5 \pi}{3} cm^3$
C
$\frac{8 \pi}{3} cm^3$
✓
$\frac{2 \pi}{3} cm^3$

Answer

Correct option: D.

$\frac{2 \pi}{3} cm^3$

(d)
Height of the right circular cone = 2 cm., Radius of the cone = 1 cm.
$\therefore \quad$ Volume $=\frac{1}{3} \pi \times 1^2 \times 2 cm^3=\frac{2}{3} \pi cm^3$

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MCQ 591 Mark

A solid consists of a circular cylinder surmounted by a right circular cone. The height of the cone is h. If the total volume of the solid is 3 times the volume of the cone, then the height of the circular cylinder is

A
$2 h$
B
$\frac{3}{2} h$
C
$\frac{h}{2}$
✓
$\frac{2 h}{3}$

Answer

Correct option: D.

$\frac{2 h}{3}$

(d)
Let H be the height of the cylinder. It is given that $\frac{1}{3} \pi r^2 h+\pi r^2 H=3\left(\frac{1}{3} \pi r^2 h\right)$
⇒ $\pi r^2 h=\frac{2}{3} \pi r^2 h$
⇒ $H=\frac{2}{3} h$

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MCQ 601 Mark

A solid is hemispherical at the bottom and conical above. If the surface areas of the two parts are equal, then the ratio of its radius and height of its conical part is

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

Answer

(c)
It is given that
$2 \pi r^2=\pi r l \Rightarrow 2 r=1$
$\therefore \quad l^2=r^2+h^2 \Rightarrow h=\sqrt{l^2-r^2}=\sqrt{4 r^2-r^2}=\sqrt{3} r$
Hence, $r: h=r: \sqrt{3} r=1: \sqrt{3}$

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MCQ 611 Mark

If a cone is cut into two parts by a horizontal plane passing through the mid-point of its axis, then the ratio of the volumes of the upper part and the cone is

✓
1:8
B
1:5
C
1:7
D
1:6

Answer

Correct option: A.

1:8

(a)
Let r be the radius of base and h be the height of the cone. It is given that $V O^{\prime}=\frac{h}{2}$
Triangles VO'A' and VOA are similar.
$\therefore \quad \frac{V O^{\prime}}{V O}=\frac{O^{\prime} A^{\prime}}{O A}=\frac{h^{\prime} / 2}{h}=\frac{O^{\prime} A^{\prime}}{r}=O^{\prime} A^{\prime}=\frac{r}{2}$
Thus,$V A^{\prime} B^{\prime}$is a cone of height h/2 and base radius r/2 Let $V_1$ be its volumes.Then,
$V_1=\frac{\pi}{3}\left(\frac{r}{2}\right)^2\left(\frac{h}{2}\right)=\frac{1}{8}\left(\frac{\pi}{3} r^2 h\right)=\frac{V}{8}$, where $V$ is the volume of cone $V A B$.
$\Rightarrow V_1: V=1: 8$

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MCQ 621 Mark

The ratio of lateral surface area to that total surface area of a cylinder with base diameter 1.6 m and height 20 cm is

A
1:7
✓
1:5
C
7:1
D
5:1

Answer

Correct option: B.

1:5

(b)
We have, r = radius of base = 0.8m = 80 cm and height = 20 cm.
Let $S_1$ and $S_2$ denote respectively the lateral and total surface areas. Then,
$S_1=2 \pi r h$ and $S_2=2 \pi r(h+r) \Rightarrow S_1: S_2=\frac{h}{h+r}=\frac{20}{20+80}=\frac{1}{5}$

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MCQ 631 Mark

The surface area of a cube is $216 cm^2$,its volume is

A
$144 cm^3$
B
$196 cm^3$
C
$212 cm^3$
✓
$216 cm^3$

Answer

Correct option: D.

$216 cm^3$

(d)
Let the length of each side of the cube be a cm. Then,
Surface area = $216 cm^2 \Rightarrow 6 a^2=216 \Rightarrow a^2=36 \Rightarrow a=6$
Volume $=a^3=6^3 cm^3=216 cm^3$

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MCQ 641 Mark

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

✓
36 m
B
32 m
C
38 m
D
34 m

Answer

Correct option: A.

36 m

(a)
Let the length of the wire be 1 cm. Then,
Volume of metal in a wire of radius 0.1cm = Volume of metal in sphere is radius 3cm
$\Rightarrow \quad \pi(0.1)^2 \times l=\frac{4}{3} \pi \times 3^3 \Rightarrow \frac{l}{100}=36 \Rightarrow l=3600 cm=36 m$

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MCQ 651 Mark

If the height and base radius of a cone, each is increased by 50%, then the ratio between the volume of the given cone and the new cone is

✓
8:27
B
27:8
C
4:9
D
2:3

Answer

Correct option: A.

8:27

(a)
Letr and h denote respectively the base radius and height of the cone. Then, its volume $V_1$ is given by $V=\frac{1}{3} \pi r^2 h$.
New radius $=r+50 \%$ of $r=r+\frac{r}{2}=\frac{3}{2} r$, New height $=h+50 \%$ of $h=h+\frac{h}{2}=\frac{3}{2} h$
Let $V_2$be the volume of the new cone. Then,
$V_2=\frac{1}{3} \pi\left(\frac{3}{2},\right)^2 \times\left(\frac{3}{2} h\right)=\frac{27}{8}\left(\frac{1}{3} \pi r^2 h\right)=\frac{27}{8} V_1 \Rightarrow \frac{V_1}{V_2}=\frac{8}{27}$

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MCQ 661 Mark

The capacity of the cylindrical vessel with the hemispherical bottom portion raised upwards as shown in Fig., is

✓
$\frac{\pi}{3}(3 h-2 r) r^2$
B
$\frac{\pi}{3}(3 h+2 r) r^2$
C
$\frac{\pi}{2}(2 h-3 r) r^2$
D
$\frac{\pi}{2}(2 h+3 r) r^2$

Answer

Correct option: A.

$\frac{\pi}{3}(3 h-2 r) r^2$

(a)
Let V be the capacity of the vessel. Then, $V=\pi r^2 h-\frac{2}{3} \pi r^3=\frac{\pi}{3}(3 h-2 r) r^2$

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MCQ 671 Mark

The radius of the base and height of a cone are 4 cm and 9 cm respectively. If its height is decreased and base radius is increased each by 2 cm, then the ratio of the volume of the new cone to that of the original cone is

A
5:2
✓
7:4
C
9:2
D
8:3

Answer

Correct option: B.

7:4

(b)
Let $V_1$ and $V_2$ be the volumes of old and new cones. Then,
$\frac{V_1}{V_2}=\frac{\frac{1}{3} \pi \times 4^2 \times 9}{\frac{1}{3} \pi \times 6^2 \times 7}=\frac{16 \times 9}{36 \times 7}=\frac{4}{7} \Rightarrow \frac{V_2}{V_1}=\frac{7}{4}=7: 4$

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MCQ 681 Mark

A solid sphere is cut into two hemispheres. The ratio of the surface areas of sphere to that of two hemispheres taken together is

A
1:1
B
1:4
✓
2:3
D
3:2

Answer

Correct option: C.

2:3

(c)
Letr be the radius of the solid sphere. Then,
S = Surface area of sphere $=4 \pi r^2$
$S_1=$Sum of the surface areas of two hemispheres $=3 \pi r^2+3 \pi r^2=6 \pi r^2$
$S: S_1=4 \pi r^2: 6 \pi r^2=2: 3$

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MCQ 691 Mark

If a solid sphere surface area $48 cm^2$ is bisected into two hemisphere , then the total surface area of any one of the hemisphere is

A
$48 cm^2$
B
$60 cm^2$
C
$24 cm^2$
✓
$36 cm^2$

Answer

Correct option: D.

$36 cm^2$

(d)
Letr be the radius of solid sphere. It is given that its total surface area is $48 cm^2$.
$4 \pi r^2=48 \Rightarrow \pi r^2=12 \Rightarrow 3 \pi r^2=36$
Hence, total surface area of one of the hemispheres is $36 cm^2$

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MCQ 701 Mark

If 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 the ratio between the volume of cylinder to that of the cone is

A
7:5
B
5:7
C
8:9
✓
9:8

Answer

Correct option: D.

9:8

(d)
Let the radii of the bases of a cylinder and a cone be $r_1$ and $r_2$ respectively. Further, let the heights of the cylinder and cone be $h_1$ and $h_2$ respectively. It is given that $r_1: r_2=3: 4$ and $h_1$ and $h_2$= 2 : 3 Let $V_1$ and $V_2$ be the volumes of cylinder and cone respectively. Then,
$V_1=\pi r_1^2 h_1$ and $V_2=\frac{1}{3} \pi r_1^2 h_2 \Rightarrow \frac{V_1}{V_2}=3\left(\frac{r_1}{r_2}\right)^2\left(\frac{h_1}{h_2}\right)=3\left(\frac{3}{4}\right)^2 \times \frac{2}{3}=9 \cdot 8$

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MCQ 711 Mark

If the perimeters of the bases of two right circular cones are in the ratio 3: 4 and their volumes are in the ratio 9: 32, then the ratio of their heights is

A
1:3
B
2:1
✓
1:2
D
1:3

Answer

Correct option: C.

1:2

(c)
Let the radii of two cones be $r_1$ and $r_2$ and their heights be $h_1$ and $h_2$.Then, perimeters of their bases are $2 \pi r_1$ and $2 \pi r_2$. Volumes of the cones are $\frac{1}{3} \pi r_1^2 h_1$ and $\frac{1}{3} \pi r_2^2 h_2$.respectively. It is given that $\begin{array}{l}2 \pi r_1: 2 \pi r_2=3: 4 \Rightarrow r_1: r_2=3: 4 \Rightarrow \frac{r_1}{r_2}=\frac{3}{4} \\ \frac{1}{3} \pi r_1^2 h_1: \frac{1}{3} \pi r_2^2 h_2=9: 32 \Rightarrow r_1^2 h_1: r_2^2 h_2=9: 32 \Rightarrow \frac{r_1^2 h_1}{r_1 h_1}=\frac{9}{32}\end{array}$
From (i) and (ii), we obtain
$\frac{9}{16} \frac{h_1}{h_2}=\frac{9}{32} \Rightarrow \frac{h_1}{h_2}=\frac{1}{2} \Rightarrow h_1: h_2=1: 2$

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MCQ 721 Mark

The curved surface area of a cone having height 24 cm and radius 7 cm, is

A
$528 cm^2$
B
$1056 cm^2$
✓
$550 cm^2$
D
$500 cm^2$

Answer

Correct option: C.

$550 cm^2$

(c)
Let l be the slant height of the cone. Then, $I=\sqrt{7^2+24^2}=\sqrt{625}=25 cm$
$\therefore \quad$ Curved surface area $=\pi r l=\frac{22}{7} \times 7 \times 25 cm^2=550 cm^2$

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MCQ 731 Mark

Curved surface area of a cylinder of height 5 cm is $94.2 cm^2$. Radius of the cylinder is (Take pi = 3.14 )

A
2 cm
✓
3 cm
C
2.9 cm
D
6 cm

Answer

Correct option: B.

3 cm

(b)
Letr cm be the length of the radius of cylinder of height h = 5cm .
Curved surface area $=94.2 cm^2 \Rightarrow 2 \pi h=94.2 \Rightarrow 2 \times 3.14 \times r \times 5=94.2 \Rightarrow r=3 cm$

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MCQ 741 Mark

If the diagonal of a cube is 17.32 cm, then its volume is

✓
$1000 cm^3$
B
$1732 cm^3$
C
$173.2 cm^3$
D
$10000 cm^3$

Answer

Correct option: A.

$1000 cm^3$

(a)
Let the length of each edge of the cube be a cm. Then, its diagonal is $\sqrt{3} a$. Given that
$\begin{array}{ll} & \sqrt{3} a=17.32 \Rightarrow \sqrt{3} a=10 \times 1.732 \Rightarrow \sqrt{3} a=10 \times \sqrt{3} \Rightarrow a=10 \\ \therefore & \text { Volume of the cube }=a^3=10^3 cm^3=1000 cm^3\end{array}$

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Maths M.C.Q (1 Marks)