MCQ Questions on Mensuration 3D for Competitive Exams

11

An inverted conical shaped vessel is filled with water to its brim. The height of the vessel is 8 cm and radius of the open end is 5 cm. When a few solid spherical metallic balls each of radius $$\frac{1}{2}$$ cm are dropped in the vessel, 25% water is overflowed. The number of balls is:

A) 100 B) 400 C) 200 D) 150

Discuss

Answer & Solution

Answer: Option A

Solution:

$$\eqalign{ & 25\% {\text{ of }}\left( {{\text{Volume of cone}}} \right) = {\text{ }}x \times {\text{ volume of sphere}} \cr & \frac{1}{4} \times \frac{1}{3}\pi {r^2}h = x \times \frac{4}{3}\pi {R^3} \cr & \frac{1}{4} \times \frac{1}{3}\pi \times 5 \times 5 \times 8 = x \times \frac{4}{3}\pi \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \cr & x = 100 \cr} $$

12

A solid wooden toy is in the shape of a right circular cone mounted on a hemisphere. If the radius of the hemisphere is 4.2 cm and the total height of the toy is 10.2 cm. Find the volume of wooden toy (nearly)?

A) 104 cm³ B) 162 cm³ C) 421 cm³ D) 266 cm³

Discuss

Answer & Solution

Answer: Option D

Solution:

$$\eqalign{ & {\text{Height of the cone}} = 10.2 - 4.2 = 6{\text{ cm}} \cr & {\text{Volume of the toy}} = \frac{1}{3}\pi {r^2}h + \frac{2}{3}\pi {r^3} \cr & = \frac{1}{3}\pi {r^2}\left( {h + 2r} \right) \cr & = \frac{1}{3} \times \frac{{22}}{7} \times {\left( {4.2} \right)^2}\left( {2 \times 4.2 + 6} \right) \cr & = \frac{1}{3} \times \frac{{22}}{7} \times {\left( {4.2} \right)^2} \times 14.4 \cr & = 266{\text{ c}}{{\text{m}}^3}{\text{ }}\left( {{\text{appx}}{\text{.}}} \right) \cr} $$

13

The base of right pyramid is an equilateral triangle, each side of which is 20 cm. Each slant edge is 30 cm. The vertical height (in cm) of the pyramid is:

A) $$10\sqrt 3 $$ B) $$10\sqrt {\frac{{23}}{3}} $$ C) $$5\sqrt {\frac{{23}}{3}} $$ D) $$5\sqrt 3 $$

Discuss

Answer & Solution

Answer: Option B

Solution:

$$\eqalign{ & {\text{Length of eaeh side}} = 20\,{\text{cm }} \cr & {\text{Cireumradius}} = OC = \frac{a}{{\sqrt 3 }} = \frac{{20}}{{\sqrt 3 }} \cr & {\text{Slant edge}} = DC = 30{\text{ cm}} \cr & {\text{In }}\Delta ODC \cr & D{C^2} = O{D^2} + O{C^2} \cr & {30^2} = O{D^2} + {\left( {\frac{{20}}{{\sqrt 3 }}} \right)^2} \cr & O{D^2} = 900 - \frac{{400}}{3} \cr & OD = \sqrt {\frac{{2300}}{3}} \cr & OD = 10\sqrt {\frac{{23}}{3}} \cr} $$

14

The base, of a right prism is a regular hexagon of side 5 cm. If its height is 12√3 cm, then its volume (in cm³) is:

A) 1800 B) 900 C) 1350 D) 675

Discuss

Answer & Solution

Answer: Option C

Solution:

$$\eqalign{ & V = {\text{Base area}} \times {\text{Height}} \cr & V = 6 \times \frac{{\sqrt 3 }}{4} \times {5^2} \times 12\sqrt 3 \cr & V = 18 \times 3 \times 25 \cr & V = 1350 \cr} $$

15

A hemispherical cup of radius 4 cm is filled to the brim with coffee. The coffee is then poured into a vertical cone of radius 8 cm and height 16 cm. The percentage of the volume of the cone that remains empty is:

A) 87.5% B) 80.5% C) 81.6% D) 88.2%

Discuss

Answer & Solution

Answer: Option A

Solution:

$$\eqalign{ & {\text{Volume of coffee}} = \frac{2}{3}\pi {r^3} \cr & = \frac{2}{3} \times \frac{{22}}{7} \times {\left( 4 \right)^3} \cr & = \frac{{128}}{3}\pi {\text{ c}}{{\text{m}}^3} \cr & {\text{Volume of cone}} = \frac{1}{3}\pi {r^2} \times h \cr & = \frac{1}{3}\pi {\left( 8 \right)^2} \times 16 \cr & = \frac{{1024}}{3}\pi \cr & \therefore {\text{Required percentage}} = \frac{{\frac{{1024}}{3} - \frac{{128}}{3}}}{{\frac{{1024}}{3}}} \times 100 \cr & = \frac{{896}}{{1024}} \times 100 \cr & = 87.5\% \cr} $$

16

The radius of the base of a cylinder is 14 cm and its volume is 6160 cm³. The curved surface area (in cm²) is: $$\left( {\pi = \frac{{22}}{7}} \right)$$

A) 880 B) 940 C) 778 D) 660

Discuss

Answer & Solution

Answer: Option A

Solution:

r = 14 cm
πr²h = 6160 cm³
Curved surface area = ?
$$\frac{{22}}{7}$$ × 14 × 14 × h = 6160
h = 10
Curved surface area = 2πrh
= 2 × $$\frac{{22}}{7}$$ × 14 × 10
= 880 cm²

17

A solid metallic sphere of radius 15 cm is melted and recast into spherical balls of radius 3 cm each. What is the ratio of the surface area of the original sphere and the sum of the surface areas of all balls?

A) 1 : 5 B) 5 : 27 C) 1 : 10 D) 3 : 40

Discuss

Answer & Solution

Answer: Option A

Solution:

R³ = nr³
15 × 15 × 15 = n × 3 × 3 × 3
n = 125
(n = number of small spherical balls)
$$\eqalign{ & \frac{{{S_1}}}{{{S_2}}} = \frac{{{R^2}}}{{n{r^2}}} \cr & \frac{{{S_1}}}{{{S_2}}} = \frac{{15 \times 15}}{{125 \times 3 \times 3}} \cr & {S_1}:{S_2} = 1:5 \cr} $$

18

Two solid right cones of equal height and of radii r₁ and r₂ are melted and made to form a solid sphere of radius R. Then the height of the cone is

A) $$\frac{{4{R^2}}}{{r_1^2r_2^2}}$$ B) $$\frac{{4R}}{{{r_1}{r_2}}}$$ C) $$\frac{{4{R^3}}}{{r_1^2 + r_2^2}}$$ D) $$\frac{{{R^2}}}{{r_1^2r_2^2}}$$

Discuss

Answer & Solution

Answer: Option C

Solution:

$$\eqalign{ & {\text{Let the height be }}H \cr & \Rightarrow \frac{1}{3}\pi r_1^2H + \frac{1}{3}\pi r_2^2H = \frac{4}{3}\pi {R^3} \cr & \Rightarrow \frac{1}{3}\pi H\left( {r_1^2 + r_2^2} \right) = \frac{4}{3}\pi {R^3} \cr & \Rightarrow H = \frac{{4{R^3}}}{{r_1^2 + r_2^2}} \cr} $$

19

The curved surface area of a right cylinder is 3696 cm². Its height is three times its radius. What is the capacity (in liters) of the cylinder? $$\left( {{\text{Take }}\pi = \frac{{22}}{7}} \right)$$

A) 25.872 B) 30.87 C) 29.75 D) 19.008

Discuss

Answer & Solution

Answer: Option A

Solution:

$$\eqalign{ & r:h = 1:3 \cr & 2\pi rh = 3696 \cr & 2 \times \frac{{22}}{7} \times x \times 3x = 3696 \cr & 3{x^2} = \frac{{3696 \times 7}}{{44}} \cr & {x^2} = \frac{{1232 \times 7}}{{44}} \cr & {x^2} = \frac{{112 \times 7}}{4} \cr & {x^2} = 28 \times 7 \cr & x = 14 \cr & {\text{Volume}} = \pi {r^2}h \cr & = \frac{{22}}{7} \times {14^2} \times 42 \cr & = \frac{{22}}{7} \times 14 \times 14 \times 42 \cr & = 25872{\text{ c}}{{\text{m}}^3} \cr & = 25.872{\text{ L}} \cr} $$

20

Water flows into a tank which is 200 m long and 150 m wide through a pipe of cross-section 0.3 m × 0.2 m at 20 km/hour. Then the time (in hours) for the water level in the tank to reach 8 m is

A) 50 B) 120 C) 150 D) 200

Discuss

Answer & Solution

Answer: Option D

Solution:

Let the number of hours be x
⇒ (0.3 × 0.2 × 20000) × x = 200 × 150 × 8
⇒ x = $$\frac{{200 \times 150 \times 8}}{{3 \times 2 \times 200}}$$
⇒ x = 200 hrs.