MCQ Questions on Mensuration 3D for Competitive Exams

1

The radius and height of a right circular cone are in the ratio 3 : 4. If its curved surface area (in cm²) is 240π. Then its volume (in cm³) is:

A) 2304π B) 384π C) 1536π D) 768π

Discuss

Answer & Solution

Answer: Option D

Solution:

r : h = 3 : 4
$$l$$ : r = 5 : 3
Mensuration 3D mcq question image

$$\eqalign{ & {\text{Curved surface area}} = \pi rl \cr & 240\pi = \pi \times 3x \times 5x \cr & {x^2} = 16 \cr & x = 4 \cr & \therefore r = 3 \times 4 = 12 \cr & {\text{Volume}} = \frac{1}{3}\pi {r^2}h \cr & = \frac{1}{3}\pi \times 144 \times 16 \cr & = 768\pi \cr} $$

2

A cylindrical pencil of diameter 1.2 cm has one of its end sharpened into a conical shape of height 1.4 cm. The volume of the material removed is

A) 1.057 cm³ B) 4.224 cm³ C) 1.056 cm³ D) 42.24 cm³

Discuss

Answer & Solution

Answer: Option C

Solution:

$$\eqalign{ & {\text{Required volume removed}} \cr & = \pi {r^2}h - \frac{1}{3}\pi {r^2}h \cr & = \frac{2}{3}\pi {r^2}h \cr & = \frac{2}{3} \times \frac{{22}}{7} \times .6 \times .6 \times 1.4 \cr & = 1.056{\text{ c}}{{\text{m}}^3} \cr} $$

3

A sphere and a hemisphere have the same volume. The ratio of their curved surface area is:

A) $${2^{\frac{3}{2}}}:1$$ B) $${2^{\frac{2}{3}}}:1$$ C) $${4^{\frac{2}{3}}}:1$$ D) $${2^{\frac{1}{3}}}:1$$

Discuss

Answer & Solution

Answer: Option D

Solution:

Let the radius of hemisphere and sphere be 'r' and 'R'
$$\eqalign{ & \Rightarrow \frac{4}{3}\pi {R^3} = \frac{2}{3}\pi {r^3} \cr & \frac{{{R^3}}}{{{r^3}}} = \frac{1}{2} \cr & \frac{R}{r} = \frac{1}{{\root 3 \of 2 }} \cr & \Rightarrow {\text{Ratio of curved surface area}} \cr & = \frac{{4\pi {R^2}}}{{2\pi {r^2}}} \cr & = \frac{{2{R^2}}}{{{r^2}}} \cr & = \frac{{2 \times 1}}{{{{\left( {\root 3 \of 2 } \right)}^2}}} \cr & = \frac{2}{{{{\left( 2 \right)}^{\frac{2}{3}}}}} \cr & \Rightarrow \frac{R}{r} = \frac{{{2^{\frac{1}{3}}}}}{1} \cr} $$

4

A tank 40 m long, 30 m broad and 12 m deep is dug in a field 1000 m long and 30 m wide. By how much will the level of the field rise if the earth dug out of the tank is evenly spread over the field?

A) 2 metre B) 1.2 metre C) 0.5 metre D) 5 metre

Discuss

Answer & Solution

Answer: Option C

Solution:

Volume of earth taken out = 40 × 30 × 12 = 14400 m³
Area of rectangular field = 1000 × 30 = 30000 m²
Area of region of tank = 40 × 30 = 1200 m²
Remaining area = 30000 - 1200 = 28800 m²
Increase in height $$ = \frac{{14400}}{{28800}} = 0.5{\text{ m}}$$

5

A tank is in the form of a cuboid with length 12 m. If 18 kilolitre of water is removed from it, the water level goes down by 30 cm. What is the width (in m) of the tank?

A) 5 B) 4.5 C) 4 D) 5.5

Discuss

Answer & Solution

Answer: Option A

Solution:

Volume of cuboid = $$l$$bh
12 × b × h - 18 = 12 × b × (h - 0.3)
12bh - 18 = 12bh - 3.6b
b = 5 m

6

A solid brass sphere of radius 2.1 dm is converted into a right circular cylindrical rod of length 7 cm. The ratio of total surface areas of the rod to the sphere is

A) 3 : 1 B) 1 : 3 C) 7 : 3 D) 3 : 7

Discuss

Answer & Solution

Answer: Option C

Solution:

$$\eqalign{ & \frac{4}{3}\pi {R^3} = \pi {r^2}h \cr & \frac{4}{3}\pi \times 21 \times 21 \times 21 = \pi \times {R^2} \times 7 \cr & {R^2} = 21 \times 21 \times 4 \cr & R = 42{\text{ cm}} \cr & \frac{{{\text{total surface area of rod}}}}{{{\text{total surface area of sphere}}}} \cr & = \frac{{2\pi r\left( {h + r} \right)}}{{4\pi {R^2}}} \cr & = \frac{{42\left( {42 + 7} \right)}}{{2 \times 21 \times 21}} \cr & = \frac{{49}}{{21}} \cr & = \frac{7}{3} \cr & = 7:3 \cr} $$

7

A solid cylinder has radius of base 14 cm and height 15 cm. 4 identical cylinders are cut from each base as shown in the given figure. Height of small cylinder is 5 cm. What is the total surface area (in cm²) of the remaining part?
Mensuration 3D mcq question image

A) 8338 B) 8951 C) 3432 D) 7979

Discuss

Answer & Solution

Answer: Option C

Solution:

Total surface area of remaining part 6
= 16πrh + 2πR² - 8πr² + 2πRH + 8πr²
= 2π(8rh + R² - 4r² + RH + 4r²)
= 2π(8 × 3.5 × 5 + (14)² - 4 × 3.5 × 3.5 + 14 × 15 + 4 × (3.5)²)
= 3432

8

A cylindrical vessel whose base is horizontal and is of internal radius 3.5 cm contains sufficient water so that when a solid sphere is placed inside, water just covers the sphere. The sphere fits in the Cylinder exactly. The depth of water in the vessel before the sphere was put, is

A) $$\frac{{35}}{3}{\text{ cm}}$$ B) $$\frac{{17}}{3}{\text{ cm}}$$ C) $$\frac{7}{3}{\text{ cm}}$$ D) $$\frac{{14}}{3}{\text{ cm}}$$

Discuss

Answer & Solution

Answer: Option C

Solution:

Height of water after ball is immersed = 3.5 × 2 = 7 cm
$$\eqalign{ & \Rightarrow {\text{Volume of water}} = \pi {r^2}h - \frac{4}{3}\pi {r^3} \cr & = \pi {r^2}\left( {h - \frac{4}{3}r} \right) \cr & = \frac{{22}}{7} \times 3.5 \times 3.5\left( {7 - \frac{4}{3} \times 3.5} \right) \cr & = 11 \times 3.5\left( {\frac{7}{3}} \right) \cr & = \frac{{269.5}}{3}{\text{ c}}{{\text{m}}^3} \cr} $$
Volume of water before ball was immersed
$$\eqalign{ & \Rightarrow \pi {\left( {3.5} \right)^2} \times h = \frac{{269.5}}{3} \cr & h = \frac{{269.5 \times 7}}{{3 \times 3.5 \times 3.5 \times 22}} \cr & h = \frac{7}{3}{\text{ cm}} \cr} $$

9

A hemispherical depression of diameter 4 cm is cut out from each face of a cubical block of sides 10 cm. Find the surface area of the remaining solid (in cm²). $$\left( {{\text{Use }}\pi = \frac{{22}}{7}} \right)$$

A) $$900\frac{4}{7}$$ B) $$112\frac{4}{7}$$ C) $$675\frac{3}{7}$$ E) $$713\frac{1}{7}$$

Discuss

Answer & Solution

Answer: Option C

Solution:

Side of cubical block = 10 cm
∴ Area of each face of cubical block = 10² = 100 cm²
Radius of Hemisphere = $$\frac{4}{2}$$ = 2 cm
Surface area of hemisphere
$$\eqalign{ & = 2\pi {r^2} \cr & = 2 \times \frac{{22}}{7} \times {2^2}{\text{ c}}{{\text{m}}^2} \cr & = \frac{{176}}{7}{\text{ c}}{{\text{m}}^2} \cr} $$
Total surface area of hemisphere $$ = 6 \times \frac{{176}}{7} = \frac{{1056}}{7}{\text{ c}}{{\text{m}}^2}$$
Remaining surface area of each face of cubical block
$$\eqalign{ & = {10^2} - 2\pi r \cr & = 100 - 2 \times \frac{{22}}{7} \times 2{\text{ c}}{{\text{m}}^2} \cr & = 100 - \frac{{88}}{7}{\text{ c}}{{\text{m}}^2} \cr & = \frac{{612}}{7}{\text{ c}}{{\text{m}}^2} \cr} $$
∴ Total surface area of 6 remaining cubical block $$ = 6 \times \frac{{612}}{7} = \frac{{3672}}{7}{\text{ c}}{{\text{m}}^2}$$
∴ Surface area of remaining solid
$$\eqalign{ & = \left( {\frac{{1056}}{7} + \frac{{3672}}{7}} \right){\text{c}}{{\text{m}}^2} \cr & = \frac{{4728}}{7}{\text{ c}}{{\text{m}}^2} \cr & = 675\frac{3}{7}{\text{ c}}{{\text{m}}^2} \cr} $$

10

The lateral surface area of frustum of a right circular cone if the area of its base is 16π cm² and the diameter of circular upper surface is 4 cm and slant height 6 cm, will be

A) 30π cm² B) 48π cm² C) 36π cm² D) 60π cm²

Discuss

Answer & Solution

Answer: Option C

Solution:

Base Area = 16π
πR² = 16π
R = 4
Given r = 2
∵ ΔABC ≅ ΔADE
$$\eqalign{ & \therefore \frac{{{\text{BC}}}}{{{\text{DE}}}} = \frac{{{\text{AC}}}}{{{\text{AE}}}} \cr & \frac{4}{8} = \frac{6}{{{\text{AE}}}} \cr & {\text{AE}} = 12 = {\text{L}} \cr} $$
Surface Area of frustum
= πRL - πr$$l$$
= π × 4 × 12 - π × 2 × 6
= 48π - 12π
= 36π