MCQ Questions on Mensuration 3D for Competitive Exams

91

The sum of the radius of spheres A and B is 14 cm, the radius of A being larger than that of B. The difference between their surface areas is 112π. What is the ratio of the volumes of A and B?

A) 125 : 64 B) 64 : 27 C) 27 : 8 D) 8 : 1

Discuss

Answer & Solution

Answer: Option B

Solution:

Let
Radius of A = R
Radius of B = r
R + r = 14 cm
4π(R² - r²) = 112π
(R - r)( R + r) = 28
14(R - r) = 28
R - r = 2
R = 8
r = 6
V → R³ : r³ = 4³ : 3³ = 64 : 27

92

If V₁, V₂ and V₃ be the volumes of a right circular cone. A sphere and a right circular cylinder having the same radius and same height then

A) $${V_1} = \frac{{{V_2}}}{4} = \frac{{{V_3}}}{3}$$ B) $$\frac{{{V_1}}}{2} = \frac{{{V_2}}}{3} = {V_3}$$ C) $$\frac{{{V_1}}}{3} = \frac{{{V_2}}}{2} = {V_3}$$ D) $$6{V_1} = 3{V_2} = 2{V_3}$$

Discuss

Answer & Solution

Answer: Option D

Solution:

$$\eqalign{ & {\text{Volume of cone }}{V_1} = \frac{1}{3}\pi {r^2}h \cr & {\text{Volume of sphere }}{V_2} = \frac{4}{3}\pi {r^3} \cr & = \frac{2}{3}\pi {r^2}\left( {2r} \right)\,\,\,\,\,\left[ {\therefore h = 2r} \right] \cr & = \frac{2}{3}\pi {r^2}h \cr & {\text{Volume of cylinder }}{V_3} = \pi {r^2}h \cr & {\text{Cone}}\left( {\pi {r^2}h} \right) = 3{V_1} \cr & {\text{Sphere}}\left( {\pi {r^2}h} \right) = \frac{3}{2}{V_2} \cr & {\text{Cylinder}}\left( {\pi {r^2}h} \right) = {V_3} \cr & 3{V_1} = \frac{3}{2}{V_2} = {V_3} \cr & 6{V_1} = 3{V_2} = 2{V_3} \cr} $$

93

How much iron sheet (in m²) will be needed to construct a rectangular tank measuring 10 m × 8 m × 6 m, if a circular opening of radius one metre is to be left at the top of the tank? (correct to one decimal place)

A) 371.6 B) 370.4 C) 372.9 D) 370.8

Discuss

Answer & Solution

Answer: Option C

Solution:

2($$l$$b + bh + h$$l$$) - πr²
= 2(10 × 8 + 8 × 6 + 6 ×10) - π
= 2(80 + 48 + 60)
= 2 × 188 - $$\frac{{22}}{7}$$
= 376 - 3.1
= 372.9 m²

94

The height of a solid right circular cylinder is 6 metres and three times the sum of the area of its two end faces is twice the area of its curved surface. The radius of its base (in metre) is

A) 4 B) 2 C) 8 D) 10

Discuss

Answer & Solution

Answer: Option A

Solution:

Height of cylinder h = 6 metre
Let radius of cylinder = r metre
∴ curved surface area = 2πrh
Area of end face = πr²
⇒ Total area of two end faces
⇒ Mensuration 3D mcq question image

⇒ 2πr²
Given that 3 × 2πr² = 2 × 2πrh
3r = 2h
r = 4 cm
∴ Radius of base = 4 cm

95

5 persons live in a tent. If each person requires 16 m² of floor area and 100 m³ space for air then the height of the cone of smallest size to accommodate these persons would be?

A) 16 m B) 18.75 m C) 10.25 m D) 20 m

Discuss

Answer & Solution

Answer: Option B

Solution:

Let the height of cone h metre
⇒ Total area of ground will be required = 5 × 16 m² = 80 m²
⇒ Total volume of air is needed = 100 × 5 m³ = 500 m³
According to the question
⇒ Volume of cone = 500 m³
⇒ $$\frac{1}{3}$$ × area of ground × height = 500
⇒ $$\frac{1}{3}$$ × πr² × h = 500
⇒ $$\frac{1}{3}$$ × 80 × h = 500
⇒ Height = $$\frac{{500 \times 3}}{{80}}$$
⇒ Height of cone = 18.75 metres

96

A circus tent is cylindrical up to a height of 3 m and conical above it. If its diameter is 105 m and the slant height of the conical part is 63 m, then the total area of the canvas required to make the tent is $$\left( {{\text{take }}\pi = \frac{{22}}{7}} \right)$$

A) 11385 m² B) 10395 m² C) 9900 m² D) 990 m²

Discuss

Answer & Solution

Answer: Option A

Solution:

∴ Radius of cone $$ = \frac{{105}}{2}{\text{m}}$$
Slant height of cone = 63 m
⇒ Curved surface area of cone
$$\eqalign{ & = \pi rl \cr & = \frac{{22}}{7} \times \frac{{105}}{2} \times 63 \cr & = 10395{\text{ }}{{\text{m}}^2} \cr} $$
⇒ Radius of cylinder $$ = \frac{{105}}{2}{\text{m}}$$
Height = 3 m (given)
∴ Curved surface area of cylinder
$$\eqalign{ & = 2\pi rh \cr & = 2 \times \frac{{22}}{7} \times \frac{{105}}{2} \times 3 \cr & = 990{\text{ }}{{\text{m}}^2} \cr} $$
∴ Total curved area of structure
⇒ Curved area of cone + Curved area of cylinder
= 10395 + 990
= 11385 m²
∴ Total area of canvas = 11385 m²

97

A cone, a hemisphere and a cylinder stand on equal bases and have the same height. The ratio of their respective volume is

A) 1 : 2 : 3 B) 2 : 1 : 3 C) 1 : 3 : 2 D) 3 : 1 : 2

Discuss

Answer & Solution

Answer: Option A

Solution:

In this case height of cylinder and cone is equal to the radius of hemisphere
⇒ h = r
Ratio of volumes
\[\begin{array}{*{20}{c}} {}&{{\text{Cone}}}&{}&{{\text{Hemisphere}}}&{}&{{\text{Cylinder}}} \\ = &{\frac{1}{3}\pi {r^2} \times r}&:&{\frac{2}{3}\pi {r^3}}&:&{\pi {r^2} \times r} \\ = &1&:&2&:&3 \end{array}\]

98

Six cubes each edge 2 cm, are joined end to end. What is the total surface area of the resulting cuboid in cm²?

A) 96 B) 144 C) 104 D) 128

Discuss

Answer & Solution

Answer: Option C

Solution:

$$l$$ = 12
b = 2
h = 2
Surface Area = 2($$l$$b + bh + h$$l$$)
= 2(12 × 2 + 2 × 2 + 2 × 12)
= 104 cm

99

The radius of the base of a conical tent is 16 metres. If $$427\frac{3}{7}$$ sq. metre canvas is required to construct the tent, then the slant height of the tent is: $$\left( {{\text{take }}\pi = \frac{{22}}{7}} \right)$$

A) 17 metre B) 15 metre C) 19 metre D) 8.5 metre

Discuss

Answer & Solution

Answer: Option D

Solution:

Radius of cone (r) = 16 metre (given)
Let slant height = $$l$$ metre
Curved surface area
$$\eqalign{ & \pi rl = 427\frac{3}{7}{\text{ }}{{\text{m}}^2}\,\,\left( {{\text{given}}} \right) \cr & \frac{{22}}{7} \times 16 \times l = \frac{{2992}}{7} \cr & l = \frac{{2992}}{{22 \times 16}} \cr & l = 8.5{\text{ metre}} \cr} $$

100

The length of a metallic pipe is 7.56 m. Its external and internal radii are 2.5 cm and 1.5 cm respectively. If 1 cm³ of the metal weight 7.5 g, then the weight of the pipe is: $$\left( {{\text{Take }}\pi = \frac{{22}}{7}} \right)$$

A) 72.82 kg B) 70.14 kg C) 71.28 kg D) 69.68 kg

Discuss

Answer & Solution

Answer: Option C

Solution:

V = π(R² - r²) × h
V = $$\frac{{22}}{7}$$ × 756(2.5² - 1.5²)
V = 22 × 108 × 4 × 1
V = 9484 cm³
The weight of the pipe
$$\eqalign{ & = \frac{{9484 \times 7.5}}{{1000}} \cr & = 71.28{\text{ kg}} \cr} $$