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MCQ Questions on Mensuration 3D for Competitive Exams

Ace your competitive exams with our MCQ questions on Mensuration 3D. Learn to solve volume and surface area problems with questions designed to test your knowledge effectively. Page-8 section-1

Each section contains 100 MCQs on Mensuration 3D .
71
A cone-shaped storage tank's height and radius are 9 and 7 feet, respectively. Determine how much liquid the tank can contain. (Take π = 3.14)
Discuss
Answer & Solution
Answer: Option D
Solution:
Height of cone (h) = 9 ft
Radius (r) = 7 ft
Volume = $$\frac{1}{3}$$πr2h
= $$\frac{1}{3}$$ × 3.14 × 7 × 7 × 9
= 461.58 cube feet
72
The areas of three adjacent faces of a cuboidal solid block of wax are 216 cm2, 96 cm2 and 144 cm2. It is melted and 8 cubes of the same size are formed from it. What is the lateral surface area (in cm2) of 3 such cubes?
Discuss
Answer & Solution
Answer: Option C
Solution:
Areas of three adjacent faces of a cuboidal solid block
L × B = 216
B × H = 96
H × L = 144
Multiply all
(LBH)2 = 216 × 96 × 144
LBH = 36 × 4 × 12
LBH = 1728
LBH = 8 × a3
1728 = 8 × a3
a3 = 216
a = 6
Lateral surface area of 3 such cubes
= 3 × 4a2
= 12 × 6 × 6
= 432 cm2
73
Balls of marbles of diameter 1.4 cm are dropped into a cylindrical beaker containing some water and fully submerged. The diameter of the beaker is 7 cm. Find how many marbles have been dropped in it if the water rises by 5.6 cm?
Discuss
Answer & Solution
Answer: Option B
Solution:
Diameter of beaker = 7 cm
Radius = $$\frac{7}{2}$$ cm
Level of water rises = 5.6 cm
Diameter of a marble = 1.4 cm
∴ Radius = $$\frac{{1.4}}{2}$$ = 0.7 cm
Let n marbles are dropped so,
Volume of n marbles $$ = n \times \frac{4}{3}\pi \times {\left( {0.7} \right)^3}$$
$$\eqalign{ & \Rightarrow n \times \frac{4}{3}\pi \times {\left( {0.7} \right)^3} = n \times {\left( {\frac{7}{2}} \right)^2} \times 5.6 \cr & \Rightarrow n \times \frac{4}{3} \times \frac{7}{{10}} \times \frac{7}{{10}} \times \frac{7}{{10}} = \frac{7}{2} \times \frac{7}{2} \times \frac{{56}}{{10}} \cr & \Rightarrow n = 150 \cr} $$
74
A cylindrical vessel with radius 6 cm and height 5 cm is to be made by melting a number of spherical metal balls of diameter 2 cm. The minimum number of balls needed is:
Discuss
Answer & Solution
Answer: Option B
Solution:
$$\eqalign{ & \pi {r^2}h = \frac{4}{3}\pi {R^3} \times n \cr & {6^2} \times 5 = \frac{4}{3} \times {1^3} \times n \cr & n = 135 \cr} $$
75
2 cm of rain has fallen on a square km of land. Assuming that 50% of the raindrops could have been collected and contained in a pool having a 100 m × 10 m base, by what level would the water level in the pool have increased?
Discuss
Answer & Solution
Answer: Option B
Solution:
Let the increase in level = $$x$$ m
$$\eqalign{ & \Rightarrow \left( {1000 \times 1000 \times \frac{2}{{100}}} \right) \times \frac{1}{2} = 100 \times 10 \times x \cr & \Rightarrow x = 10{\text{ m}} \cr} $$
76
A large solid sphere is melted and moulded to form identical right circular cones with base radius and height same as the radius of the sphere. One of these cones is melted and moulded to form a smaller solid sphere. Then the ratio of the surface area of the smaller sphere to the surface area of the larger sphere is
Discuss
Answer & Solution
Answer: Option D
No explanation is given for this question. Let's Discuss on Board
77
Three solid iron cubes of edges 4 cm, 5 cm and 6 cm are melted together to make a new cube. 62 cm3 of the melted material is lost due to improper handling. The area (in cm2) of the whole surface of the newly formed cube is
Discuss
Answer & Solution
Answer: Option A
Solution:
Volume of all three cube
= 43 + 53 + 63
= 64 + 125 + 216
= 405 cm3
Now, 62 cm3 is lost due to improper handling.
∴ Volume of new cube
= 405 - 62
= 343
(Side of new cube)3 = 343
Side of new cube $$ = \root 3 \of {343} = 7$$
Total surface area of new cube
= 6 × (side)2
= 6 × (7)2
= 6 × 49
= 294 cm2
78
If the radius of a sphere is increased by 2.5 decimetre (dm), then its surface area increases by 110 dm2. What is the volume (in dm3) of the sphere? $$\left( {{\text{take }}\pi = \frac{{22}}{7}} \right)$$
Discuss
Answer & Solution
Answer: Option D
Solution:
$$\eqalign{ & 4\pi \left[ {{{\left( {x + 2.5} \right)}^2} - {x^2}} \right] = 110 \cr & \Rightarrow 4 \times \frac{{22}}{7}\left[ {5x + 6.25} \right] = 110 \cr & \Rightarrow 20x = 10 \cr & \Rightarrow x = \frac{1}{2} \cr & \Rightarrow {\text{Volume}} = \frac{4}{3} \times \frac{{22}}{7} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{{11}}{{21}} \cr} $$
79
Let x cm2 be the surface area and y cm3 be the volume of a sphere such that y = 14x. What is the radius (in cm) of the sphere?
Discuss
Answer & Solution
Answer: Option B
Solution:
Surface area of sphere = 4πr2 = x cm2
Volume of sphere = $$\frac{4}{3}$$πr3 = y cm3
According to the question,
y = 14x
⇒ $$\frac{4}{3}$$πr3 = 14 × 4πr2
⇒ r = 42 cm
80
A conical tent has to accommodate 25 persons. Each person must have 4 m2 of space on the ground and 80 m3 of air to breathe. Find the height of the tent.
Discuss
Answer & Solution
Answer: Option A
Solution:
Mensuration 3D mcq question image
$$\eqalign{ & \pi {r^2} = 25 \times 4 \cr & \pi {r^2} = 100{\text{ }}{{\text{m}}^2} \cr & {r^2} = \frac{{100}}{\pi } \cr & {\text{Volume}} = \frac{1}{3}\pi {r^2}h \cr & 80 \times 25 = \frac{1}{3} \times \pi \times \frac{{100}}{\pi } \times h \cr & h = 60{\text{ m}} \cr} $$