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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-6 section-1

Each section contains 100 MCQs on Mensuration 3D .
51
The curved surface area and volume of a cylindrical object are 88 cm2 and 132 cm3, respectively. The height (in cm) of the cylindrical object is:
Discuss
Answer & Solution
Answer: Option B
Solution:
Curved area of cylinder = 88 cm2
2πrh = 88
2 × $$\frac{{22}}{7}$$ × r × h = 88
r × h = 14
Volume of cylinder = πr2h = 132 cm3
$$\frac{{22}}{7}$$ × r2h = 132
r2h = 42
r × rh = 42
r × 14 = 42
r = 3
r × h = 14
h $$ = \frac{{14}}{3} = 4\frac{2}{3}$$
Height $$ = 4\frac{2}{3}{\text{ cm}}$$
52
The ratio of weights of two spheres of different materials is 8 : 17 and the ratio of weights per 1 cc of materials of each is 289 : 64. The ratio of radii of the two spheres is
Discuss
Answer & Solution
Answer: Option A
Solution:
Mensuration 3D mcq question image
Ratio of volume of sphere × ratio of weight per 1 cc of material of each = Ratio of weight of two sphere
$$\eqalign{ & \frac{{\frac{4}{3}\pi r_1^3}}{{\frac{4}{3}\pi r_2^3}} \times \frac{{289}}{{64}} = \frac{8}{{17}} \cr & \frac{{r_1^3}}{{r_2^3}} = \frac{{8 \times 64}}{{17 \times 289}} \cr & \frac{{{r_1}}}{{{r_2}}} = \frac{{8 \times 8 \times 8}}{{17 \times 17 \times 17}} \cr & \frac{{{r_1}}}{{{r_2}}} = \frac{8}{{17}} \cr & {r_1}:{r_2} = 8:17 \cr} $$
53
The base of a right prism is a trapezium. The length of the parallel sides are 8 cm and 14 cm and the distance between the parallel sides is 8 cm. If the volume of the prism is 1056 cm3, then the height of the prism is
Discuss
Answer & Solution
Answer: Option C
Solution:
Area of trapezium
= $$\frac{1}{2}$$ × h(AB + CD)
= $$\frac{1}{2}$$ × 8 × (8 + 14)
= 4 × 22
= 88 cm2
Volume of prism = Height of prism × Area of base
⇒ height × 88 = 1056 (given)
⇒ height = $$\frac{{1056}}{{88}}$$
⇒ height = 12 cm
54
Determine the total surface area of a hemisphere closed at bottom. Radius of the hemisphere is $$\sqrt {\frac{{25}}{\pi }} $$  unit.
Discuss
Answer & Solution
Answer: Option A
Solution:
Total surface area (TSA) of hemisphere = 3πr2
Radius of hemisphere = $$\sqrt {\frac{{25}}{\pi }} $$
So, TSA = 3 × π × $${\left( {\sqrt {\frac{{25}}{\pi }} } \right)^2}$$
= 75 unit2
55
A cylindrical tube, open at both ends, is made of a metal sheet which is 0.5 cm thick. Its outer radius is 4 cm and length is 2 m. How much metal (in cm3) has been used in making the tube?
Discuss
Answer & Solution
Answer: Option A
Solution:
Mensuration 3D mcq question image
Outer radius = 4 cm
Internal radius = 4 - 0.5 = 3.5 cm
Length = 2m = 200 cm
Volume of cylinder = π × (R2 - r2)h
= π × (42 - 3.52) × 200
= π × 7.5 × 0.5 × 200
= 750π
56
If the height of a cylinder is 4 times its circumference, the volume of the cylinder in terms of its circumference, $$c$$ is
Discuss
Answer & Solution
Answer: Option B
Solution:
Here h = 4c,
Volume of cylinder = πr2h
$$ = \frac{{4\pi \times \pi {r^2}h}}{{4\pi }}$$
(Multiply 4π both in Numerator & Denominator)
$$\eqalign{ & = \frac{{{{\left( {2\pi r} \right)}^2} \times \left( {4c} \right)}}{{4\pi }} \cr & = \frac{{{c^3}}}{\pi } \cr} $$
57
The areas of three consecutive faces of a cuboid are 12 cm2, then the volume (in cm3) of the cuboid is
Discuss
Answer & Solution
Answer: Option D
Solution:
Let the three sides of the cuboid be $$l$$, b and h
⇒ $$l$$b = bh = h$$l$$ = 12
⇒$$l$$2b2h2 = 12 × 12 × 12
⇒$$l$$2b2h2 = 1728
⇒ $$l$$bh = $$\sqrt {1728} $$
⇒ $$l$$bh = $$12\sqrt {12} $$
⇒ $$l$$bh = $$24\sqrt 3 {\text{ c}}{{\text{m}}^3}$$
58
The ratio of the height and the diameter of a right circular cone is 6 : 5 and its volume is $$\frac{{2200}}{7}{\text{ c}}{{\text{m}}^3}.$$   What is its slant height? $$\left( {{\text{Take }}\pi = \frac{{22}}{7}} \right)$$
Discuss
Answer & Solution
Answer: Option B
No explanation is given for this question. Let's Discuss on Board
59
A sphere of maximum volume is cut out from a solid hemisphere. What is the ratio of the volume of the sphere to that of the remaining solid?
Discuss
Answer & Solution
Answer: Option B
Solution:
Mensuration 3D mcq question image
$$\eqalign{ & {\text{Sphere }}\left( {\text{V}} \right) = \frac{4}{3}\pi {\left( {\frac{R}{2}} \right)^3} = \frac{4}{3}\pi \times \frac{{{R^3}}}{8} \cr & {\text{Hemisphere }}\left( {\text{V}} \right) = \frac{2}{3}\pi {R^3} \cr & {\text{Sphere}}:{\text{Hemisphere}} = \frac{{4{R^3}}}{{3 \times 8}}:\frac{{2{R^3}}}{3} = 1:4 \cr & {\text{Sphere}}:{\text{Remaining Solid}} = 1:3 \cr} $$
60
A cube of edge 6 cm is painted on all sides and then cut into unit cubes. The number of unit cubes with no sides painted is
Discuss
Answer & Solution
Answer: Option B
Solution:
Number of cubes with no side painted = (n - 2)3
Where n is the side of the bigger cube
Required number = (6 - 2)3 = 64