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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-3 section-1

Each section contains 100 MCQs on Mensuration 3D .
21
A solid metallic cube of side 9 cm and a solid metallic cuboid having dimensions 5 cm, 13 cm, 31 cm are melted to form a single cube. How much (in Rs.) is the cost to polish the new at a rate of Rs. 10 per cm2?
Discuss
Answer & Solution
Answer: Option C
Solution:
Volume of cube = 93 = 729
Volume of cuboid = 5 × 31 × 13 = 2015
Total volume = 729 + 2015 = 2744
Let volume of new cube = a3
a3 = 2744
a = 14
T.S.A = 6 × 142 = 6 × 196
The cost of polish = 6 × 196 × 10 = Rs. 11760
22
A wall 12 m long, 9 m high and 90 cm thick is to be constructed. Given that one pack contains 450 bricks, how many packs of bricks will be required if each brick is of dimensions 10 cm × 7.5 cm × 7 cm and the cement and sand mixture occupies one-eighteenth volume of the wall? (Rounded off to the nearest integer)
Discuss
Answer & Solution
Answer: Option D
Solution:
Volume of wall = 12 m × 9 m × 90 cm
Volume of Bricks in wall = (12 m × 9 m × 90 cm) × $$\frac{{17}}{{18}}$$
Let, number of bricks = n
Now, n × 450 × 10 cm × 7.5 cm × 7 cm = (12 m × 9 m × 90 cm) × $$\frac{{17}}{{18}}$$
$$\eqalign{ & {\text{n}} = \frac{{1200 \times 900 \times 90}}{{450 \times 10 \times 7.5 \times 7}} \times \frac{{17}}{{18}} \cr & {\text{n}} = \frac{{160 \times 17}}{7} \cr & {\text{n}} = \frac{{2720}}{7} \cr & {\text{n}} = 388.5 \cr & {\text{n}} = 389{\text{ Answer}} \cr} $$
23
If a right circular cone is separated into solids of volumes V1, V2, V3 by two planes parallel to the base which also trisect the altitude, then V1 : V2 : V3 is
Discuss
Answer & Solution
Answer: Option D
Solution:
Ratio of volume of bigger cone and smaller cone
= (Ratio of altitude)2
= (1 : 2 : 3)3
= (1 : 8 : 27)
∴ Ratio of parts
= 1 : 8 - 1 : 27 - 8
= 1 : 7 : 19
24
A sector of radius 10.5 cm with the central angle 120° is folded to form a cone by joining the two bounding radii of the sector. What is the volume (in cm3) of the cone so formed?
Discuss
Answer & Solution
Answer: Option A
Solution:
Mensuration 3D mcq question image
$$\eqalign{ & {\text{Length of Arc}} = L \cr & L = 2 \times \frac{{22}}{7} \times \frac{{r \times 120}}{{360}} \cr & L = 2 \times \frac{{22}}{7} \times 10.5 \times \frac{1}{3} \cr & L = 22 \cr} $$
Mensuration 3D mcq question image
$$\eqalign{ & 2\pi R = 22 \cr & 2 \times \frac{{22}}{7}R = 22 \cr & R = 3.5 \cr & {h^2} + {R^2} = {10.5^2} \cr & {h^2} + {3.5^2} = {10.5^2} \cr & h = 7\sqrt 2 \cr & V = \frac{1}{3}\pi {r^2}h \cr & V = \frac{1}{3}\pi \times 3.5 \times 3.5 \times 7\sqrt 2 \cr & V = \frac{{343\sqrt 2 }}{{12}}\pi \cr} $$
25
A sphere of diameter 6 cm is dropped in a right circular cylindrical vessel partly filled with water. The diameter of the cylindrical vessel is 12 cm. If the sphere is just completely submerged in water, then the rise of water level in the cylindrical vessel is:
Discuss
Answer & Solution
Answer: Option B
Solution:
Mensuration 3D mcq question image
$$\eqalign{ & {\text{Volume of sphere}} = \frac{4}{3}\pi {r^3} \cr & = \frac{4}{3}\pi \times 3 \times 3 \times 3 \cr & = 36\pi {\text{ c}}{{\text{m}}^3} \cr & {\text{If the water level rises by H cm}} \cr & \pi {R^2}H = 36\pi \cr & 6 \times 6 \times H = 36 \cr & H = 1{\text{ cm}} \cr} $$
26
A solid cube has side 8 cm. It is cut along diagonals of top face to get 4 equal parts. What is the total surface (in cm2) of each part?
Discuss
Answer & Solution
Answer: Option A
Solution:
Mensuration 3D mcq question image
Total surface area of each part is = 2 × Area of base + Perimeter of base × Height
Mensuration 3D mcq question image
$$\eqalign{ & = 2 \times \frac{1}{2} \times 4\sqrt 2 \times 4\sqrt 2 + \left( {8 + 8\sqrt 2 } \right) \times 8 \cr & = 32 + 64 + 64\sqrt 2 \cr & = 96 + 64\sqrt 2 \cr} $$
27
The base of a right pyramid is a square of side 8√2 cm and each of its slant edge is of length 10 cm. What is the volume (in cm3) of the pyramid?
Discuss
Answer & Solution
Answer: Option A
Solution:
Mensuration 3D mcq question image
Length of each side = 8√2 cm
Diagonal = √2a
= √2 × 8√2
= 16
OC $$ = \frac{{\text{d}}}{2} = \frac{{16}}{2} = 8$$
Slant edge = EC = 10 cm
In ΔOEC
EC2 = OE2 + OC2
102 = OE2 + 82
OE2 = 62
Height = OE = 6 cm
Volume of the pyramid
= $$\frac{1}{3}$$ × Area of base × Height
= $$\frac{1}{3}$$ × (8√2)2 × 6
= 256
28
The base of a solid right prism of height 10 cm is a square and its volume is 160 m3. What is the total surface area of the prism (in cm2)?
Discuss
Answer & Solution
Answer: Option C
Solution:
10 cm ⇒ height
Volume of Prism = Base Area × height
160 = a2 × 10
a = 4
Total surface area = 2Base Area + Base per meter × length
= 2 × 16 + 16 × 10
= 32 + 160
= 192
29
A solid lead sphere of radius 11 cm is melted and recast into small solid spheres of radius 2 cm each. How many maximum number (in integer) of such spheres can be made?
Discuss
Answer & Solution
Answer: Option B
Solution:
$$\eqalign{ & {R^3} = n \times {r^3} \cr & 11 \times 11 \times 11 = n \times {2^3} \cr & \frac{{1331}}{8} = n \cr & n = 166 \cr & n = 166{\text{ }}\left( {{\text{integer}}} \right) \cr} $$
30
A hemisphere is kept on top of a cube. Its front view is shown in the given figure. The total height of the figure is 21 cm. The ratio of curved surfaces area of hemisphere and total surface area of cube is 11 : 42. What is the total volume (in cm3) of figure?
Mensuration 3D mcq question image
Discuss
Answer & Solution
Answer: Option B
Solution:
$$\frac{{{\text{Curved surfaces area of hemisphere}}}}{{{\text{Total surface area of cube}}}} = \frac{{2\pi {r^2}}}{{6{a^2}}}$$
Mensuration 3D mcq question image
3r = 21
r = 7 cm
Total volume of figure = Volume of hemisphere + Volume of cube
= $$\frac{2}{3}$$πr3 + a3
= $$\frac{2}{3}$$ × $$\frac{{22}}{7}$$ × 7 × 7 × 7 + 143
= 718.66 + 2744
= 3462.67 cm3