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Volume And Surface Area - Aptitude MCQ Questions and Solutions with Explanations

Volume And Surface Area MCQ Questions and answers with easy and logical explanations.Arithmetic Ability provides you all type of quantitative and competitive aptitude mcq questions on Volume And Surface Area with easy and logical explanations. Volume And Surface Area MCQ is important for exams like Banking exams,IBPS,SCC,CAT,XAT,MAT etc. Page-9 section-1

Each section contains 100 MCQs on Volume and Surface Area .
81
A circular well with a diameter of 2 metres, is dug to a depth of 14 metres. What is the volume of the earth dug out ?
Discuss
Answer & Solution
Answer: Option D
Solution:
Volume :
$$\eqalign{ & = \pi {r^2}h \cr & = \left( {\frac{{22}}{7} \times 1 \times 1 \times 14} \right){m^3} \cr & = 44\,{m^3} \cr} $$
82
If the radius of the base of a right circular cylinder is halved, keeping the height same, what is the ratio of the volume of the reduced cylinder to that of the original one ?
Discuss
Answer & Solution
Answer: Option B
Solution:
Let original radius = R
Then, new radius = $$\frac{{\text{R}}}{2}$$
$$\eqalign{ & \therefore \frac{{{\text{Volume of reduced cylinder }}}}{{{\text{Volume of original cylinder}}}} \cr & = \frac{{\pi \times {{\left( {\frac{R}{2}} \right)}^2} \times h}}{{\pi \times {R^2} \times h}} \cr & = \frac{1}{4}\,Or\,1:4 \cr} $$
83
The number of circular pipes with an inside diameter of 1 inch which will carry the same amount of water as a pipe with an inside diameter of 6 inches is :
Discuss
Answer & Solution
Answer: Option C
Solution:
Let the length of each pipe be $$l$$ inches
Then, volume of water in thinner pipe :
$$\eqalign{ & = \left[ {\pi \times {{\left( {\frac{1}{2}} \right)}^2} \times 1} \right] \text{cu.inch} \cr & = \left( {\frac{{\pi l}}{4}} \right)\text{cu.inch} \cr} $$
Volume of water in thinker pipe :
$$\eqalign{ & = \left( {\pi \times {3^2} \times l} \right)\text{cu.inch} \cr & = \left( {9\pi l} \right)\text{cu.inch} \cr} $$
∴ Required number of pipes :
$$\eqalign{ & = \frac{{9\pi l}}{{\left( {\frac{{\pi l}}{4}} \right)}} \cr & = 36 \cr} $$
84
A right triangle with sides 3 cm, 4 cm and 5 cm is rotated about the side of 3 cm to form a cone. The volume of the cone so formed is :
Discuss
Answer & Solution
Answer: Option A
Solution:
Clearly, we have r = 3 cm and h = 4 cm
∴ Volume :
$$\eqalign{ & = \frac{1}{3}\pi {r^2}h \cr & = \left( {\frac{1}{3} \times \pi \times {3^2} \times 4} \right)\pi {r^3} \cr & = 12\pi {\text{ c}}{{\text{m}}^3} \cr} $$
85
The radius of the base and height of a metallic solid cylinder are r cm and 6 cm respectively. It is melted and recast into a solid cone of the same radius of base. The height of the cone is :
Discuss
Answer & Solution
Answer: Option B
Solution:
Let the height of the cone be h cm
Then,
$$\eqalign{ & \pi \times {r^2} \times 6 = \frac{1}{3} \times \pi \times {r^2} \times h \cr & \Rightarrow h = 18\,cm \cr} $$
86
For a sphere of radius 10 cm, What percent of the numerical value of its volume would be the numerical value of the surface area ?
Discuss
Answer & Solution
Answer: Option C
Solution:
Volume of the sphere :
$$ = \left[ {\frac{4}{3}\pi {{\left( {10} \right)}^3}} \right]{\text{ c}}{{\text{m}}^3}$$
Surface area of the sphere :
$$ = \left[ {4\pi {{\left( {10} \right)}^2}} \right]{\text{ c}}{{\text{m}}^2}$$
∴ Required percentage :
$$\eqalign{ & = \left[ {\frac{{4\pi {{\left( {10} \right)}^2}}}{{\frac{4}{3}\pi {{\left( {10} \right)}^3}}}} \times 100 \right]\% \cr & = 30\% \cr} $$
87
How many lead shots each 3 mm in diameter can be made from a cuboid of dimensions 9 cm × 11 cm × 12 cm ?
Discuss
Answer & Solution
Answer: Option D
Solution:
Volume of each lead shot :
$$\eqalign{ & = \left[ {\frac{4}{3}\pi \times {{\left( {\frac{{0.3}}{2}} \right)}^3}} \right]{\text{ c}}{{\text{m}}^3} \cr & = \left( {\frac{4}{3} \times \frac{{22}}{7} \times \frac{{27}}{{8000}}} \right){\text{ c}}{{\text{m}}^3} \cr & = \frac{{99}}{{7000}}{\text{ c}}{{\text{m}}^3} \cr} $$
∴ Number of lead shots :
$$\eqalign{ & = \left( {9 \times 11 \times 12 \times \frac{{7000}}{{99}}} \right) \cr & = 84000 \cr} $$
88
A metallic sphere of radius 5 cm is melted to make a cone with base of the same radius. What is the height of the cone ?
Discuss
Answer & Solution
Answer: Option D
Solution:
Let the height of the cone be h cm
Then,
$$\eqalign{ & \frac{4}{3}\pi \times {\left( 5 \right)^3} = \frac{1}{3}\pi \times {\left( 5 \right)^2} \times h \cr & \Rightarrow h = 20\,cm \cr} $$
89
A hemisphere of lead of radius 6 cm is cast into a right circular cone of height 75 cm. The radius of the base of the cone is :
Discuss
Answer & Solution
Answer: Option C
Solution:
Let the radius of the cone be R cm
Then,
$$\eqalign{ & \frac{1}{3}\pi \times {R^2} \times 75 = \frac{2}{3}\pi \times 6 \times 6 \times 6 \cr & \Rightarrow {R^2} = \left( {\frac{{2 \times 6 \times 6 \times 6}}{{75}}} \right) \cr & \Rightarrow {R^2} = \frac{{144}}{{25}} \cr & \Rightarrow {R^2} = \frac{{{{\left( {12} \right)}^2}}}{{{{\left( 5 \right)}^2}}} \cr & \Rightarrow R = \frac{{12}}{5} \cr & \Rightarrow R = 2.4\,cm \cr} $$
90
Base of a right prism is a rectangle, the ratio of whose length and breadth is 3 : 2. If the height of the prism is 12 cm and total surface area is 288 sq.cm the volume of the prism is :
Discuss
Answer & Solution
Answer: Option B
Solution:
Let the length of base be 3a cm and breadth be 2a cm
Total surface area of prism :
= [Perimeter of base × height] + [2 × Area of base]
= [2 (3a + 2a) × 12 + 2 × 3a × 2a] sq.cm
= (120a + 12a2) sq.cm
According to the question,
120a + 12a2 = 288
⇒ a2 + 10a = 24
⇒ a2 + 10a - 24 = 0
⇒ a2 + 12a - 2a - 24 = 0
⇒ a (a + 12) - 2 (a + 12) = 0
⇒ (a - 2)(a + 12) = 0
⇒ a = 2 because a $$ \ne $$ -12
∴ Volume of prism :
= Area of base × Height
= (3a × 2a × 12)cu.cm
= 72a2 cu.cm
= (72 × 2 × 2)cu.cm
= 288 cu.cm