Mathematics · Quantitative Aptitude

Mensuration of Solids

194 Questions

Mensuration of solids focuses on calculating the volume and surface area of three dimensional shapes like cylinders and cones. These geometry problems require applying standard mathematical formulas. They frequently appear in state public service and engineering exams.

Cylinder volume and areaCone slant heightSurface area ratiosHollow pipe calculationsDimensional optimization

Mensuration of Solids Questions

Multiple choice general knowledge math & puzzles

A cube with edge length x and a right cylinder with circular base have the same volume. Find x ,in cm, if the radius of the cylinder is 5 cm and its height is 5 cm.

  1. 5(Pi)^1/3

  2. 5Pi

  3. 125Pi

  4. 125
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The volume of the cylinder is V = πr²h = π(5)²(5) = 125π cubic cm. For a cube with edge length x, the volume is V = x³. Setting the volumes equal: x³ = 125π, which gives x = (125π)^(1/3) = 5π^(1/3) cm. Option B (5π) would give volume 125π³, Option C (125π) would give volume (125π)³, and Option D (125) would give volume 1,953,125 - all incorrect.

Multiple choice general knowledge math & puzzles

A cylinder of radius 5 cm is inserted within a cylinder of radius 10 cm. The two cylinders have the same height of 20 cm. What is the volume of the region between the two cylinders?

  1. 100Pi

  2. 500Pi

  3. 1000Pi

  4. 1500Pi
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Volume of region between cylinders = π(R² - r²)h = π(10² - 5²)(20) = π(100 - 25)(20) = π(75)(20) = 1500π. This uses the formula for volume of a cylinder (V=πr²h) and subtracts the inner cylinder volume from the outer.

Multiple choice general knowledge math & puzzles

A cone made of cardboard has a vertical height of 8 cm and a radius of 6 cm. If this cone is cut along the slanted height to make a sector, what is the central angle, in degrees, of the sector?

  1. 216

  2. 180

  3. 90

  4. 36
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Slant height l = √(r² + h²) = √(6² + 8²) = √(36 + 64) = √100 = 10cm. Sector radius = 10cm, arc length = 2πr = 12π. Central angle = (arc length / radius) × (360/2π) = (12π/10) × (180/π) = 216°.

Multiple choice

The base radius of a cylinder is 1 2/3 times its height. The cost of painting is CSA at 2 paise/cm2 is Rs. 92.40. The volume of liquid is

  1. 80850 cm3

  2. 80580 cm3

  3. 80508 cm3

  4. none
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given CSA cost = Rs. 92.40 at 2 paise/cm², CSA area = 9240/2 = 4620 cm². With radius r = 5h/3 (since 1 2/3 = 5/3), CSA = 2πrh = 2π(5h/3)h = (10π/3)h². Solving: h² = 4620 × 3/(10π) ≈ 441, so h ≈ 21 cm and r = 35 cm. Volume = πr²h = π(35)²(21) ≈ 80850 cm³. Option A is correct. Options B and C are digit permutations, D is incorrect.

Multiple choice maths area of complex plane figures 2d and 3d figures

A square sheet of paper is converted into a cylinder by rolling it along its length. What is the ratio of the base radius to side of the square ?

  1. $\displaystyle \frac{1}{2\pi}$
  2. $\displaystyle \frac{\sqrt{2}}{\pi}$
  3. $\displaystyle \frac{1}{\sqrt{2\pi }}$
  4. $\displaystyle \frac{1}{\pi}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let length of each side of the square $=a$

Base radius of cylinder $=r$
Surface area of sheet $={ a }^{ 2 }$
Surface area of cylinder $=2\pi rh$
But height $h=a$, because the square sheet is rolled it along its length.
Thus, surface area of cylinder $=2\pi ra$
Therefore, ${ a }^{ 2 }=2\pi ra\Rightarrow a=2\pi r\Rightarrow \dfrac { r }{ a } =\dfrac { 1 }{ 2\pi  } $

Multiple choice maths area of complex plane figures 2d and 3d figures

The ratio of the slant height of two right cones of equal base is 3 : 2 then the ratio of their volumes is

  1. $1:4$
  2. $9:4$
  3. $3:2$

  4. None
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the radius of both the cones be r, slant height of first cone $ = 3a $; slant height of second cone $ = 2a $

For a cone, l $ = \sqrt { { h }^{ 2 }+  {r}^{ 2 } } $ where h is the height. 

Hence, for cone 1,  $ 3a = \sqrt { { h 1 }^{ 2 }+  {r}^{ 2 } } $ 

$ 9{a}^{2} = { h 1 }^{ 2 } + {r}^{ 2 }$

$ { h 1 }^{ 2 } = 9{a}^{2} - {r}^{ 2 } $

$ h 1 = \sqrt {9{a}^{2} - {r}^{ 2 }} $

For cone 1,  $ 2a = \sqrt { { h 2 }^{ 2 }+  {r}^{ 2 } } $ 

$ 4{a}^{2} = { h 2 }^{ 2 } + {r}^{ 2 }$

$ { h 2 }^{ 2 } =4{a}^{2} - {r}^{ 2 } $

$ h 2 = \sqrt {4{a}^{2} - {r}^{ 2 }} $
Now, ratio of their volumes is calculated as:
$V _{ 1 } : V _{ 2 }$
$ \frac { 1 }{ 3 }\pi {r }^{ 2 }h 1 = \frac { 1 }{ 3 } \pi { r }^{2 }h 2 $
$ h 1 : h 2 $
$ \sqrt {9{a}^{2} - {r}^{ 2 }} : \sqrt {4{a}^{2} - {r}^{ 2 }}  $
$9{a}^{2} - {r}^{ 2 }: 4{a}^{2} - {r}^{ 2 } $
$ 9{a}^{2} : 4{a}^{2} $
$ 9:4 $