Quantitative Aptitude

Mensuration

2,507 Questions
All Direct Formula Application 1067 Shape Transformations 343 Mensuration Ratio Equations 277 Combined & Hollow Solids 247 Successive Percentage Changes 178 Inscribed & Circumscribed Figures 165 Cost Estimation Problems 100 Liquid Pour & Submergence 67 Unit Conversions 35 Path & Border Calculations 17 Time & Work Integration 7 Scale & Model Proportions 2

Mensuration Questions

Multiple choice

The height of the cylinder of maximum volume inscribed in a sphere of radius '$a$' is
  1. $\dfrac { 3a }{ 2 } $
  2. $\dfrac { \sqrt { 2 } a }{ 3 } $
  3. $\dfrac { a }{ \sqrt { 3 } } $
  4. $\dfrac{2a}{\sqrt{3}} $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Volume of cylinder V = pi * r^2 * h. In a sphere of radius a, r^2 + (h/2)^2 = a^2, so r^2 = a^2 - h^2/4. V = pi * (a^2 - h^2/4) * h = pi * (a^2*h - h^3/4). Differentiating with respect to h: dV/dh = pi * (a^2 - 3h^2/4) = 0. 3h^2/4 = a^2, h^2 = 4a^2/3, h = 2a/sqrt(3).

Multiple choice

The diameter of one of the bases of a truncated cone is $100\ mm$. If the diameter of this base is increased by $21\%$ such that it still remains a truncated cone with the height and the other base unchanged, the volume also increases by $21\%$. The radius of the other base (in $mm$) is
  1. $65$
  2. $55$
  3. $45$
  4. $35$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Volume of a truncated cone V = (1/3)pi*h(R^2 + r^2 + R*r). If R increases by 21%, R' = 1.21R. Since V increases by 21%, V' = 1.21V. The equation (R'^2 + r^2 + R'r) = 1.21(R^2 + r^2 + Rr) must hold. Substituting R=50 (radius is half diameter) and solving for r yields r=55.

Multiple choice

A hollow cylinder has mass M. outside radius ${ R }{ 2 }$ and inside radius ${ R }{ 1 }$ Its moment of inertia about an or axis parallel to its symmetry axis and tangential to tho outer surface is equal to :
  1. $\dfrac { M } { 2 } \left( R _ { 2 } ^ { 2 } + R _ { 1 } ^ { 2 } \right)$
  2. $\dfrac { M } { 2 } \left( R ^ { 2 } - R _ { 1 } ^ { 2 } \right)$
  3. $\dfrac { M } { 4 } \left( R _ { 2 } + R _ { 1 } \right) ^ { 2 }$
  4. $\dfrac { M } { 2 } \left( 3 R _ { 2 } ^ { 2 } + R _ { 1 } ^ { 2 } \right)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The moment of inertia of the hollow cylinder about its symmetry axis is M/2 x (R2^2 + R1^2). The tangential axis is displaced by R2, so the parallel-axis theorem adds M R2^2, giving M/2 x (3R2^2 + R1^2).

Multiple choice

A solid cylinder of mass $0.1\, kg$ and radius $0.025\, metre$ is rolling on a horizontal smooth table with uniform velocity of $0.1\,ms^{-1}$. Its total energy will be :
  1. $7.5 \times 10^{-2}\, Joule$
  2. $7.5 \times 10^{-3}\, Joule$
  3. $7.5 \times 10^{-4}\, Joule$
  4. $0.07 \times 10^{-4}\, Joule$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Total energy of a rolling cylinder = Translational KE + Rotational KE = 1/2 * m * v^2 + 1/2 * I * w^2. For a solid cylinder, I = 1/2 * m * r^2 and w = v/r. So, Total Energy = 1/2 * m * v^2 + 1/2 * (1/2 * m * r^2) * (v/r)^2 = 1/2 * m * v^2 + 1/4 * m * v^2 = 3/4 * m * v^2. Energy = 0.75 * 0.1 * (0.1)^2 = 0.75 * 0.1 * 0.01 = 0.00075 = 7.5 * 10^-4 Joule.

Multiple choice

A solid cylinder is released from the top of an inclined plane of height $h$. What will be the speed of its centre of mass on reaching the bottom?
  1. $\sqrt{2 \,gh}$
  2. $\sqrt{4 \,gh}$
  3. $\left[\dfrac{4}{3}gh \right]^{1/2}$
  4. $\left[\dfrac{2}{3}gh \right]^{1/2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using conservation of energy: mgh = (1/2)mv^2 + (1/2)Iw^2. For a solid cylinder, I = (1/2)mR^2 and w = v/R. So mgh = (1/2)mv^2 + (1/2)(1/2)mR^2(v^2/R^2) = (1/2)mv^2 + (1/4)mv^2 = (3/4)mv^2. v^2 = (4/3)gh. v = sqrt(4gh/3).

Multiple choice

A sphere of radius R and made of material of relative density $\sigma$ has a concentric cavity of radius r.It just floats when placed in a tank full water.The value of the ratio R/r will be
  1. $\left( \dfrac { \sigma }{ \sigma -1 } \right) ^{ 1/3 }$
  2. $\left( \dfrac { \sigma -1 }{ \sigma } \right) ^{ 1/3 }$
  3. $\left( \dfrac { \sigma \div 1 }{ \sigma } \right) ^{ 1/3 }$
  4. $\left( \frac { \sigma -1 }{ \sigma \div 1 } \right) ^{ 1/3 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For floating equilibrium, the weight of the material equals the weight of the displaced water. Thus, sigma(R^3 - r^3) = R^3, which gives R^3/r^3 = sigma/(sigma - 1). Taking cube roots gives the stated ratio.

Multiple choice

The force required to drag a circular flat plate of radius $5\mathrm { cm }$ on the surface of water is $( \mathrm { ST }$ of water is $75 dyne/cm).$
  1. $30 dyne$
  2. $60 dyne$
  3. $750 dyne$
  4. $750 \pi dyne$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The force required to lift a circular plate from a liquid surface is F = Surface Tension * Perimeter. The perimeter of the plate is 2 * pi * r. Given r = 5 cm and ST = 75 dyne/cm, F = 75 * 2 * pi * 5 = 750 * pi dyne.

Multiple choice

An object shaped as a hemisphere rests with its flat surface on a table. The object has radius r and density $\rho$. The volume of a sphere is $\dfrac{4}{3}\pi r^3$. Which average pressure does the object exert on the table?
  1. $\dfrac{1}{3}\rho r^2$
  2. $\dfrac{1}{3}\rho r^2g$
  3. $\dfrac{2}{3}\rho r$
  4. $\dfrac{2}{3} \rho rg$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Pressure = Force / Area. Force = Weight = mass * g = (density * volume) * g. Volume of hemisphere = (2/3)*pi*r^3. Area of base = pi*r^2. Pressure = (rho * (2/3)*pi*r^3 * g) / (pi*r^2) = (2/3)*rho*r*g.