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

A capillary tube of radius 0.25 mm is dipped vertically in a liquid of density $800 kg m^{-3}$ and of surface tension $3\times 10^{-2} Nm^{-2}$.The angle of contact of liquid-glass is given by $cos\theta = 0.3$. If $g=10ms^{-2}$, the rise of liquid in the capillary tube is______ $( Cm)$:
  1. $9$
  2. $0.9$
  3. $9\ x\ 10^ {-3}$
  4. $0.09$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The formula for capillary rise is h = (2 * T * cos(theta)) / (r * rho * g). T = 3 * 10^-2, cos(theta) = 0.3, r = 0.25 * 10^-3 m, rho = 800, g = 10. h = (2 * 3 * 10^-2 * 0.3) / (0.25 * 10^-3 * 800 * 10) = (1.8 * 10^-2) / (2) = 0.9 * 10^-2 m = 0.9 cm.

Multiple choice

Volume of the solid obtained by revolving the area of the ellipse $\cfrac { { x }^{ 2 } }{ { a }^{ 2 } } +\cfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$ about major and minor axes are in the ratio:
  1. ${b}^{2}:{a}^{2}$
  2. ${a}^{2}:{b}^{2}$
  3. $a:b$
  4. $b:a$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Volume of ellipsoid revolved about major axis (a) is 4/3 * pi * a * b^2. About minor axis (b) is 4/3 * pi * b * a^2. Ratio = (4/3 * pi * a * b^2) / (4/3 * pi * b * a^2) = b / a.

Multiple choice

A square wire of side $2.0 cm$ is placed $20 cm$ in front of a concave mirror of focal length $10 cm$ with its center on the axis of the mirror and its plane normal to the axis. The area enclosed by the image of wire is
  1. $7.5{ cm }^{ 2 }$
  2. $6{ cm }^{ 2 }$
  3. $2{ cm }^{ 2 }$
  4. $4{ cm }^{ 2 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Mirror formula: 1/v + 1/u = 1/f. u = -20, f = -10. 1/v - 1/20 = -1/10 -> 1/v = -1/20 -> v = -20. Magnification m = -v/u = -(-20)/(-20) = -1. The image is the same size as the object. Area = 2*2 = 4 cm^2.

Multiple choice

The work done in blowing a soap bubble of radius $0.2 m$, given that the surface tension of soap solution is $\displaystyle 60\times { 10 }^{ -3 }{ N }/{ m }$ is :
  1. $\displaystyle 24\pi \times { 10 }^{ -2 }J$
  2. $\displaystyle 4\pi \times { 10 }^{ -4 }J$
  3. $\displaystyle 96\pi \times { 10 }^{ -4 }J$
  4. $\displaystyle 1.92\pi \times { 10 }^{ -2 }J$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Work done in blowing a soap bubble is W = 2 * (Surface Area * Surface Tension) because a bubble has two surfaces. W = 2 * (4 * pi * r^2) * T. W = 8 * pi * (0.2)^2 * 60 * 10^-3 = 8 * pi * 0.04 * 60 * 10^-3 = 19.2 * pi * 10^-2 J.

Multiple choice

A metallic rod of cross-sectional area $20 cm^2$, with the lateral surface insulated to prevent heat loss, has one end immersed in boiling water and the other in ice water mixture. The heat conducted through the rod melts the ice at the rate of $1$ gm for every $84$ sec. The thermal conductivity of the rod is $160 Wm^{1}K^{1}$. Latent heat of $ice = 80 cal/gm, 1 cal = 4.2 joule$. What is the length (in m) of the rod?

  1. 4 m

  2. 5 m

  3. 8 m

  4. 10 m
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Melting 1 g of ice requires 80 × 4.2 = 336 J, so the heat current is 336/84 = 4 W. Using Q/t = kAΔT/L with A = 0.002 m^2 and ΔT = 100 K gives L = 8 m.

Multiple choice

A vessel has a small hole at its bottom. If water can be poured into it upto a height of $7 cm$ without leakage ($g=10 ms^{-2}$), the radius of the hole is (surface tension of water is $0.07 Nm^{-1}$).
  1. $2\ mm$
  2. $0.2\ mm$
  3. $0.1\ mm$
  4. $0.4\ mm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The pressure difference due to surface tension at the hole is 2T/r. This must balance the hydrostatic pressure rho*g*h. 2T/r = rho*g*h. r = 2T / (rho*g*h). Using T=0.07, rho=1000, g=10, h=0.07m: r = 2*0.07 / (1000*10*0.07) = 2 / 10000 = 0.0002 m = 0.2 mm.

Multiple choice

Consider a hollow cylinder of inner radius $r$ and thickness of wall $t$ and length $l$. The volume of the above cylinder is given by
  1. $2\pi l t \left ( r^2-l^2 \right )$
  2. $ \displaystyle 2\pi rl t \left( \frac{t}{2r}+1 \right )$
  3. $2\pi l t \left ( r^2+l^2 \right )$
  4. $2\pi rl t\left ( r+t \right )$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Volume = Outer Cylinder Volume - Inner Cylinder Volume. Outer radius = r+t. Volume = pi * (r+t)^2 * l - pi * r^2 * l = pi * l * ((r+t)^2 - r^2) = pi * l * (r^2 + 2rt + t^2 - r^2) = pi * l * (2rt + t^2) = pi * l * t * (2r + t) = 2 * pi * r * l * t * (1 + t/(2r)).

Multiple choice

The radius of a sphere is $2r$, then its volume will be:
  1. $\dfrac{4}{3} \pi r^{3}$
  2. $4 \pi r^{3}$
  3. $\dfrac{8 \pi r^{3}}{3}$
  4. $\dfrac{32}{3} \pi r^{3}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Volume of a sphere = (4/3) * pi * R^3. Here R = 2r. Volume = (4/3) * pi * (2r)^3 = (4/3) * pi * 8r^3 = (32/3) * pi * r^3.

Multiple choice

$ABCD$ is a square with side $a$. With centres $A, B, C$ and $D$ four circles are drawn such that each circle touches externally two of the remaining three circles. Let $\delta$ be the area of the region in the interior of the square and exterior of the circles. Then the maximum value $\delta$ is:
  1. $a^2(1 - \pi)$
  2. $\displaystyle{a^2\left(\frac{4 - \pi}{4}\right)}$
  3. $a^2(\pi - 1)$
  4. $\displaystyle{\frac{\pi a^2}{4}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The area of the square is a^2. Each circle has radius a/2. The area covered by the four sectors inside the square is 4 * (1/4 * pi * (a/2)^2) = pi * a^2 / 4. The area of the region in the interior of the square and exterior of the circles is a^2 - pi * a^2 / 4 = a^2(1 - pi/4) = a^2((4 - pi)/4).