Question

A massive stone pillar 20m high and of uniform cross section rests on a rigid base and supports a vertical load of $5.0 \times {10^5}N$ at its upper end. If the compressive stress in the pillar does not exceed $16 \times {10^5}N/m$, what is the minimum cross-sectional area of the pillar? (Density of the stone $ = 2.5 \times {10^3}kg/{m^3}$. Take $g = 10N/kg$)
$
  {\text{A}}{\text{. 0}}{\text{.15}}{{\text{m}}^2} \\
  {\text{B}}{\text{. 0}}{\text{.25}}{{\text{m}}^2} \\
  {\text{C}}{\text{. 0}}{\text{.35}}{{\text{m}}^2} \\
  {\text{D}}{\text{. 0}}{\text{.45}}{{\text{m}}^2} \\
$

Answer 1

Hint: Stress on a body is defined as the force applied on it per unit cross-sectional area. The total weight or force on the base is equal to the sum of weight of pillar and the additional vertical load on the upper end of the pillar.

Formula used:
The stress on a body due to an applied force is given as follows:
$Stress = \dfrac{F}{A}$
where stress S is equal to the force F per unit cross-sectional area A.

Complete step-by-step answer:
We are given a pillar which has 20 m height and the cross section of its base is A.
$l = 20m$
Therefore, volume of the pillar can be written as
$V = l \times A$
Density of the stone of the pillar is given as
$\rho = 2.5 \times {10^3}kg/{m^3}$
Using this information, we can calculate the mass of the pillar which is equal to the product of density and volume of the pillar.
$m = \rho V$
Inserting the known values, we get the mass of pillar to be
$
  m = 2.5 \times {10^3} \times 20 \times A \\
   = 5 \times {10^4} \times A \\
$
We are given $g = 10N/kg$. Using this value of g, we can calculate the weight of pillar as follows:
$
  W = mg = 5 \times {10^4} \times A \times 10 \\
   = 5 \times {10^5} \times A \\
$
The pillar supports an additional vertical load of $5.0 \times {10^5}N$
Therefore the total weight on the base of the pillar is equal to the sum of weight of pillar and the additional weight supported on top of the pillar.
$
  W' = 5 \times {10^5} \times A + 5 \times {10^5} \\
   = 5 \times {10^5} \times \left( {A + 1} \right) \\
$
We are given the maximum compressive stress of pillar to be
$S = 16 \times {10^5}N/m$
The stress is equal to ratio of force applied to the cross-sectional area of the base of the pillar given as
$S = \dfrac{{W'}}{A}$
Inserting the known values, we get
$
  16 \times {10^5} = \dfrac{{5 \times {{10}^5} \times \left( {A + 1} \right)}}{A} \\
   \Rightarrow \dfrac{{A + 1}}{A} = \dfrac{{16 \times {{10}^5}}}{{5 \times {{10}^5}}} = 3.2 \\
   \Rightarrow 1 + \dfrac{1}{A} = 3.2 \\
   \Rightarrow \dfrac{1}{A} = 3.2 - 1 = 2.2 \\
   \Rightarrow A = \dfrac{1}{{2.2}} = 0.45{m^2} \\
$
Hence, the correct answer is option D.

Note: The maximum compressive stress of the pillar implies the maximum amount of force that the base of the pillar can support without breaking apart. The obtained cross sectional area is the minimum cross-sectional area with which the pillar can support the given load.