A massive stone pillar $20\,m$ high and uniform cross section rests on a rigid base and supports a vertical load of $5\, \times \,{10^5}\,N$ at its upper end. If the compressive stress in the pillar is not to exceed $1.6\, \times \,{10^6}\,N{m^{ - 2}}$. What is the minimum cross section area of the pillar? Density of the stone = $2.5\, \times \,{10^3}\,kg{m^{ - 3}}$. (Take $g = 10\,Nk{g^{ - 1}}$)
A. $0.15\,{m^2}$
B. $0.25\,{m^2}$
C. $0.35\,{m^2}$
D. $0.45\,{m^2}$

Hint: The force acting on the unit area of a material is known as stress. The stress can deform the body. The total force on the base is equal to the sum of vertical load on the upper end of the pillar and the weight of the pillar.

Formula used:
The stress on a body due to the applied force is given as,
$S\, = \,\dfrac{F}{A}$
Where,
$S$ is the stress in the pillar
$F$ is the force acting on the pillar
$A$ is the area of the pillar

Complete step by step solution:
Given, Height of the pillar $l$ = $20\,m$
Pillar supports an additional vertical load $m = 5\, \times \,{10^5}\,N$
Maximum compressive stress of pillar $s = 16\, \times \,{10^5}\,N{m^{ - 1}}$
Density of the stone $\rho = 2.5 \times {10^3}\,kg{m^{ - 3}}$
Acceleration due to gravity $g = 10\,Nk{g^{ - 1}}$
Let the cross section of pillar’s base is $A$,
$l = 20\,m$
So, the volume of the pillar can be written as,
$V = l \times A$
Density of the pillar is given as
$\rho = 2.5 \times {10^3}\,kg{m^{ - 3}}$
By using this information, we can calculate the mass of the pillar, which is equal to the product of density and volume of the pillar.
$m = \rho \times V$
Substituting the value of density and volume of the pillar
$m = 2.5 \times {10^3} \times 20 \times A$
$m = 5 \times {10^4} \times A$
By using the value of $g$ we can calculate the weight of the pillar as follows
$
  W = m \times g \\
  W = 5 \times {10^4} \times A \times 10 \\
  W = 5 \times {10^5} \times A \\
 $
As the pillar supports an additional vertical load of $5 \times {10^5}\,N$
So, the total weight on the base of the pillar is equal to the sum of the weight of pillars and the additional weight supported on the top of the pillar,
$
  W = 5 \times {10^5} \times A + 5 \times {10^5} \\
\Rightarrow W = 5 \times {10^5} \times \left( {A + 1} \right) \\
 $
The maximum compressive stress of pillar is given as
$S = 16 \times {10^5}\,N{m^{ - 1}}$
Stress is equal to the ratio of force applied to the cross-sectional area of the base of the pillar as
$S = \dfrac{W}{A}$
Substituting the known values,
$
\Rightarrow 16 \times {10^5} = \dfrac{{5 \times {{10}^5} \times \left( {A + 1} \right)}}{A} \\
\Rightarrow \dfrac{{\left( {A + 1} \right)}}{A} = \dfrac{{16 \times {{10}^5}}}{{5 \times {{10}^5}}} \\
\Rightarrow \dfrac{{\left( {A + 1} \right)}}{A} = 3.2 \\
\Rightarrow 1 + \dfrac{1}{A} = 3.2 \\
\Rightarrow \dfrac{1}{A} = 3.2 - 1 \\
 \Rightarrow \dfrac{1}{A} = 2.2 \\
 \Rightarrow A = \dfrac{1}{{2.2}} \\
 \Rightarrow A = 0.45\,{m^2} \\
 $
$\therefore$ The minimum cross-sectional area of the pillar is $0.45\,{m^2}$. Thus, the option (D) is correct.

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