A composite rod is $1000{{mm}}$ long, its two ends are $40{{m}}{{{m}}^{{2}}}$ and $30{{m}}{{{m}}^{{2}}}$ in area and length are $300{{mm}}$ and $200{{mm}}$ respectively. The middle portion of the rod is $20{{m}}{{{m}}^{{2}}}$ in area and $500{{mm}}$ long. If the rod is subjected to an axial tensile load of $1000{{N}}$, find its total elongation(in mm).$\left( {E = 200{{GPa}}} \right)$
a. $0.165$
b. $0.111$
c. $0.196$
d. None of the above

Hint: The force given is the same for three different regions of the rod. And the elongation of each region is proportional to the force loaded. This different proportionality constant wants to be found and the equivalent elongation can be found.

Complete step by step answer:
Given an axial tensile load of $1000{{N}}$ is applied to the rod and the modulus of elasticity is $E = 200{{GPa}}$.
The expression for modulus of elasticity is given as,
$E = \dfrac{{F \times L}}{{A \times dl}}$
Where, $F$ is the force applied, $L$ is the length, $A$ is the area and $dl$ is the elongation.
From the above equation,
$
  F = \dfrac{{E \times A}}{L}dl \\
  F = K \times dl \\
$
The elongation is proportional to the force. And the factor $K$ is different for each region.

Let’s find that factor, $K = \dfrac{{E \times A}}{L}$
Given length of the first region, ${L_1} = 300{{mm}}$ and area of that region,${A_1} = 40{{m}}{{{m}}^{{2}}}$.
Therefore,
${K_1} = \dfrac{{E \times {A_1}}}{{{L_1}}}$
Substitute the values in the above expression.
$
  {K_1} = \dfrac{{E \times 40{{m}}{{{m}}^{{2}}}}}{{300{{mm}}}} \\
 \Rightarrow\dfrac{{E \times 40}}{{300}} \times {10^{ - 3}}{{m}} \\
  {{ = }}\dfrac{{4E}}{3} \times {10^{ - 4}}{{m}} \\
$
For the second region, the length is given as, length ${L_2} = 500{{mm}}$ and area, ${A_2} = 20{{m}}{{{m}}^{{2}}}$.
Therefore, ${K_2} = \dfrac{{E \times {A_2}}}{{{L_2}}}$

Substitute the values in the above expression.
$
  {K_2} = \dfrac{{E \times 20{{m}}{{{m}}^{{2}}}}}{{500{{mm}}}} \\
 \Rightarrow\dfrac{{E \times 20}}{{500}} \times {10^{ - 3}}{{m}} \\
  {{ = }}\dfrac{{2E}}{5} \times {10^{ - 4}}{{m}} \\
$
For the third region length is given as, ${L_3} = 200{{mm}}$ and area is ${A_3} = 30{{m}}{{{m}}^{{2}}}$

Therefore, ${K_3} = \dfrac{{E \times {A_3}}}{{{L_3}}}$
Substitute the values in the above expression.
$
  {K_3} = \dfrac{{E \times 30{{m}}{{{m}}^{{2}}}}}{{200{{mm}}}} \\
 \Rightarrow\dfrac{{E \times 30}}{{200}} \times {10^{ - 3}}{{m}} \\
  {{ = }}\dfrac{{3E}}{2} \times {10^{ - 4}}{{m}} \\
$
Since the three regions are in series. Then the equivalent constant factor is given as,
$\dfrac{1}{{{K_{eq}}}} = \dfrac{1}{{{K_1}}} + \dfrac{1}{{{K_2}}} + \dfrac{1}{{{K_3}}}$
Substituting the values in the above expression,
$
\dfrac{1}{{{K_{eq}}}} = \dfrac{{3 \times {{10}^4}}}{{4E}} + \dfrac{{5 \times {{10}^4}}}{{2E}} + \dfrac{{2 \times {{10}^4}}}{{3E}} \\
 \Rightarrow\dfrac{{\left( {9 + 30 + 8} \right) \times {{10}^4}}}{{12E}} \\
 \Rightarrow\dfrac{{47 \times {{10}^4}}}{{12E}} \\
$
Therefore, ${K_{eq}} = \dfrac{{12E}}{{47 \times {{10}^4}}}$

We know that, $F = {K_{eq}} \times dl$
Therefore the total elongation is given as,
$
  dl = \dfrac{F}{{{K_{eq}}}} \\
 \Rightarrow1000{{N}} \times \dfrac{{47 \times {{10}^4}}}{{12E}} \\
 \Rightarrow1000{{N}} \times \dfrac{{47 \times {{10}^4}}}{{12 \times 200 \times {{10}^9}{{Pa}}}} \\
 \Rightarrow1.956 \times {10^{ - 4}}{{m}} \\
  {{dl = 0}}{{.1956}}{{mm}} \\
  {{dl = 0}}{{.196}}{{mm}} \\
$

Thus the total elongation of the rod is ${{0}}{{.196}}{{mm}}$.
Hence, the correct answer is option (C).

Note: We have to note that the equivalent sum of the constant factor will be a smaller value since they are in series. But the total elongation will be greater since the common force is divided by the equivalent constant factor.