Question

A thin rod of negligible mass, length $0.5\text{m}$ and area of cross-section $4\times {10}^{-6}{\text{m}}^2$, suspended vertically from one end. The rod is cooled down and it can contract by $5\times {10}^{-4}\text{m}$ but prevented from contracting by attaching a mass at the lower end. Choose the correct option(s).
(Given: Take young's modulus of rod $Y={10}^{11}{\text{N/m}}^2,g=10{\text{m/s}}^2)$

• A. Value of mass is $80\text{kg}$
• B. Value of mass is $40\text{kg}$
• C. Energy stored in the rod is $0.1\text{J}$
• D. Energy stored in the rod is $0.2\text{J}$

Answer 1

Answer: (B), (C)

The correct options are
B Value of mass is $40\text{kg}$
C Energy stored in the rod is $0.1\text{J}$

Area of cross-section, $A=4\times {10}^{-6}{\text{m}}^2$
Length of rod, $L=0.5\text{m}$
Possible contraction in rod, $\Delta L=5\times {10}^{-4}\text{m}$
According to question,


Force applied by attached mass $\left(m\right)$,
$\Rightarrow F=mg=10m....\left(1\right)$
For Young's Modulus we can write,
$Y=\frac{FL}{A\Delta L}$
$\Rightarrow F=\frac{YA\Delta L}{L}$$F=\frac{{10}^{11}\times 4\times {10}^{-6}\times 5\times {10}^{-4}}{0.5}$
$\therefore F=400\text{N}....\left(2\right)$
For the external load $\left(F\right)$ on rod, equating Eq.$\left(1\right)$ and $\left(2\right)$:
$10m=400$
$\Rightarrow m=40\text{kg}$
Energy stored in the rod,
$U=\frac{1}{2}\times \frac{YA}{L}\times \Delta L^2$
$U=\frac{1}{2}\times \frac{{10}^{11}\times 4\times {10}^{-6}}{0.5}\times (5\times {10}^{-4}{)}^2$
$\therefore U=0.1\text{J}$