Question

On applying force on a rod along its length, the rod gets elongated by $0.04\text{m}$. If the length of rod is doubled and diameter is also doubled and same force is applied along the length, and it is found that the rod gets elongated by $x\times {10}^{-2}\text{m}$. Find the value of $x$.

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is A $2$


Let the length of the rod is $L$ and cross sectional area of the rod is $A$.
We know that, Young's modulus
$Y=\frac{\text{Stress}}{\text{Strain}}=\frac{\frac{F}{A}}{\frac{\Delta L}{L}}$
$\Rightarrow Y=\frac{FL}{A\Delta L}$
$\Rightarrow \frac{F}{A}=\frac{Y\Delta L}{L}$
$\Rightarrow \frac{F}{A}=\frac{Y\times 0.04}{L}\dots \dots \left(1\right)$

Now, length and diameter is doubled,
$\Rightarrow \frac{F}{4A}=Y\times \frac{\Delta L}{2L}[A=\frac{\pi d^2}{4}]$
$\Rightarrow \frac{Y\times 0.04}{L\times 4}=\frac{Y\times \Delta L}{2L}$ [from 1]
$\Rightarrow \Delta L=0.02\text{m}$
$\Rightarrow \Delta L=2\times {10}^{-2}\text{m}=x\times {10}^{-2}\text{m}$ (given)
$\Rightarrow x=2$