Question

In Searl's experiment diameter of wire is measured with screw gauge of least count $0.001 \text{ cm}$ and its value is $0.08 \text{ cm}$. Length of wire is $150 \text{ cm}$ with least count $0.1 \text{ cm}$ and elongation $0.5 \text{ cm}$ with least count $0.001 \text{ cm}$. Weight suspended is $100 \text{ N}$ then absolute error in Young's modulus is $N \times 10^9 \text{ N/m}^2$. Then $N$ is :

• A. 1.64
• B. 1.65
• C. 1.63
• D. 1.66

Answer 1

The Correct Option is B

Step 1: Write the formula for Young's modulus $Y$.
$Y = \frac{4WL}{\pi d^2 \ell}$.

Step 2: Calculate the value of $Y$.
$Y \approx 5.97 \times 10^{10} \text{ N/m}^2$.

Step 3: Express the relative error in $Y$.
$\frac{\Delta Y}{Y} = \frac{\Delta L}{L} + 2\frac{\Delta d}{d} + \frac{\Delta \ell}{\ell}$.

Step 4: Substitute the given values of absolute errors (least counts) and measurements.
$\Delta L = 0.1$, $L = 150$.
$\Delta d = 0.001$, $d = 0.08$.
$\Delta \ell = 0.001$, $\ell = 0.5$.
$\frac{\Delta Y}{Y} = \frac{0.1}{150} + 2\left(\frac{0.001}{0.08}\right) + \frac{0.001}{0.5} = 0.00067 + 0.025 + 0.002 = 0.02767$.

Step 5: Calculate absolute error $\Delta Y$.
$\Delta Y = 0.02767 \times 5.97 \times 10^{10} = 1.652 \times 10^9 \text{ N/m}^2$.
Comparing with $N \times 10^9$, we find $N = 1.65$.

Final Answer: Option (2).