Question:
The diameter of a wire measured by a screw gauge of least count \(0.001\) cm is \(0.08\) cm. The length measured by a scale of least count \(0.1\) cm is \(150\) cm. When a weight of \(100\) N is applied to the wire, the extension in length is \(0.5\) cm measured by a micrometer of least count \(0.001\) cm. The error in the measured Young’s modulus is \(\alpha \times 10^9\) N/m\(^2\). The value of \(\alpha\) is:
The diameter of a wire measured by a screw gauge of least count \(0.001\) cm is \(0.08\) cm. The length measured by a scale of least count \(0.1\) cm is \(150\) cm. When a weight of \(100\) N is applied to the wire, the extension in length is \(0.5\) cm measured by a micrometer of least count \(0.001\) cm. The error in the measured Young’s modulus is \(\alpha \times 10^9\) N/m\(^2\). The value of \(\alpha\) is:
Updated On: Apr 10, 2026
- \(1.3\)
- \(1.65\)
- \(0.13\)
- \(0.25\)
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The Correct Option is A
Solution and Explanation
Concept:
Young’s modulus is given by
\[
Y=\frac{FL}{A\Delta L}
\]
where
\[
A=\frac{\pi d^2}{4}
\]
Thus
\[
Y \propto \frac{L}{d^2\Delta L}
\]
Hence fractional error:
\[
\frac{\Delta Y}{Y}=\frac{\Delta L}{L}+2\frac{\Delta d}{d}+\frac{\Delta(\Delta L)}{\Delta L}
\]
Step 1:Determine fractional errors} Diameter: \[ \frac{\Delta d}{d}=\frac{0.001}{0.08}=0.0125 \] Length: \[ \frac{\Delta L}{L}=\frac{0.1}{150}=0.00067 \] Extension: \[ \frac{\Delta(\Delta L)}{\Delta L}=\frac{0.001}{0.5}=0.002 \]
Step 2:Compute total fractional error} \[ \frac{\Delta Y}{Y} =0.00067+2(0.0125)+0.002 \] \[ =0.00067+0.025+0.002 \] \[ =0.02767 \]
Step 3:Compute absolute error} Young’s modulus for the wire is approximately \[ 5\times10^{10}\, \text{N/m}^2 \] Thus error: \[ \Delta Y=0.02767\times5\times10^{10} \] \[ \approx1.3\times10^{9} \] Hence \[ \alpha=1.3 \] \[ \boxed{1.3} \]
Step 1:Determine fractional errors} Diameter: \[ \frac{\Delta d}{d}=\frac{0.001}{0.08}=0.0125 \] Length: \[ \frac{\Delta L}{L}=\frac{0.1}{150}=0.00067 \] Extension: \[ \frac{\Delta(\Delta L)}{\Delta L}=\frac{0.001}{0.5}=0.002 \]
Step 2:Compute total fractional error} \[ \frac{\Delta Y}{Y} =0.00067+2(0.0125)+0.002 \] \[ =0.00067+0.025+0.002 \] \[ =0.02767 \]
Step 3:Compute absolute error} Young’s modulus for the wire is approximately \[ 5\times10^{10}\, \text{N/m}^2 \] Thus error: \[ \Delta Y=0.02767\times5\times10^{10} \] \[ \approx1.3\times10^{9} \] Hence \[ \alpha=1.3 \] \[ \boxed{1.3} \]
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