A student performs an experiment to determine the Young's modulus of a wire, exactly 2 m long, by Searle's method. In a particular reading, the student measures the extension in the length of the wire to be 0.8 rom with an uncertainty of ± 0.05 mm at a load of exactly 1.0 kg. The student also measures the diameter of the wire to be 0.4 mm with an uncertainty of ± 0.01mm. Take g =9.8 m/s² (exact). The Young's modulus obtained from the reading is

The correct option is (2.0 ±0.2)x 10¹¹ N/m².

Step 1: Formula for Young's Modulus
Young's modulus ($Y$) is defined as the ratio of tensile stress to tensile strain:
$Y=\frac{\text{Stress}}{\text{Strain}}=\frac{F/A}{\Delta L/L}=\frac{FL}{A\Delta L}$
For a wire of circular cross-section with diameter $d$, the cross-sectional area is $A=\frac{\pi }{4}{d}^2$. The force applied is due to a load of mass $M$, so $F=Mg$. Substituting these values into the formula gives:
$Y=\frac{4MgL}{\pi d^2\Delta L}$

Step 2: Calculate the nominal value of Young's Modulus
From the given data, we have:
Length, $L=2\text{m}$ (exact)
Mass, $M=1.0\text{kg}$ (exact)
Acceleration due to gravity, $g=9.8{\text{m/s}}^2$ (exact)
Diameter, $d=0.4\text{mm}=0.4\times {10}^{-3}\text{m}$
Extension, $\Delta L=0.8\text{mm}=0.8\times {10}^{-3}\text{m}$
Substituting these values:
$Y=\frac{4\times 1.0\times 9.8\times 2}{\pi \times {\left(0.4\times {10}^{-3}\right)}^2\times \left(0.8\times {10}^{-3}\right)}$
$Y=\frac{78.4}{\pi \times 0.16\times {10}^{-6}\times 0.8\times {10}^{-3}}$
$Y=\frac{78.4}{0.128\pi \times {10}^{-9}}\approx 1.95\times {10}^{11}{\text{N/m}}^2\approx 2.0\times {10}^{11}{\text{N/m}}^2$

Step 3: Error Analysis for Young's Modulus
Since the variables $M$, $g$, and $L$ are exact, the uncertainty in $Y$ depends only on the measurements of diameter $d$ and extension $\Delta L$. The relative error is given by:
$\frac{\delta Y}{Y}=2\frac{\delta d}{d}+\frac{\delta \left(\Delta L\right)}{\Delta L}$
Given uncertainties:
Uncertainty in diameter, $\delta d=0.01\text{mm}$
Uncertainty in extension, $\delta \left(\Delta L\right)=0.05\text{mm}$
Substituting the values:
$\frac{\delta Y}{Y}=2\left(\frac{0.01}{0.4}\right)+\frac{0.05}{0.8}$
$\frac{\delta Y}{Y}=2\left(0.025\right)+0.0625=0.05+0.0625=0.1125$

Step 4: Calculate the absolute uncertainty
Using the nominal value of $Y\approx 2.0\times {10}^{11}{\text{N/m}}^2$:
$\delta Y=Y\times 0.1125=2.0\times {10}^{11}\times 0.1125=0.225\times {10}^{11}{\text{N/m}}^2\approx 0.2\times {10}^{11}{\text{N/m}}^2$
Therefore, the value of Young's modulus with the uncertainty limit is:
$Y=\left(2.0\pm 0.2\right)\times {10}^{11}{\text{N/m}}^2$