A student performs an experiment to determine the Young’s modulus of a wire, exactly 2m  long, by Searle’s method. In a particular reading, the student measures the extension in the length of the wire to be 0.8mm  with an uncertainty of  $\pm 0.05\text{\hspace{0.17em}mm}$ at a load of exactly  1.0kg. The student also measures the diameter of the wire to be  0.4mm with an uncertainty of  $\pm 0.01\text{\hspace{0.17em}mm}$. Take $g=9.8\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}$ (exact). The Young’s modulus obtained from the reading is[ Take $\text{π}=3.14$ ]

Solution:

Given Length of wire $l=2\text{\hspace{0.17em}m}$
Extension= $0.8\text{\hspace{0.17em}mm}$
Uncertainty =0.05 mm
Diameter of the wire   $d=0.4\text{\hspace{0.17em}mm}$
Mass  $m=1.0\text{\hspace{0.17em}kg}$
Acceleration due to gravity  
Using Hooke’s Law
$Y=\frac{FL}{Al}=\frac{4FL}{\pi d^{2}l}\mathrm{.....}\left(1\right)$
Also   $Y=\frac{FL}{Al}=\frac{mgL}{\text{π}{r}^{2}l}\left(\begin{array}{l}\because F=mg\\ \because A=\pi r^2\left(or\right)\pi \frac{d^2}{4}\end{array}\right)$
Where  L= length of the wire
l= elongation of the wire
d= diameter of the wire
 $\begin{array}{l}Y=\frac{1.0\times 9.8\times 2}{3.14\times {\left(0.2\right)}^2\times 0.8}\\ Y=194.96\times {10}^9\\ =1.95\times {10}^{11}\text{\hspace{0.17em}N}/{\text{m}}^{\text{2}}\end{array}$
From equation (1)
$Y=\frac{FL}{Al}=\frac{4FL}{\pi d^{2}l}$
Taking log on both sides
$\Rightarrow \log Y=\log 4FL-\log \text{π}{d}^{2}l$
Now partially differentiating
$\begin{array}{c}\Rightarrow \frac{\mathrm{\Delta }Y}{Y}=2\frac{\mathrm{\Delta }d}{d}+\frac{\mathrm{\Delta }l}{l}\\ =2\left(\frac{0.01}{0.4}\right)+\frac{0.05}{0.8}=\frac{9}{80}\\ \Rightarrow \mathrm{\Delta }Y=\frac{9}{80}\times Y\\ =\frac{9}{80}\times 1.95\times {10}^{11}\\ =2.19\times {10}^{11}\mathrm{N}/{\mathrm{m}}^2\\ Y=(2\pm 0.219)\times {10}^{11}\mathrm{N}/{\mathrm{m}}^2\\ \therefore Y\approx (2\pm 0.2)\times {10}^{11}\mathrm{N}/{\mathrm{m}}^2\end{array}$
Therefore, the correct answer is (B).