A student performs an experiment to determine 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 $\pm 0.05mm$ at a load of exactly $1.0kg$. The student also measures the diameter of the wire to be $0.4mm$ with an uncertainty of $\pm 0.01mm$. Take $g=9.8m{s}^{-2}$ (exact). The Young’s modulus obtained from the reading is

The correct option is B

$\left(2.0\pm 0.2\right)\times {10}^{11}N{m}^{-2}$

Step 1. Given data

Exact length of the wire, $L=2m$

Extension in the length of the wire, $l=0.8mm$

Uncertainty in the length of the wire, $∆l=\pm 0.05mm$

Diameter of the wire, $D=0.4mm$

Uncertainty in the diameter of the wire, $∆D=\pm 0.01mm$

Mass of the load, $m=1.0kg$

Step 2. Finding the Young's modulus, $Y$

By using the formula of Young's modulus,

$Y=\frac{4mgL}{{\mathrm{\pi D}}^2\mathrm{l}}$

$Y=\frac{4\times 1\times 9.8\times 2}{3.14\times {\left(0.4\times {10}^{-3}\right)}^2\times 0.8\times {10}^{-3}}$ [$1mm={10}^{-3}m$, $g=9.8m{s}^{-2},\mathrm{\pi }=3.14$]

$Y=2\times {10}^{11}N{m}^{-2}$

Step 3. Finding the uncertainty in the young's modulus, $∆Y$

By using the formula,

$\frac{∆Y}{Y}=2\times \frac{∆D}{D}+\frac{∆l}{l}$

$\frac{∆Y}{2\times {10}^{11}}=2\times \frac{0.01}{0.4}+\frac{0.05}{0.8}$

$∆Y=0.22\times {10}^{11}N{m}^{-2}$

Therefore, the Young's modulus obtained from the reading is $\left(2.0\pm 0.2\right)\times {10}^{11}N{m}^{-2}$.

Hence, the correct option is $\left(B\right)$.