In the Searle’s method to determine the Young’s modulus of a wire, a steel wire of length 156 cm and diameter 0.054 cm is taken as experimental wire. The average increase in length for $1\frac{1}{2}$kg wt is found to be 0.050 cm. Then the Young’s modulus of the wire.

Given,

length of steel wire =156cm

diameter = 0.054cm

change in length$\mathrm{\Delta L}=0.050 \mathrm{cm}$

mass of the steel = 1.5 kgwt

$$\begin{gathered} \mathrm{radius}=\frac{\mathrm{d}}{2} \\ \mathrm{r}=\frac{0.054\times {10}^{-2}}{2} \\ \mathrm{r}=2.7\times {10}^{-4} \end{gathered}$$

Thus the young's modulus is,

$$\begin{gathered} \mathrm{Force}=\mathrm{mg} \\ \mathrm{F}=1.5\times 9.8; \end{gathered}$$

$$\begin{gathered} \mathrm{F}=14.7\mathrm{N}; \\ \mathrm{Y}=\frac{\mathrm{FL}}{\mathrm{A}∆\mathrm{l}}; \end{gathered}$$

$$\begin{gathered} \mathrm{Y}=\frac{14.7\times 1.56}{3.14\times 2.7\times {10}^{-4}\times 5\times {10}^{-4}}; \\ \mathrm{Y}=2.002\times {10}^{-11}\mathrm{N}/{\mathrm{m}}^2; \end{gathered}$$

Hence the correct answer is $2.002\times {10}^{-11}N/m^2.$