Question
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 $$0.8 \,mm$$ with an uncertainty of $$ \pm 0.05\,mm$$ 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 $$ \pm 0.01\,mm.$$ Take $$g = 9.8\,m/{s^2}$$ (exact). The Young's modulus obtained from the reading is-
A.
$$\left( {2.0 \pm 0.3} \right) \times {10^{11}}N/{m^2}$$
B.
$$\left( {2.0 \pm 0.2} \right) \times {10^{11}}N/{m^2}$$
C.
$$\left( {2.0 \pm 0.1} \right) \times {10^{11}}N/{m^2}$$
D.
$$\left( {2.0 \pm 0.05} \right) \times {10^{11}}N/{m^2}$$
Answer :
$$\left( {2.0 \pm 0.2} \right) \times {10^{11}}N/{m^2}$$
Solution :
$$\eqalign{
& Y = \frac{{4mgL}}{{\pi {D^2}\ell }} \cr
& = \frac{{4 \times 1 \times 9.8 \times 2}}{{\pi {{\left( {0.4 \times {{10}^{ - 3}}} \right)}^2} \times \left( {0.8 \times {{10}^{ - 3}}} \right)}} \cr
& = 2.0 \times {10^{11}}\,N/{m^2} \cr
& {\text{Now, }}\frac{{\Delta Y}}{Y} = \frac{{2\Delta D}}{D} + \frac{{\Delta \ell }}{\ell } \cr
& \left[ {\because {\text{ the value of }}m,{\text{ }}g{\text{ and }}L{\text{ are exact}}} \right] \cr
& = 2 \times \frac{{0.01}}{{0.4}} + \frac{{0.05}}{{0.8}} \cr
& = 2 \times 0.025 + 0.0625 \cr
& = 0.05 + 0.0625 \cr
& = 0.1125 \cr
& \Rightarrow \Delta Y = 2 \times {10^{11}} \times 0.1125 \cr
& = 0.225 \times {10^{11}} \cr
& = 0.2 \times {10^{11\,}}N/{m^2} \cr} $$
Note: We can also take value of $$y$$ from options given without calculating it as it is same in all options.
$$\therefore Y = \left( {2 \pm 0.2} \right) \times {10^{11}}N/{m^2}$$