Question

A student performs an experiment to determine the Young's modulus of a wire exactly 2cm 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 ? 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 ? 0.01 mm. Take g = 9.8 m/s² (exact). The Young's modulus obtained from the reading is -

• A. $\ (\text{2.0 } \pm \ 0.3)\ \cdot \ 10^{11}\ Ν/m^{2}$
• B. $(\text{2.0 } \pm \ 0.2)\ \cdot \ 10^{11}\ Ν/m^{2}\$
• C. $(\text{2.0 } \pm \ 0.1)\ \cdot \ 10^{11}\ Ν/m^{2}$
• D. $(\text{2.0 } \pm \ 0.05)\ \cdot \ 10^{11}\ Ν/m^{2}$

Answer 1

Answer: (B)

$(\text{2.0 } \pm \ 0.2)\ \cdot \ 10^{11}\ Ν/m^{2}\$

Answer 2

$\frac{\Delta Y}{Y}$ = $\frac{2\Delta D}{D}$ + $\frac{\Delta\mathcal{l}}{\mathcal{l}}$

$\frac{\Delta Y}{Y}$ = 2 $\left( \frac{0.01}{0.4} \right)$ + $\left( \frac{0.05}{0.8} \right)$

= 2× 0.025 + 0.0625

$\frac{\Delta Y}{Y}$ = 0.05 + 0.0625 = 0.1125

$\Delta Y\ = \ 2 \times \ 10^{11}\ \times \ 0.\ 1125\ = \ 0.225\ \times \ 10^{11}$So $(\text{2 } \pm \text{ }0.2)\ \times 10^{11}\ Ν/m^{2}$