General Physics Ques 33
Question 12
- A student performs an experiment to determine the Young’s modulus of a wire, exactly $2 \mathrm{~m}$ long, by Searle’s method. In a particular reading, the student measures the extension in the length of the wire to be $0.8 \mathrm{~mm}$ with an uncertainty of $\pm 0.05 \mathrm{~mm}$ at a load of exactly $1.0 \mathrm{~kg}$. The student also measures the diameter of the wire to be $0.4 \mathrm{~mm}$ with an uncertainty of $\pm 0.01 \mathrm{~mm}$. Take $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ (exact). The Young’s modulus obtained from the reading is close to
(2007, 3M) (a) $(2.0 \pm 0.3) 10^{11} \mathrm{~N} / \mathrm{m}^{2}$ (b) $(2.0 \pm 0.2) \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}$ (c) $(2.0 \pm 0.1) \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}$ (d) $(2.0 \pm 0.05) \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}$
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Answer:
Correct Answer: 33.(b)
Solution:
Formula:
$$ \begin{aligned} Y & =\frac{F L}{A l}=\frac{4 F L}{\pi d^{2} l}=\frac{(4)(1.0 \times 9.8)(2)}{\pi\left(0.4 \times 10^{-3}\right)^{2}\left(0.8 \times 10^{-3}\right)} \ & =2.0 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2} \end{aligned} $$
Further $\frac{\Delta Y}{Y}=2 \frac{\Delta d}{d}+\frac{\Delta l}{l}$
$\therefore \quad \Delta Y=2 \frac{\Delta d}{d}+\frac{\Delta l}{l} \quad Y$
$=2 \times \frac{0.01}{0.4}+\frac{0.05}{0.8} \times 2.0 \times 10^{11}$
$=0.225 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}$
$=0.2 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}$
(By rounding off)
or $(Y+\Delta Y)=(2 \pm 0.2) \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}$
BETA
