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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
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Verified Answer
The correct answer is:
$(2.0 \pm 0.2) \times 10^{11} \mathrm{~N} / \mathrm{m}^2$
$(2.0 \pm 0.2) \times 10^{11} \mathrm{~N} / \mathrm{m}^2$
$\begin{aligned} Y & =\frac{F L}{A l}=\frac{4 F L}{\pi d^2 l} \\ & =\frac{(4)(1.0 \times 9.8)(2)}{\pi\left(0.4 \times 10^{-3}\right)^2\left(0.8 \times 10^{-3}\right)}=20 \times 10^{11} \mathrm{~N} / \mathrm{m}^2\end{aligned}$
Further $\quad \frac{\Delta Y}{Y}=2\left(\frac{\Delta d}{d}\right)+\left(\frac{\Delta l}{l}\right)$
$$
\begin{aligned}
\therefore \quad \Delta Y & =\left\{2\left(\frac{\Delta d}{d}\right)+\left(\frac{\Delta l}{l}\right)\right\} Y=\left\{2 \times \frac{0.01}{0.4}+\frac{0.05}{0.8}\right\} \times 2.0 \times 10^{11} \\
& =0.2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2 \text { or }(Y+\Delta Y)=(2+0.2) \times 10^{11} \mathrm{~N} / \mathrm{m}^2
\end{aligned}
$$
The correct option is (b).
Further $\quad \frac{\Delta Y}{Y}=2\left(\frac{\Delta d}{d}\right)+\left(\frac{\Delta l}{l}\right)$
$$
\begin{aligned}
\therefore \quad \Delta Y & =\left\{2\left(\frac{\Delta d}{d}\right)+\left(\frac{\Delta l}{l}\right)\right\} Y=\left\{2 \times \frac{0.01}{0.4}+\frac{0.05}{0.8}\right\} \times 2.0 \times 10^{11} \\
& =0.2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2 \text { or }(Y+\Delta Y)=(2+0.2) \times 10^{11} \mathrm{~N} / \mathrm{m}^2
\end{aligned}
$$
The correct option is (b).
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