JEE Advanced · Physics · 1. Math in Physics
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
- 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 \((2.0 \pm 0.05) \times 10^{11} \mathrm{~N} / \mathrm{m}^2\)
Answer & Solution
Correct Answer
(B) \((2.0 \pm 0.2) \times 10^{11} \mathrm{~N} / \mathrm{m}^2\)
Step-by-step Solution
Detailed explanation
\(Y =\frac{F L}{A l}=\frac{4 F L}{\pi d^2 l} \)
\( =\frac{(4)(1.0 \times 9.8)(2)}{\pi\left(0.4 \times 10^{-3}\right)^2\left(0.8 \times 10^{-3}\right)}=20 \times 10^{11} \mathrm{~N} / \mathrm{m}^2\)
Further \(\quad \frac{\Delta Y}{Y}=2\left(\frac{\Delta d}{d}\right)+\left(\frac{\Delta l}{l}\right)\)
\(\therefore \Delta Y =\) \(\left\{2\left(\frac{\Delta d}{d}\right)+\left(\frac{\Delta l}{l}\right)\right\} Y=\left\{2 \times \frac{0.01}{0.4}+\frac{0.05}{0.8}\right\}\) \(\times~ 2.0 \times 10^{11} \)
\( =0.2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2 \text { or }(Y+\Delta Y)=(2+0.2)\) \(\times~ 10^{11} \mathrm{~N} / \mathrm{m}^2\)
The correct option is (b).
\( =\frac{(4)(1.0 \times 9.8)(2)}{\pi\left(0.4 \times 10^{-3}\right)^2\left(0.8 \times 10^{-3}\right)}=20 \times 10^{11} \mathrm{~N} / \mathrm{m}^2\)
Further \(\quad \frac{\Delta Y}{Y}=2\left(\frac{\Delta d}{d}\right)+\left(\frac{\Delta l}{l}\right)\)
\(\therefore \Delta Y =\) \(\left\{2\left(\frac{\Delta d}{d}\right)+\left(\frac{\Delta l}{l}\right)\right\} Y=\left\{2 \times \frac{0.01}{0.4}+\frac{0.05}{0.8}\right\}\) \(\times~ 2.0 \times 10^{11} \)
\( =0.2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2 \text { or }(Y+\Delta Y)=(2+0.2)\) \(\times~ 10^{11} \mathrm{~N} / \mathrm{m}^2\)
The correct option is (b).
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