A student performs an experiment to determine the Young's modulus… - Vidyadip

MCQ

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
$(2.0 \pm 0.3) \times 10^{11} \mathrm{~N} / \mathrm{m}^2$
✓
$(2.0 \pm 0.2) \times 10^{11} \mathrm{~N} / \mathrm{m}^2$
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

Correct option: B.

$(2.0 \pm 0.2) \times 10^{11} \mathrm{~N} / \mathrm{m}^2$

b
$Y =\frac{4 Fl }{\pi D ^2 e }$

Taking long both the sides

$\log Y =\log 4 Fl -\log \pi D ^2 e$

Now, partially differentiating,

$\frac{\Delta Y }{ Y }=-\left(\frac{2 \Delta D }{ D }+\frac{\Delta e }{ e }\right)$

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

$\frac{\Delta Y }{ Y }=-0.1125$

$\text { Also, } Y =\frac{ Fl }{ Ae }=\frac{9.8 \times 2}{\pi(0.2)^2 \times 0.8}=194.96 \times 10^9 \approx 2 \times 10^{11}$

$\Delta Y =- Y \times\left(\frac{2 \Delta r }{ r }+\frac{\Delta e }{ e }\right)$

$\Delta Y =0.225 \times 10^{11}$

$Y =(2 \pm 0.2) \times 10^{11} Nm ^{-2}$

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