JEE MAIN - Physics (2022 - 26th July Morning Shift - No. 21)
In an experiment to determine the Young's modulus of wire of a length exactly $$1 \mathrm{~m}$$, the extension in the length of the wire is measured as $$0.4 \mathrm{~mm}$$ with an uncertainty of $$\pm\, 0.02 \mathrm{~mm}$$ when a load of $$1 \mathrm{~kg}$$ is applied. The diameter of the wire is measured as $$0.4 \mathrm{~mm}$$ with an uncertainty of $$\pm \,0.01 \mathrm{~mm}$$. The error in the measurement of Young's modulus $$(\Delta \mathrm{Y})$$ is found to be $$x \times 10^{10}\, \mathrm{Nm}^{-2}$$. The value of $$x$$ is _________________. $$\left(\right.$$take $$\mathrm{g}=10 \mathrm{~ms}^{-2}$$ )
Answer
2
Explanation
$${{F/A} \over {l/L}} = Y,\,A = \pi {D^2}$$
$${{\Delta Y} \over Y} = {{\Delta F} \over F} + {{2\Delta D} \over D} + {{\Delta l} \over e} + {{\Delta L} \over L}$$
$$ = 2 \times {{0.01} \over {0.4}} + {{0.02} \over {0.4}}$$
$$ = {{0.04} \over {0.4}} = {1 \over {10}}$$
$$Y = {{Fl} \over {A\Delta l}}$$
$$ = {{10 \times 1} \over {\pi {{(0.1\,mm)}^2} \times 0.4\,mm}}$$
$$ = 1.988 \times {10^{11}}$$
$$ \approx 2 \times {10^{11}}$$
$${{\Delta y} \over y} = {1 \over {10}}$$
$$\Delta y = {y \over {10}} = 2 \times {10^{10}}$$
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