JEE MAIN - Physics (2023 - 25th January Morning Shift - No. 23)
As shown in the figure, in an experiment to determine Young's modulus of a wire, the extension-load curve is plotted. The curve is a straight line passing through the origin and makes an angle of 45$$^\circ$$ with the load axis. The length of wire is 62.8 cm and its diameter is 4 mm. The Young's modulus is found to be $$x\times10^4$$ Nm$$^{-2}$$. The value of $$x$$ is ___________.
_25th_January_Morning_Shift_en_23_1.png)
Answer
5
Explanation
Given, Length, $L=62.8 \mathrm{~cm}$
diameter, $d=4 \mathrm{~mm}$
radius, $r=2 \mathrm{~mm}$
$y=x \times 10^4 \mathrm{~N} / \mathrm{m}^2$
_25th_January_Morning_Shift_en_23_2.png)
According to graph, $\Delta l / f=\tan 45^{\circ}=1$
$Y=\frac{F L}{A \cdot \Delta L}$
$Y=\frac{1 \times 62.8 \times 10^{-2}}{3.14 \times 2 \times 2 \times 10^{-6}}=5 \times 10^4 \mathrm{~N} / \mathrm{m}^2$
$$ \therefore $$ x = 5
diameter, $d=4 \mathrm{~mm}$
radius, $r=2 \mathrm{~mm}$
$y=x \times 10^4 \mathrm{~N} / \mathrm{m}^2$
According to graph, $\Delta l / f=\tan 45^{\circ}=1$
$Y=\frac{F L}{A \cdot \Delta L}$
$Y=\frac{1 \times 62.8 \times 10^{-2}}{3.14 \times 2 \times 2 \times 10^{-6}}=5 \times 10^4 \mathrm{~N} / \mathrm{m}^2$
$$ \therefore $$ x = 5
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