JEE MAIN - Physics (2023 - 30th January Evening Shift - No. 4)
A force is applied to a steel wire 'A', rigidly clamped at one end. As a result elongation in the wire is $0.2 \mathrm{~mm}$. If same force is applied to another steel wire ' $\mathrm{B}$ ' of double the length and a diameter $2.4$ times that of the wire ' $\mathrm{A}$ ', the elongation in the wire ' $\mathrm{B}$ ' will be (wires having uniform circular cross sections)
$6 .9 \times 10^{-2} \mathrm{~mm}$
$6.06 \times 10^{-2} \mathrm{~mm}$
$2.77 \times 10^{-2} \mathrm{~mm}$
$3.0 \times 10^{-2} \mathrm{~mm}$
Explanation
$$\because$$ $$\Delta l = {{Fl(4)} \over {Y\pi {d^2}}}$$
$${{\Delta {l_1}} \over {\Delta {l_2}}} = {{\Delta {l_1}} \over {d_1^2}} \times {{d_2^2} \over {{l_2}}}$$
$${{0.2} \over {\Delta {l_2}}} = {1 \over 2} \times {(2.4)^2}$$
$$\Delta {l_2} = {{2 \times 0.2} \over {{{(2.4)}^2}}}$$
$$ = 6.9 \times {10^{ - 2}}$$ mm
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