JEE MAIN - Physics (2022 - 28th July Morning Shift - No. 6)
The force required to stretch a wire of cross-section $$1 \mathrm{~cm}^{2}$$ to double its length will be : (Given Yong's modulus of the wire $$=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}$$)
$$1 \times 10^{7} \mathrm{~N}$$
$$1.5 \times 10^{7} \mathrm{~N}$$
$$2 \times 10^{7} \mathrm{~N}$$
$$2.5 \times 10^{7} \mathrm{~N}$$
Explanation
$$A = 1$$ cm2
$$Y = {{Fl} \over {A\Delta l}}$$
$$F = {{YA\Delta l} \over l} = {{2 \times {{10}^{11}} \times {{10}^{ - 4}} \times l} \over l}$$
$$ = 2 \times {10^7}$$ N
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