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JEE Main 2022 (29 Jun, Shift 1) · Wire Combinations & Special Cases

Wire Combinations & Special Cases question from JEE Main 2022 (29 Jun, Shift 1)

Q15JEE Main 2022 · 29 JUN · SHIFT 1Medium

Wire Combinations & Special Cases

A wire of length L is hanging from a fixed support. The length changes to L1 and L2 when masses 1 kg and 2 kg are suspended respectively from its free end. Then the value of L is equal to:

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Answer: (c)

Extension $\propto$ load: $L_1 - L = k(1)$ and $L_2 - L = k(2)$
So $L_2 - L = 2(L_1 - L)$
$L = 2L_1 - L_2$ , so the answer is (c).

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