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Numerical · Q39

Q.A wire gets stretched by 4 mm due to a certain load. If the same load is applied to a wire of the same material with half the length and double the diameter of the first wire, what will be the change in its length?

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
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Step 1. For a wire under a fixed load FF: elongation l=FLAYl = \dfrac{FL}{AY}, and since A=π(d/2)2∝d2A = \pi (d/2)^2 \propto d^2, we get l∝Ld2l \propto \dfrac{L}{d^2} for the same material (same YY) and same FF.

Step 2. Original wire: length LL, diameter dd, elongation l1=4l_1 = 4 mm. New wire: length L′=L/2L' = L/2, diameter d′=2dd' = 2d, so area A′=4AA' = 4A. …

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