Medium MCQ +4 / -1 PYQ · JEE Mains 2022

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 :

  1. A $\sqrt {{L_1}{L_2}}$
  2. B ${{{L_1} + {L_2}} \over 2}$
  3. C $2{L_1} - {L_2}$ Correct answer
  4. D $3{L_1} - 2{L_2}$

Solution

<p>$y = {{FL} \over {A\Delta L}}$</p> <p>$\Rightarrow \Delta L = {{FL} \over {Ay}}$</p> <p>$\Rightarrow {L_1} = L + {{(1g)L} \over {Ay}}$ ..... (i)</p> <p>and ${L_2} = L + {{(2g)L} \over {Ay}}$ ..... (ii)</p> <p>$\Rightarrow L = 2{L_1} - {L_2}$</p>

About this question

Subject: Physics · Chapter: Properties of Solids and Liquids · Topic: Elasticity

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