JEE Main 2022 · 29 Jun · Shift 1 · Physics Q36

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PhysicsSingle correct+4 / -1Question 36 of 90

A wire of length L is hanging from a fixed support. The length changes to $L_1$ and $L_2$ when masses 1 kg and 2 kg are suspended respectively from its free end. Then the value of L is equal to :

\sqrt{L_1L_2}

\dfrac{L_1+L_2}2

2L_1-L_2

3L_1-2L_2

Answer and worked solution

Official answer: C

Hooke's law. $L_1=L+k$ , $L_2=L+2k$ , so $L=2L_1-L_2$ .

(C)

Topics:Hooke's law and elastic moduli

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JEE Main 2022 · 29 Jun · Shift 1, Physics question 36