JEE MAIN - Physics (2022 - 29th June Morning Shift - No. 5)
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 :
$$\sqrt {{L_1}{L_2}} $$
$${{{L_1} + {L_2}} \over 2}$$
$$2{L_1} - {L_2}$$
$$3{L_1} - 2{L_2}$$
Explanation
$$y = {{FL} \over {A\Delta L}}$$
$$ \Rightarrow \Delta L = {{FL} \over {Ay}}$$
$$ \Rightarrow {L_1} = L + {{(1g)L} \over {Ay}}$$ ..... (i)
and $${L_2} = L + {{(2g)L} \over {Ay}}$$ ..... (ii)
$$ \Rightarrow L = 2{L_1} - {L_2}$$
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