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  • One end of a horizontal thick copper wire of length 2L and radius 2R is welded to an end of another horizontal thin copper wire of length L and radius R. When the arrangement is stretched by applyi

One end of a horizontal thick copper wire of length 2L and radius 2R is welded to an end of another horizontal thin copper wire of length L and radius R. When the arrangement is stretched by applying forces at two ends, the ratio of the elongation in the thin wire to that in the thick wire will be:

Option: 1

4


Option: 2

2


Option: 3

1


Option: 4

5


Answers (1)

best_answer

The elongation in the thin wire:

\therefore \Delta l = \frac{FL}{Ay} = \frac{FL}{(\Pi r^2)y} \implies \Delta l \alpha \frac{L}{A}

\therefore \frac{\Delta l1}{\Delta l2} = \frac{\frac{L}{R^2}}{\frac{2L}{(2R)^2}} = 2

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