A wire of length 'L' and radius 'r' is clamped rigidly at one end. When the other end of the wire is pulled by a force f, its length increases by 'l'. Another wire of same material of length '2L' and radius '2r' is pulled by a force '2f'. Then the increase in its length will be
- l
- 2l
- 4l
- \(\frac{l}{2}\)
The Correct Option is A
Solution and Explanation
By Hooke's law, we have the relation for the elongation of a wire due to force:
\( \gamma = \frac{F}{A} = \frac{\Delta L}{L} \)
where \( A \) is the cross-sectional area and \( L \) is the length of the wire. The elongation is directly proportional to \( F \) and \( L \), and inversely proportional to the area. The change in length is given by:
\( \Delta L = \frac{F L}{A} = \frac{F L}{\pi r^2} \)
For the second wire, we can write:
\( \frac{\Delta L_2}{\Delta L_1} = \frac{F_2 L_2}{F_1 L_1} \times \frac{A_1}{A_2} \)
Given that \( L_2 = 2L \), \( r_2 = 2r \), and \( F_2 = 2F \), we have:
\( \frac{\Delta L_2}{\Delta L_1} = \frac{2F \times 2L}{F \times L} \times \frac{\pi r^2}{\pi (2r)^2} \)
\( \frac{\Delta L_2}{\Delta L_1} = \frac{4F L}{F L} \times \frac{1}{4} = 1 \)
Thus, the elongation of the second wire is equal to that of the first wire.
Therefore, the ratio \(\frac{\Delta L_2}{\Delta L_1}\) is 1.
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