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Question: The ratio of diameters of two wires of same material is n : 1. The length of each wire is 4 m. On ap...

The ratio of diameters of two wires of same material is n : 1. The length of each wire is 4 m. On applying the same load, the increase in length of thin wire will be (n > 1) -

A

n2 times

B

n times

C

2n times

D

(2n + 1) times

Answer

n2 times

Explanation

Solution

Y = FaΔll=FlaΔl\frac{\frac{F}{a}}{\frac{\Delta\mathcal{l}}{\mathcal{l}}} = \frac{F\mathcal{l}}{a\Delta\mathcal{l}}, Y = Fl×4πD2×Δl\frac{F\mathcal{l \times}4}{\pi D^{2} \times \Delta\mathcal{l}}

or Dl µ 1D2\frac{1}{D^{2}} or Δl2Δl1=D12D22=n21\frac{\Delta\mathcal{l}_{2}}{\Delta\mathcal{l}_{1}} = \frac{D_{1}^{2}}{D_{2}^{2}} = \frac{n^{2}}{1}