Question

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 1

Answer: (A)

n² times

Answer 2

Y = $\frac{\frac{F}{a}}{\frac{\Delta\mathcal{l}}{\mathcal{l}}}$ = $\frac{F\mathcal{l}}{a\Delta\mathcal{l}}$= $\frac{F\mathcal{l \times}4}{\pi D^{2} \times \Delta\mathcal{l}}$ or Dl µ $\frac{1}{D^{2}}$

or $\frac{\Delta\mathcal{l}_{2}}{\Delta\mathcal{l}_{1}}$ = $\frac{D_{1}^{2}}{D_{2}^{2}}$= $\frac{n^{2}}{1}$