Question

What is the percentage increase in length of a wire of diameter 12.5 mm stretched by a weight of 1000 kg ωt ? [Given Y = 12.5 × 10¹¹ dyne/sq.cm]

Answer 1

Area =\(\frac{πd^2}{4}\) = \(\frac {π(1.25)^2}{4} cm^2\)

F = 1000 kg ωt

= 1000 × 10³ × 980

F = 9.8 × 10⁸ dyne

We know that,

= \(\frac{9.8 \times 4 \times 10^{-3}}{π \times (12.5) (1.25)^2}\)

= 0.638 × 10^-3

Percentage increase = 0.064%.