What is the percentage increase in length of a wire of diameter 3.0 mm stratched by a force 150 kg wt ? Young's modulus of elasticity of wire is `12.5 xx 10^(11) dyn e cm ^(-2)`

Answer

0.0017

Explanation

Here, Diameter of wire, D = 3.0 mm =0.3 cm Stretching force, `F = 150 kg wt = 150 xx1000 xx980 dyn e` ,
`Y = 12.5 xx 10^(11) dyn e// cm^2`
As, `Y = (F)/(A) (l)/(Delta l)` or `(Delta l)/(l) = (F)/(AY)= (F)/((pi D^2 // 4)Y)` % increase in length is ` = (Delta l)/(l)xx100 = (4Fxx100)/(pi D^2Y)`
`=(4xx(150xx1000xx980)xx100)/((22//7)xx(0.3)^2xx(12.5xx10^(11)) =0.17 %`