Question

a wire of length L and radius R is fixed at one end and a force f is applied to the other end Producers and extension I the extension produced is another wire of the same material of 2 l length and radius 2r by a force 4F is​

Answer 1

L

Deep Uncovered Truth For A Simple Answer:-

$y = \frac{fl}{al}$

Y - Young's Modulus

F - Applied Force

A - Cross Section Area

l - Elongation

For 1st Wire

$y = \frac{fl}{(pie)(2 {}^{2} ) {(r}^{2})z } \: \: \: m$

For 2nd Wire

z - elongation in 2nd wire

$y = \frac{2f2l}{(pie)( {2}^{2})( {r}^{2})z }$

Deviding M by N.

$= 1 = \frac{z}{l} \\ therefore \: z = l$