A) 1 / 8
B) 1 / 6
C) 1 / 10
D) 1 / 4
Correct Answer: A
Solution :
Let l be the original length of wire and x be its length stretched uniformly such that final length is 1.5 l Then \[4R=\rho \frac{(l-x)}{A}+\rho \frac{(0.5l+x)}{A'}\]where \[A'=\frac{x}{(0.5l+x)}A\] \ \[4\rho \frac{l}{A}=\rho \frac{l-x}{A}+\rho \frac{{{(0.5l+x)}^{2}}}{xA}\] or \[4l=l-x+\frac{1}{4}\frac{{{l}^{2}}}{x}+\frac{{{x}^{2}}}{x}+\frac{lx}{x}\] or \[\frac{x}{l}=\frac{1}{8}\]You need to login to perform this action.
You will be redirected in
3 sec