A uniform cylindrical rod of length L and cross-sectional area by forces as shown in figure.

The elongation produced in the rod is

Options

A. `(3F L)/(8 AY)`
B. `(3FL)/(5AY)`
C. `(8FL)/(3AY)`
D. `(5FL)/(3AY)`

Answer

C

Explanation

(c) Elongation production if the `(2)/(3)` part of length L of rod is stretched by a force of 3F, is given by
`DeltaL_(1)=(F_(1)L_(1))/(AY)=(3Fxx(2)/(3)L)/(AY)=2(FL)/(AY)`
Total elongation produced in the rod is
`DeltaL=DeltaL_(1)+DeltaL_(2)=2(FL)/(AY)+(2)/(3)(FL)/(AY)`
`therefore DeltaL=(8)/(3)(FL)/(AY)`