A `5 kg` rod of square cross section `5 cm` on a side and `1m` long is pulled along a smooth horizontal surface by a force applied at one end. The rod has a constant acceleration of `2 ms^(-12)`. Determine the elongation in the rod. (Young's modulus of the material of the rod is `5 xx 10^(3) N//m^(9)`).

Options

A. Zero, as for elongation to be there, equal and opposite forces must act on the rod
B. Non-zero but cannot be determine from the give, situation
C. `0.4 mum`
D. `16 mum`

Answer

C

Explanation

Let the force applied be `p` then `P=ma`

`impliesp=5xx2=10N`
For the element shown in figure `T=m/l(l-x)xxa`
`implies impliesT=5/I(1-x)xx2=10(1-x)`
Elongation in `dx` is `(/_\(dx))/(dX)=(T//A)/A`
`/_\(dx)=(10(1-x))/((5xx10^(-2))^(2))xx1/(5xx10^(9))dx`
Total elongation `/_\l=int_(0)^(1)(10(1-x))/(5xx10^(-4))xx1/(5xx10^(9))dx=0.4xx10^(-6)m`