question_answer

A uniform rod of length L is placed over a horizontal frictionless surface. It is pulled over the surface by applying a force F on one of its end. Area of rod is A

Stress in the rod at a distance L/3 from the end, where the force is applied is

A) \[\frac{F}{A}\]

B) \[\frac{1F}{3A}\]

C) \[\frac{2F}{3A}\]          

D) \[\frac{3F}{4A}\]

Correct Answer: C

Solution :

Acceleration of rod \[=\frac{F}{M}\]

Let T = tension in the rope at a distance x from the end where force F is applied.

For \[L-x\]length, T is only force, so

T = Mass \[\times \] Acceleration

\[=\frac{M}{L}\times (L-x)\times \frac{F}{M}\]

= Mass per unit length \[\times \] Length of remaining part \[\times \] Acceleration

\[=\left( \frac{L-x}{L} \right)F=\left( \frac{L-\frac{L}{3}}{L} \right)F=\frac{2}{3}F\]

\[\therefore \]  Stress, \[\sigma =\frac{2}{3}\frac{F}{A}\]