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