A system consist of block of mass m, rod of length $l$ and mass 2m connected with the help of string, which passes through an ideal pulley is released from rest as shown in the figure. Tension in the rod at a distance $\frac{l}{3}$ from the upper end of the rod is

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$\frac{2 m g}{3}$

$\frac{4 m g}{9}$

$\frac{2 m g}{9}$

$\frac{8 m g}{9}$

answer is C.

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Detailed Solution

$a = \frac{2 m - m}{3 m} g = \frac{g}{3}$
Now apply Newton’s 2nd law on lower 2/3rd part of rod.

We get $\frac{4 m g}{3} - T = \frac{4 m}{3} a = \frac{4 m}{3} \times \frac{g}{3}$

$T = \frac{4 m g}{3} - \frac{4 m g}{9}$ $T = \frac{8 m g}{9}$

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