Two blocks each of the mass `m` are attached to the ends, a massless rod which pivots as shown in figure. Initial the rod is held in the horizontal position and then release, Calculate the net torque on this system above pivot.

Options

A. `(ml_(2)g-ml_(1)g)hatk`
B. `(ml1g-ml_(2)g)hatk`
C. `(ml_(1)g+ml_(2)g)hatk`
D. `-(ml1g+ml_(2)g)hatk`

Answer

B

Explanation

Torque due to weight of left `m`
`vectau_(1)=mgl_(1)hatk`(this is anticlockwise)
Torque due to weight of right `m`
`vectau_(2)=-mgl_(2)hatk`(this is clockwise)
So net torque `vectau=vectau+vectau_(2)=(mgl_(1)-mgl_(2))hatk`