A thin uniform rod of mass and length l is in equilibrium as shown in figure. P is a pivot. When a current I flows along the rod, it rotates from its place about point P. Assume that there is no effect of current on the spring and the spring remains horizontal. Find the torque experienced by the rod about P and spring force, when the rod makes an angle $θ$ with the horizontal.

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τ = \frac{1}{2} B I l^{2} + K l^{2} \sin θ \cos θ - \frac{mg l}{2} \cos θ

τ = \frac{1}{2} B I l^{2} + K l^{2} \sin θ \cos θ + \frac{mg l}{2} \cos θ

τ = - \frac{1}{2} B I l^{2} - K l^{2} \sin θ \cos θ - \frac{mg l}{2} \cos θ

τ = \frac{1}{2} B I l^{2} - K l^{2} \sin θ \cos θ + \frac{mg l}{2} \cos θ

answer is A.

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Detailed Solution

From fig $x = l \cos θ$

So spring force $F = k l \cos θ$

Torque about P

$τ = \int_{0}^{L} [I {Bd}_{y} . y] - (k l \cos θ) l \sin θ + \frac{mg l}{2} \cos θ$

$τ = \frac{1}{2} B I L^{2} - K L^{2} \sin θ . \cos θ + \frac{mg l}{2} \cos θ$

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