A uniformly charged non-conducting rod of mass M and linear charge density $λ (C / m)$ is rotating in horizontal plane in gravity free space with constant angular velocity $ω$ rad/sec. The rod is hinged at one end as shown in figure. A uniform magnetic field $\vec{B}$ exists in the region perpendicular to the plane. Area of cross- section of rod is A.

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Net force applied by hinge on the rod (when $ω > \frac{B λ l}{m}$ ) is zero

Find out net force applied by hinge on the rod (when $ω > \frac{B λ l}{m}$ ) is $\frac{m ω^{2} l}{2} - \frac{B λ ω l^{2}}{2}$

Tension in the rod at the midpoint of rod (When $ω = \frac{B λ l}{M}$ ) is zero

Tension in the rod at midpoint of rod (When $ω = \frac{B λ l}{M}$ ) is $\frac{B λ ω l^{2}}{8 A}$

answer is A, C.

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Detailed Solution

Considering an element of length “dx” at x distance
$T - (T + d T) + d q v B = d m ω^{2} x$ from hinge,
$\Rightarrow - \int_{0}^{T} d T + λ ω B \int_{l}^{0} x d x = μ ω^{2} \int_{l}^{0} x d x ∴ - [T - 0] - \frac{λ B ω l^{2}}{2} = - \frac{μ ω^{2} l^{2}}{2} ∴ T = \frac{m ω^{2} l}{2} - \frac{λ B ω l^{2}}{2}$
For tension at mid-point:
$- \int_{0}^{T} d T + λ B ω \int_{l}^{l / 2} x d x = μ ω^{2} \int_{l}^{l / 2} x d x$
If $ω = \frac{B λ l}{m} \Rightarrow T_{c} = 0$

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