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A rod of length $l$ , negligible resistance and mass $m$ slides on two horizontal frictionless rails of negligible resistance by hanging a block of mass $m_{1}$ with the help of insulating massless string passing through fixed massless pulley (as shown) . If a constant magnetic field $B$ acts vertically upwards perpendicular to the figure, the terminal velocity of hanging mass is

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$\frac{m_{1} g R}{B^{2} l}$ downward

$\frac{m_{1} g R}{B^{2} l^{2}}$ upwards

$\frac{m_{1} g R}{B^{2} l^{2}}$ downward

$\frac{m_{1} g R}{2 B^{2} l^{2}}$ downward

answer is B.

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Detailed Solution

The tension in the string will be

$T = m_{1} g$ since the block is descending with a steady acceleration.

Therefore

$T = F$ where F is magnetic force on the rod.

We know $ε = B l V$

$F = B i l$

$i = \frac{ε}{R} = \frac{B l V}{R}$

$\Rightarrow F = \frac{B^{2} l^{2} V}{R} = m_{1} g$

$\Rightarrow V = \frac{m_{1} g R}{B^{2} l^{2}}$

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