What is lts teminal velocity?

1.5 m / s e c

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0.8 m / s e c

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0.7 m / s e c

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1 m / s e c

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Solution

The correct option is D $1 m / s e c$
The rod becomes a seat of e.m.f $e = B l v$ when it accquire the terminal velocity the resultant force on it is zero
$m g - B i l = 0$ or $m g = B i l$
The equivalent corcuit is shown

$i = \frac{e}{\frac{R_{1} R_{2}}{R_{1} + R_{2}}}$

$= \frac{B l v (R_{1} + R_{2})}{R_{1} R_{2}}$

$m g = \frac{B B l v (R_{1} + R_{2})}{R_{1} R_{2}} \Rightarrow v = \frac{m g R}{B^{2} I^{2}}$

$e = B l v = \frac{m g R}{B l}$

$P_{1} = \frac{e^{2}}{R_{1}} = \frac{m^{2} g^{2} R^{2}}{B^{2} l^{2} R_{1}} = \frac{m^{2} g^{2}}{B^{2} l^{2} R_{1}} \times \frac{R_{1}^{2} R_{2}^{2}}{(R_{1} + R_{2})^{2}}$

$\Rightarrow \frac{(R_{1} + R_{2})^{2}}{R_{1} R_{2}^{2}} = \frac{m^{2} g^{2}}{B^{2} l^{2} P_{1}}$

$\Rightarrow \frac{1}{R_{1}} \frac{(R_{1} + R_{2})^{2}}{R_{2}} = \frac{m^{2} g^{2}}{B^{2} l^{2} P_{1}}$

Similarly, $\frac{1}{R_{2}} {(\frac{R_{1} + R_{2}}{R_{1}})}^{2} = \frac{m^{2} g^{2}}{B^{2} l^{2} P_{2}}$

Dividing $\frac{R_{2}}{R_{1}} = \frac{P_{1}}{P_{2}} \Rightarrow \frac{R_{2} + R_{1}}{R_{2}} = \frac{P_{1} + P_{2}}{P_{1}}$

$\frac{1}{R_{1}} {(\frac{P_{1} + P_{2}}{P_{1}})}^{2} = \frac{m^{2} g^{2}}{B^{2} l^{2} P_{1}} \Rightarrow R_{1} = \frac{B^{2} l^{2}}{m^{2} g^{2}} \frac{(P_{1} + P_{2})^{2}}{P_{1}}$

$R_{2} = \frac{b^{2} l^{2}}{m^{2} g^{2}} \frac{(P_{1} + P_{2})^{2}}{P_{2}} = \frac{(0.76 + 1.2)^{2}}{1.2} \times \frac{{0.6}^{2 \times 1^{2}}}{{0.2}^{2} \times {9.8}^{2}}$

$= 0.3 o h m$

$R_{1} = \frac{(0.76 + 1.2)^{2}}{0.76} \times \frac{{0.6}^{2} \times 1^{2}}{{0.2}^{2} \times {9.8}^{2}} = 0.47 o h m$

$v = \frac{m g}{B^{2} l^{2}} . \frac{R_{1} R_{2}}{(R_{1} + R_{2})} = 1 m / s e c$

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