The force between two magnetic poles is F in air. If the pole strength of first pole is doubled, that of second pole is tripled and the distance between the poles is halved the force between the poles becomes and increases by

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24F and 2300%

24F and 2400%

2F and 100%

$\frac{3}{2} F$ and 150%

answer is A.

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Detailed Solution

$F \propto \frac{m_{1} m_{2}}{d^{2}}$

$\Rightarrow \frac{F^{'}}{F} = {(\frac{d}{d^{'}})}^{2} \times (\frac{{m_{1}}^{'}}{m_{1}} \times \frac{{m_{2}}^{'}}{m_{2}})$

$\Rightarrow \frac{F^{'}}{F} = {(\frac{2}{1})}^{2} \times (\frac{2 m_{1}}{m_{1}} \times \frac{3 m_{2}}{m_{2}}) = 24$

$\Rightarrow F^{'} = 24 F$

$a g a i n, \frac{F^{'} - F}{F} = \frac{F^{'}}{F} - 1 = 23$

$\Rightarrow \frac{F^{'} - F}{F} \times 100 = 2300$

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