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Q.

In a meter bridge experiment S is a standard resistance. R is a resistance wire. It is found that balancing length is $l = 25 c m$ . If R is replaced by a wire of half length and half diameter that of R of same material, then the balancing distance $' l'$ (in cm) will now be _____

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a

10

b

20

c

40

d

30

answer is D.

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Detailed Solution

In balancing $\frac{R}{S} = \frac{25}{75}$
New resistance $R' = \frac{ρ l}{A}$
$= \frac{ρ \times \frac{l}{2}}{\frac{A}{4}} = \frac{ρ l}{2} \times 4 A R^{'} = 2 R$
$\frac{2 R}{S} = \frac{l^{'}}{100 - l^{'}}$
$2 \times \frac{1}{3} = \frac{l^{'}}{100 - l^{'}} = 3 l^{'} = 200 - 2 l^{'}$
$5 l^{'} = 200$
$l^{'} = 40$

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