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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 $l = 25 cm$ . If ‘R’ is replaced by a wire of half length and same diameter that of R of same material. Then the balancing distance $l (in cm)$ will now be

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a

13.2

b

11.1

c

14.29

d

9.5

answer is D.

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Detailed Solution

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$I n i t i a l l y, \frac{S}{R} = \frac{75}{25} =3 N o w, R= \frac{ρ l}{A} = \frac{4 ρ l}{{πd}^{2}} ∴ R^{'} = \frac{4 ρ (l / 2)}{{πd}^{2}} = \frac{R}{2} F i n a l l y, \frac{S}{R'} = (\frac{100 - l}{l}) \Rightarrow \frac{S}{(R/2)} = (\frac{100 - l}{l}) \Rightarrow \frac{2S}{R} = (\frac{100 - l}{l}) \Rightarrow 2 \times 3 = \frac{100 - l}{l} \Rightarrow l = \frac{100}{7} = 14.29 cm$

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