In a meter bridge, the balance point is obtained at 40cm. If a resistance equal to that in the left gap is shunted across itself, the new balance point is:

40 c m

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15 c m

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25 c m

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32 c m

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Solution

The correct option is B $25 c m$
Case 1 :

$\frac{R_{1}}{R_{2}} = \frac{l}{100 - l} = \frac{40}{60} = 2 / 3$

Shunting $R_{1}$ with $R_{1}$

$\Rightarrow R_{e f f} = \frac{R_{1}}{2}$

$\Rightarrow \frac{R_{1}}{2 R_{2}} = \frac{l}{100 - l}$

$\Rightarrow \frac{1}{2} (\frac{R_{1}}{R_{2}}) = \frac{l}{100 - l}$

$\Rightarrow \frac{1}{2} \frac{2}{3} = \frac{l}{100 - l}$

$3 l = 100 - l$

$l = 25 c m$

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In a meter bridge, the balance point is obtained at 40cm. If a resistance equal to that in the left gap is shunted across itself, the new balance point is:

The correct option is B 25 cmCase 1 :R1R2=l100−l=4060=2/3Shunting R1 with R1⇒Reff=R12⇒R12R2=l100−l⇒12(R1R2)=l100−l⇒1223=l100−l3l=100−ll=25cm