1 ohm and 2 ohm are connected in series across the left gap of a metre-bridge. The balancing length is 60 cm. If the resistors were connected in parallel instead of in series, the balancing length would be (in cm)

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Solution

The correct option is A 25
Case 1
series
Resistance in left gap $= 1 + 2 = 3 Ω$
Right gap $= R_{2}$

$\frac{3}{R_{2}} = \frac{l}{100 - l}$

$l = 60 c m$

$\Rightarrow \frac{3}{R_{2}} = \frac{60}{40} \Rightarrow R_{2} = 2 Ω$

case 2
Parallel

$R_{1} = \frac{1 \times 2}{1 + 2} = 2 / 3$

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

$\frac{2 / 3}{2} = \frac{l}{100 - l}$

$3 l = 100 - l \Rightarrow l = 25 c m$

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