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Byju's Answer

Standard XII

Physics

Meter Bridge

In a metre br...

Question

In a metre bridge experiment with resistance $R_{1}$ in the left gap and a resistance S in the right gap, balance point is obtained at 40 cm from zero end. With another resistance $R_{2}$ in left gap with same S in right gap, the balance point is obtained at 50 cm. If $R_{1}$ and $R_{2}$ are put in parallel in the left gap without changing S, the balancing length will be

28.5 cm

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32.5 cm

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82.5 cm

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42.5 cm

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Solution

The correct option is A 28.5 cm
In balanced condition of meter bridge,
$\frac{P}{Q} = \frac{L}{100 - L} \frac{R_{1}}{S} = \frac{40}{60} = \frac{2}{3} \to 1 \frac{R_{2}}{S} = \frac{50}{50} = 1 \to 2 \frac{(\frac{R_{1} R_{2}}{R_{1} + R_{2}})}{S} = \frac{l}{100 - l} [\begin{matrix} e q u (1) R_{1} = \frac{2}{3} S e q u (2) R_{2} = S \end{matrix}] [\frac{\frac{2}{3} S \times S}{\frac{2}{3} S + S}] \frac{1}{S} = \frac{l}{100 - l} \frac{2}{3} \times \frac{3}{5} = \frac{l}{100 - l} 200 - 2 l = 5 l 200 = 7 l l = \frac{200}{7} = 28.5 c m$

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The correct option is A 28.5 cmIn balanced condition of meter bridge, PQ=L100−LR1S=4060=23→1R2S=5050=1→2(R1R2R1+R2)S=l100−l[equ(1)R1=23Sequ(2)R2=S][23S×S23S+S]1S=l100−l23×35=l100−l200−2l=5l200=7ll=2007=28.5 cm