In a potentiometer a cell of emf $1.5$ V gives a balanced point at $32$ cm length of the wire. If the cell is replaced by another cell then the balance point shifts to $65.0$ cm then the emf of second cell is?

3.05

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2.05

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4.05

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6.05

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Solution

The correct option is B $3.05$ V
Given,
Balencing point $l_{1} = 32 c m$ when votage applied $V = 1.5$
Balencing point $l_{2} = 65 c m$
Potential gradiant $K = \frac{V}{x} = \frac{1.5}{32}$

potential diffrence in length $l_{2} = K \times l_{2} = V^{'}$

$V^{'} = \frac{1.5}{32} \times 65 = 3.05 V o l t$
Option A

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In a potentiometer a cell of emf 1.5V gives a balanced point at 32cm length of the wire. If the cell is replaced by another cell then the balance point shifts to 65.0cm then the emf of second cell is?

The correct option is B 3.05VGiven,Balencing point l1=32cm when votage applied V=1.5Balencing point l2=65cm Potential gradiant K=Vx=1.532potential diffrence in length l2=K×l2=V′V′=1.532×65=3.05 VoltOption A

In a potentiometer a cell of emf $1.5$ V gives a balanced point at $32$ cm length of the wire. If the cell is replaced by another cell then the balance point shifts to $65.0$ cm then the emf of second cell is?