A potentiometer wire of length 1 m is connected in series with 495Ω resistance and 2 V battery. If 0.2 mV / cm is the potential gradient, then the resistance of the potentiometer wire is

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Let the resistance of the potentiometer wire be $R$ .

The length of the potentiometer wire is $L = 1 m = 100 cm$ .

The potential difference across the potentiometer wire is given by $V_{p} = k \times L$ , where $k$ is the potential gradient.

$V_{p} = (0.2 mV/cm) \times 100 cm = 20 mV = 0.02 V$

The current in the circuit is $I = \frac{V}{R + R _{s}}$ , where $V = 2 V$ and $R_{s} = 495Ω$ .

The potential difference across the wire can also be written as $V_{p} = I \times R$ .

$V_{p} = \frac{V \times R}{R + R _{s}}$

Substituting the known values:

$0.02 = \frac{2 \times R}{R + 495}$

$0.02 (R + 495) = 2 R$

$0.02 R + 9.9 = 2 R$

$1.98 R = 9.9$

$R = \frac{9.9}{1.98} = 5Ω$

Answer: $5Ω$

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