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VITEEE 2009 Paper

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Question : 29 of 120

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A cell in secondary circuit gives null deflection for 2.5m length of potentiometer having 10m length of wire. If the length of the potentiometer wire is increased by 1m without changing the cell in the primary, the position of the null point now is

3.5m
3m
2.75m
2.0m

Solution:

Here length of potentiometer wire,

l = 10\,\text{m}

Resistance of potentiometer wire

R = \frac{\rho \times l}{A}

or

R = \left( \frac{\rho \times 10}{A} \right)

The value of 2.5m length wire

R' = \frac{\rho \times 10}{A \times 10} \times 2.5

⇒

R' = \left( \frac{2.5 \rho}{A \times 10} \right)

Potential,

V' = I \times R' = I \left( \frac{2.5 \rho}{A \times 10} \right)

Again the length of potentiometer wire is increase by 1m.

Resistance of null position

R^{n} = \left( \frac{\rho \times l}{11 \times A} \right)

\therefore V^{n} = I R^{n}

and

V = V'

⇒

\frac{I \times 2.5 \rho}{A \times 10} = \frac{\rho \times l}{11 \times A} \times I

or

\frac{2.5 \times 11}{10} = l = 2.75\,\text{m}

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