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  • A cell in secondary circuit gives null deflection for length of potentiometer havin

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

Option: 1

3.5\mathrm{m}


Option: 2

3\mathrm{m}


Option: 3

2.75\mathrm{m}


Option: 4

2.0\mathrm{m}


Answers (1)

best_answer

In first case, Potential gradient,\mathrm{k}=\frac{\varepsilon_0}{l}=\frac{\varepsilon_0}{10}

Where \varepsilon_0 is the emf of the battery in the potentiometer circuit.

As per questions, \varepsilon=\mathrm{k} \times 2.5=\frac{\varepsilon_0}{10} \times 2.5           (i)

In second case, Length of potentiometer wire =10+1=11 \mathrm{~m}

Potential gradient, \mathrm{k}^{\prime}=\frac{\varepsilon_0}{11}

If l^{\prime} is the new balancing length, then

\varepsilon=\mathrm{k}^{\prime} l^{\prime}=\frac{\varepsilon_0}{11} \times l^{\prime}                                                 (ii)

Equating (i) and (ii), we get

\frac{\varepsilon_0}{10} \times 2.5=\frac{\varepsilon_0}{11} \times l^{\prime} \quad \text { or } \quad l^{\prime}=\frac{2.5 \times 11}{10}=2.75 \mathrm{~m}

Hence option 3 is correct

Posted by

Irshad Anwar

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