Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

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

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

5 g of Na₂SO₄ was dissolved in x g of H₂O. The change in freezing point was found to be 3.82⁰C. If Na₂SO₄ is 81.5% ionised, the value of x (K

A capacitor is made of two square plates each of side 'a' making a very small angle $\alpha$

A solution of m-chloroaniline, m-chlorophenol and m-chlorobenzoic acid in ethyl acetate was extracted initially with a saturated solution of NaHCO₃ to give fraction A. The leftover organic phase was extracted with d

Trending Articles/News

anam-khan

JEE Main 2026: JCECEB invites fresh applications for Jharkhand state merit list correction

Latest Question