A potentiometer wire of length L and resistance 10Ω is connected in series with a battery of emf 2.5V and a resistance in its primary circuit. The null point corresponding to a cell of a emf 1V is obtained at a distance L/2. If the resistance in the primary circuit is doubled then the position of new null point will be

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

0.4L

0.5L

0.6L

0.8L

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$P o t e n t i a l g r a d i e n t = (\frac{2.5}{10 + R}) \times \frac{10}{L} = \frac{25}{(10 + R)}$

$\frac{25}{(10 + R) L} \times \frac{L}{2} = 1 \Rightarrow 10 + R = 12.5$

$R = 2.5 Ω$

$\begin{array}{l} N e w p o t e n t i a l g r a d i e n t = (\frac{2.5}{10 + 2 R}) \times \frac{10}{L} \\ = \frac{25}{(10 + 5) L} = \frac{5}{3 L} \end{array}$

$\frac{5}{3 L} x = 1 \Rightarrow x = 0.6 L$

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE