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JEE Advanced Physics Experimental Physics 2021 JEE Advanced 2021 (Paper 2)

JEE Advanced Physics Question (2021) — Solution

Question

In order to measure the internal resistance  r 1  of a cell of emf  ε ,  a meter bridge of wire resistance  R 0 = 50   Ω , a resistance  R 0 2 , another cell of emf  ε 2  (internal resistance r ) and a galvanometer  G  are used in a circuit, as shown in the figure. If the null point is found at  l = 72   cm ,  then the value of  r 1 = _ _ _ _ Ω .  

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

Resistance of potential wire is  R 0 = 50   Ω Resistance of  100   cm  wire  = 50   Ω So, Resistance of  72   cm  wire  = 50 100 × 72 = 36   Ω Current, I = ε 2 14 + 25 = ε r 1 + 75 ⇒ r 1 = 3   Ω

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