JEE Advanced2021PhysicsExperimental PhysicsActual
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 .
Correct answer
0
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