A potentiometer wire of length 300 cm is connected in series with a resistance 780 Ω and a standard cell of emf 4 V . A constant current flows through potentiometer wire. The length of the null point for cell of emf 20 mV is found to be 60 cm . The resistance of the potentiometer wire is _____ Ω .

Step-by-step solution

Let resistance of potentiometers wire is R , then the current through A B i A B = 4 R + 780 Potential difference across A B V A B = 4 R R + 780 Potential difference across AC V A C = V A B l L = 4 R R + 780 × 60 300 = 4 R 5 R + 780 This should be equal to 20   mV ⇒ 4 R 5 R + 780 = 20 × 10 - 3 = 2 × 10 - 2 ⇒ 4 R = 10 - 1 R + 780 ⇒ 4 R - R 10 = 78 ⇒ 39 R 10 = 78 ⇒ R = 20   Ω