JEE Main2017PhysicsCurrent ElectricityActual
A potentiometer P Q is set up to compare two resistances, as shown in the figure. The ammeter A in the circuit reads 1 . 0 A when the two-way key K 3 is open. The balance point is at a length l 1 cm from P when the two-way key K 3 is plugged in between 2 and 1 , while the balance point is at a length l 2 cm from P when the key K 3 is plugged in between 3 and 1 . The ratio of two resistances R 1 R 2 , is found to be
Options
- Al 1 l 1 - l 2
- Bl 2 l 2 - l 1
- Cl 1 l 1 + l 2
- Dl 1 l 2 - l 1
Correct answer
D. l 1 l 2 - l 1
Step-by-step solution
When the key is between 2 and 1 Then, emf V 1 =   I R 1 = x l 1 . When the key is between 3 and 1 Emf, V 2 = I R 1 + R 2 = x l 2 . The ratio of both emf is R 1 R 1 + R 2 = l 1 l 2 On simplifying the above relation, we get the ratio of resitances, R 1 R 2 = l 1 l 2 - l 1 .