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Potentiometer wire PQ of length 1m is connected to a standard cell E1. Another cell of E2 of emf 1.02v is connected with a resistance 'r' and with a switch 'S' as shown in the circuit diagram. With switch S is open, the null position is obtained at a distance of 49 cm from Q. Calculate the potential gradient in the potentiometer wire is:-
Option: 1 0.02 V/cm
Option: 2 0.04 V/cm
Option: 3 0.03 V/cm  
Option: 4 0.01 V/cm

Answers (1)


\\ \text{Length of potentiometer wire l}=1 \mathrm{m} \\ \mathrm{Emf} \ \text{of standard cell}=E_{1} \\ \text{Emf of cell} \ E_{2}=1.02 v / s \\ \text{Given Null point from Q is 49 cm}\\ \text {Null point from} \ P =100-49=51 \mathrm{cm}=0.51 \mathrm{m}\\ \text{Potential gradient of the wire} \ V=\frac{1.02}{0.51} V=2 V / m=0.02 V/cm

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