Question

The potentiometer wire of length 500cm has a resistance of 50Ω.It

is connected in series with a resistance of 100Ω and an accumulator of

emf 10V having negligible resistance. A source of 1.5V is balanced

against a length L of potentiometer wire. Find the value of L.​

Answer 1

Answer:

Since, 

1 m potentiometer has 10Ω of resistance.

⇒40cm length would have = 110×0.4  [R=PAL⇒RαL]

R1=4Ω resistance

At balance point, current through both driver circuit and connected cell would be same.  (I = constant)

Now from onm's Law : V=1R we get,

⇒R1E=R2V [E : Emf of primary cell]

⇒4E=56

Gives, 

⇒E=4.8V

it helps you some i think