Question

potentiometer wire of length 1 m has a resistance of 10 Ω. It is connected to a 6 V battery in series with a resistance of 5 Ω. Determine the emf of the primary cell which gives a balance point at 40 cm.

Answer 1

length of wire L =1m = 100cm
length of balanced point l' = 40cm
resistance of wire (r) = 10 ohm
potential of battery = 6V
resistance of circuit -> R = 5 ohm
therefore
current in potentiometer
I= V/(r+R)
=>6/(10+5)
=0.4 A
potential in potentiometer wire
v= iR
V = 0.4x10
V = 4 V
the EMF in the potentiometer
E = V( l' )/ L
=> 4x40/100
= 1.6 v

Answer 2

Hey !

Current flowing in potentiometer wire,

                    I = $\frac{V}{R + R'}$

Where, R is the resistance of potentiometer wire and R' = 5 Ω

                     I = $\frac{5}{10 + 5} A = 0.4 A$

Potential drop across the potentiometer wire

                        V = IR

       = 0.4 × 10 V = 4.0 V

Potential gradient k = $\frac{V}{l}$ = 4.0 $\frac{V}{m}$

∴ Unknown emf of the cell (E) = kl'

                    = 4.0 × 0.4 V

FINAL RESULT = 1.6 V

GOOD LUCK !!