Given initially a resistance P=4Ω is connected in the left gap and let Q is connected in the right gap of meter bridge and the null point is obtained at 60cm,
From the balanced condition of the Wheatstone bridge
QP=SR...(1)
Let ρ is the resistance per unit length of meter bridge wire than
For wire AN, P=ρl=40ρ(∵l=40cm)
And for NB wire
Q=ρ(100−l)Q=ρ(100−60)=40ρ
On substituting the values from equation (1)
604=40Q
⇒Q=616=38Ω
Now an unknown resistance R is connected in series with the 4Ω resistance and the null point is obtained at l=80cm
Now, P=ρl=80l
and Q=ρ(100−l)Q=ρ(100−80)=20ρ
Then from equation (1)
804+R=20Q⇒804+R=2038(∵Q=38)⇒804+R=608⇒R=320Ω
