BITSAT 2022 Paper

When S2 open.
Assume resistance of Xγ=R
Resistance of wire per unit length, x=

R

L

=RΩm−1
∵I=

E0

R

Now, the potential drop across 50 cm length is 6V, so

E0

R

×R×

50

100

=6
⇒E0=12V
When S2 closed, potential drop across 0.416cm length,
V1=

E0

R

×R×0.416=12×12×0.416≈5V
Hence, E−Ir=5V
⇒6−Ir=5
∵I=

5

10

∴6−5=

5

10

r
⇒r=2Ω