Question

any genius here.......................?​

Answer 1

Explanation:

Given,

Length of wire , l = 4 m.

Resistance of wire , R = 10 ohm.

Emf of the battery = 2V

internal resistance = 2 ohm.

We need to find potential gradient,

current flowing in the potentiometer wire , I =E/R+r

={/2/10+2

= 2/12

=1/6

Potential gradient = IR/l

= 1\6×10/4

=0.416666667

=0.416 V/m

hope it helps...

pls mark it as Brainliest..

Answer 2

GIVEN

L = 4 m

R = 10 ohm

E = 2 V

Internal Resistance = 2 ohm

External Resistance = 38 ohm

SOLUTION

→ Req = (10 + 2 + 38) = 50 ohm

→ I = E/Req

= 2V/50 ohm

→ V (across 4m long wire) = 2/50 × 10 = 2/5

→ Potential Gradient = 2/(5×4) = 1/10 = 0.1

So, the Potential Gradient along the wire is 0.1 V/m