A cell of emf 10 V and internal resistance 3 Ω is connected to a uniform wire of length 500 cm and reistance 3 Ω. The drop of potential with length in the wire is:

10 mV/cm

The correct answer is Option (3) → 10 mV/cm

Given:

EMF of cell $E = 10 \, V$

Internal resistance $r = 3 \, \Omega$

Resistance of wire $R = 3 \, \Omega$

Length of wire $L = 500 \, cm$

Total resistance $= R + r = 3 + 3 = 6 \, \Omega$

Current in the circuit: $I = \frac{E}{R+r} = \frac{10}{6} = \frac{5}{3} \, A$

Potential drop across the wire: $V_{wire} = I \times R = \frac{5}{3} \times 3 = 5 \, V$

Potential drop per unit length: $\frac{V_{wire}}{L} = \frac{5}{500} = 0.01 \, V/cm$

Final Answer: Potential drop in the wire $= 0.01 \, V/cm$