MHTCET 11 May 2024 Shift 2 Paper

Concept:For a uniform wire, the potential drops uniformly along its length. Use potential gradient or Ohm’s law.Explanation:Given total resistance of the wire is $R = 100\ \Omega$ and battery voltage is $V = 6\ \text{V}$.Current through the wire: $I = \frac{V}{R} = \frac{6}{100} = 0.06\ \text{A}$.Resistance per metre of the wire: $\frac{100}{3}\ \Omega/\text{m}$.Resistance of a 50 cm segment: $R' = \frac{100}{3} \times 0.5 = \frac{50}{3}\ \Omega$.Potential difference across the 50 cm segment: $V' = I R' = 0.06 \times \frac{50}{3} = 1\ \text{V}$.Alternatively, potential gradient $= \frac{6\ \text{V}}{3\ \text{m}} = 2\ \text{V/m}$.For length 50 cm $= 0.5\ \text{m}$: $V' = 2 \times 0.5 = 1\ \text{V}$.Answer:The potential difference between two points 50 cm apart on the wire is $1\ \text{V}$.Correct option: A. 1 V.