MHTCET 18 Apr 2026 Shift 1 Paper

Concept:A potentiometer compares unknown e.m.f. with the potential drop across its wire.The potential gradient is the voltage per unit length of the wire.Explanation:First, find the total resistance of the primary circuit.$$R_{\text{total}} = 5 + 992 + 3 = 1000\ \Omega$$Current flowing in the circuit:$$I = \frac{4}{1000} = 4 \times 10^{-3}\ \text{A}$$Potential difference across the potentiometer wire of resistance $5\ \Omega$:$$V = I \times 5 = \left(4 \times 10^{-3}\right)(5) = 20 \times 10^{-3}\ \text{V} = 20\ \text{mV}$$Since the wire length is $4\ \text{m}$, the potential gradient is:$$k = \frac{20\ \text{mV}}{4\ \text{m}} = 5\ \text{mV m}^{-1}$$For a balancing length of $0.75\ \text{m}$, the balanced e.m.f. is:$$E' = k \times l = 5 \times 0.75 = 3.75\ \text{mV}$$Answer:The correct option is C: $3.75\ \text{mV}$.