GUJCET 2026 Physics Paper

Concept:Current sensitivity is the deflection produced per unit current, while voltage sensitivity is the deflection produced per unit voltage.Explanation:Current sensitivity is given by $I_s = \frac{NAB}{k}$.Here $N$ is the number of turns, $A$ is the coil area, $B$ is the magnetic field, and $k$ is the torsional constant.If $N$ is doubled, then $I_s$ becomes $2I_s$.So the current sensitivity is doubled.Voltage sensitivity is $V_s = \frac{I_s}{R} = \frac{NAB}{kR}$, where $R$ is the coil resistance.The resistance of the coil is directly proportional to the length of the wire.Length is proportional to the number of turns, so $R \propto N$.Thus, doubling $N$ also doubles $R$.Substituting $N \to 2N$ and $R \to 2R$: $V_s = \frac{(2N)AB}{k(2R)} = \frac{NAB}{kR}$.Hence, voltage sensitivity remains unchanged.Answer:Current sensitivity is doubled and voltage sensitivity remains unchanged.Correct option: C.