The voltage sensitivity of a voltmeter can be increased by:(A) decrea…

The correct answer is Option (2) → (A), (C), (E) Only

The deflection of the coil is \( \theta = \frac{NIAB}{k} \) and the current through the voltmeter is \( I = V/R \). The voltage sensitivity is the deflection per unit voltage:

    \( \displaystyle \frac{\theta}{V} = \frac{NAB}{kR} \)

(A) Decreasing the resistance of the voltmeter — Correct. R is in the denominator, so a smaller R gives a larger deflection per volt.

(B) Increasing the resistance of the voltmeter — Incorrect. This lowers \( \theta/V \).

(C) Decreasing the torsional constant of spring — Correct. A weaker spring (smaller k) twists more for the same torque.

(D) Increasing the number of turns of the coil — Incorrect. N doubles the deflection per unit current, but the coil is longer wire, so its resistance R also rises in proportion; N and R cancel and the voltage sensitivity stays unchanged.

(E) Taking a strong permanent magnet — Correct. B is in the numerator, so a stronger field gives a larger torque.

NCERT (2026-27):
The Moving Coil Galvanometer – “… We define the voltage sensitivity as the deflection per unit voltage. From Eq. (4.26),

    \( \displaystyle \frac{\phi}{V} = \left(\frac{NAB}{k}\right)\frac{I}{V} = \left(\frac{NAB}{k}\right)\frac{1}{R} \qquad (4.28) \)

An interesting point to note is that increasing the current sensitivity may not necessarily increase the voltage sensitivity. … In Eq. (4.28), N → 2N, and R → 2R, thus the voltage sensitivity,

    \( \displaystyle \frac{\phi}{V} \to \frac{\phi}{V} \)

remains unchanged.”