If we double the number of turns of coil in a moving coil galvanometer, its voltage sensitivity will:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → remain unchanged
The voltage sensitivity of a moving coil galvanometer is the deflection per unit voltage. Its value is
\( \displaystyle \frac{\theta}{V} = \frac{N B A}{k R} \)
where \( N \) is the number of turns, \( B \) the magnetic field, \( A \) the area of the coil, \( k \) the torsion constant and \( R \) the resistance of the coil.
Doubling \( N \) doubles the numerator. But the coil now has twice as much wire, so its resistance \( R \) also doubles. The two doublings cancel, so the voltage sensitivity is unchanged.
Why the other options are incorrect
Option (1) reduce to half — Incorrect. Neither \( N \) nor \( R \) can make the ratio fall; both double, so the ratio stays the same.
Option (3) increase to twice — Incorrect. This is the change in the current sensitivity \( NBA/k \), which does double; the voltage sensitivity also divides by the doubled resistance.
Option (4) increase to 4 times — Incorrect. The number of turns appears only once in the numerator and once in the resistance, so there is no factor of 4.
Note: A similar question was also asked in CUET (UG) 2024 on 19-07-2024 (Slot 1), and in CUET (UG) 2022 on 17-08-2022 (Slot 2).
NCERT (2026-27):
The Moving Coil Galvanometer – “… An interesting point to note is that increasing the current sensitivity may not necessarily increase the voltage sensitivity. … If N → 2N, i.e., we double the number of turns, then
\( \displaystyle \frac{\phi}{I} \rightarrow 2\frac{\phi}{I} \)
Thus, the current sensitivity doubles. However, the resistance of the galvanometer is also likely to double, since it is proportional to the length of the wire. In Eq. (4.28), N → 2N, and R → 2R, thus the voltage sensitivity,
\( \displaystyle \frac{\phi}{V} \rightarrow \frac{\phi}{V} \)
remains unchanged.”