In a galvanometer, the number of turns N is doubled (so resistance also doubles). The voltage sensitivity then:
ADoubles
BRemains unchanged
CHalves
DBecomes four times
Answer & Solution
Correct answer: B. Remains unchanged
1. Voltage sensitivity: $\dfrac{\phi}{V} = \dfrac{NAB}{k}\dfrac{1}{R}$.
2. Doubling N makes the numerator $N \to 2N$.
3. But the wire length, and hence resistance, also doubles: $R \to 2R$.
4. The factors cancel: $\dfrac{2N}{2R} = \dfrac{N}{R}$, so voltage sensitivity is unchanged.
5. (Current sensitivity, $\propto N$ alone, would double — a common trap.)
_Source: syllabus Physics Ch 4 "Moving Charges and Magnetism", p.161_
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