The current sensitivity of a galvanometer increases by 10%. If its resistance also increases by 20%, the voltage sensitivity will
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → decrease by 8.3 %
The voltage sensitivity of a galvanometer is its current sensitivity divided by its resistance.
\( \displaystyle S_V = \frac{S_I}{R} \)
\( \displaystyle \frac{S_V'}{S_V} = \frac{1.10\,S_I}{1.20\,R} \times \frac{R}{S_I} = \frac{1.10}{1.20} = 0.9167 \)
The new value is \( 0.9167 \) times the old one, a fall of \( 1 - 0.9167 = 0.0833 \), that is, 8.3 %.