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(D) Current sensitivity $(I_s)$ of a moving coil galvanometer is given by $I_s = \frac{NBA}{k}$,where $N$ is the number of turns,$B$ is the magnetic f

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(D) Current sensitivity $(I_s)$ of a moving coil galvanometer is given by $I_s = \frac{NBA}{k}$,where $N$ is the number of turns,$B$ is the magnetic field,$A$ is the area,and $k$ is the torsional constant.Since $I_s \propto N$,if $I_s$ increases by $50 \%$,the new number of turns $N' = 1.5N$.Voltage sensitivity $(V_s)$ is given by $V_s = \frac{I_s}{R}$,where $R$ is the resistance of the coil.Resistance $R$ is proportional to the length of the wire,and length is proportional to the number of turns $(R \propto N)$.Therefore,$V_s = \frac{I_s}{R} \propto \frac{N}{N} = 	ext{constant}$.Since $V_s$ is independent of the number of turns $N$,the voltage sensitivity remains unchanged.Thus,the percentage change in voltage sensitivity is $0 \%$.

The number of turns of the coil of a moving coil galvanometer is increased in order to increase current sensitivity by $50 \%$. The percentage change in voltage sensitivity of the galvanometer will be $..........\%$

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