QJEEM23Phd6e5: JEE Main 2023 Physics - Class 12 JEEM23Phd6e5 Solution: The current sensitivity of moving coil galvanometer is…

The current sensitivity ( $I_s$ ) of a moving coil galvanometer is given by:

$I_s = \frac{nBA}{C}$

where $n$ is the number of turns in the coil, $B$ is the magnetic field, $A$ is the area of the coil, and $C$ is the torsional constant of the suspension wire.

We are told that the current sensitivity is increased by 25% by changing only the number of turns and the area of the cross section while keeping the resistance of the galvanometer coil constant. Let the new number of turns be $n'$ , and the new area be $A'$ .

Since the current sensitivity is increased by 25%, we have:

$I_s' = 1.25I_s = \frac{n' B A'}{C}$

The resistance ( $R$ ) of the galvanometer coil is given by:

$R = \rho \frac{l}{A}$

where $\rho$ is the resistivity of the wire and $l$ is the length of the wire. Since the resistance is kept constant, we can write:

$R' = \rho \frac{l'}{A'} = R$

Now, the voltage sensitivity ( $V_s$ ) of the galvanometer is given by:

$V_s = I_s \times R$

We want to find the percentage change in the voltage sensitivity. Let the new voltage sensitivity be $V_s'$ :

$V_s' = I_s' \times R'$

Since $R = R'$ , we can write:

$V_s' = 1.25I_s \times R = 1.25V_s$

The percentage change in the voltage sensitivity is:

$\frac{V_s' - V_s}{V_s} \times 100\% = \frac{1.25V_s - V_s}{V_s} \times 100\% = 25\%$