In the given problem, we need to find the resistance that must be added in series with the galvanometer so that the main current in the circuit remains unchanged after shunting the galvanometer with a resistance \( S \). Let's go through the steps to find the solution.
A galvanometer is typically a sensitive instrument used to detect small currents. When in use as an ammeter, it is shunted with another resistor to bypass most of the current. This helps protect the galvanometer from large currents.
When the galvanometer of resistance \( G \) is shunted by a resistance \( S \), the effective resistance (\( R_{\text{eff}} \)) of the parallel combination can be calculated as:
R_{\text{eff}} = \frac{SG}{S+G}
To ensure the main current remains unchanged, the total resistance before and after the introduction of the shunt should be the same. This means that the effective resistance of the galvanometer and its series resistance R should equal the unshunted galvanometer resistance \( G \).
Thus, the equation becomes:
G = R_{\text{eff}} + R_{\text{series}}
Substituting the expression for \( R_{\text{eff}} \):
G = \frac{SG}{S+G} + R_{\text{series}}
Solving for \( R_{\text{series}} \) gives:
R_{\text{series}} = G - \frac{SG}{S+G}
Simplifying further:
R_{\text{series}} = \frac{G(S + G) - SG}{S + G}
R_{\text{series}} = \frac{G^2}{S + G}
The resistance \( \frac{G^2}{S+G} \) needs to be added in series with the galvanometer to keep the current unchanged. Thus, the correct answer is:
\frac{G^2}{S+G}
The magnetic moment is associated with its spin angular momentum and orbital angular momentum. Spin only magnetic moment value of Cr^{3+ ion (Atomic no. : Cr = 24) is: