When a galvanometer is shunted with a `4Omega` resistance, the deflection is reduced to `1//5`. If the galvanometer is further shounted with a `2Omega` wire, the new deflection will be (assuming the main current remains the same)

Options

A. `5//13` of the deflection when shunted with `4Omega` only
B. `8//13` of the deflection when shuntedwith `4Omega`only
C. `3//4` of the deflection when shunted with `4Omega` only
D. `3//13` of the deflection when shunted with `4Omega`only

Explanation

To solve the problem step by step, we need to analyze the situation with the galvanometer and the shunt resistances.

### Step 1: Understand the Initial Condition
When the galvanometer is shunted with a `4Ω` resistance, the deflection is reduced to `1/5` of its original deflection. Let's denote:
- \( I_G \): the original current through the galvanometer,
- \( D \): the original deflection,
- \( D' = \frac{D}{5} \): the new deflection after shunting with `4Ω`.

### Step 2: Determine the Current through the Galvanometer
When the galvanometer is shunted with a `4Ω` resistor, the current through the galvanometer can be expressed as:
\[ I_g' = \frac{I_G}{5} \]
This means that the current flowing through the galvanometer is reduced to \( \frac{4I_G}{5} \) due to the shunt.

### Step 3: Calculate the Resistance of the Galvanometer
Using the relationship between the currents and resistances, we can set up the equation:
\[ I_G \cdot R_G = \left(\frac{I_G}{5}\right) \cdot 4 \]
Where \( R_G \) is the resistance of the galvanometer. Simplifying gives:
\[ R_G = 4 \cdot 4 = 16Ω \]

### Step 4: Adding a Second Shunt Resistance
Now, we shunt the galvanometer with an additional `2Ω` resistor. We need to find the equivalent resistance of the `4Ω` and `2Ω` resistors in parallel:
\[ R_{eq} = \frac{R_1 \cdot R_2}{R_1 + R_2} = \frac{4 \cdot 2}{4 + 2} = \frac{8}{6} = \frac{4}{3}Ω \]

### Step 5: Calculate the New Current through the Galvanometer
The total current \( I_G \) will now split between the equivalent resistance \( R_{eq} \) and the galvanometer. Let \( I_g'' \) be the new current through the galvanometer:
\[ I_G = I_g'' + I_{eq} \]
Using Ohm's law:
\[ I_{eq} = \frac{V}{R_{eq}} = \frac{I_g'' \cdot R_G}{R_G} \cdot \frac{R_{eq}}{R_G + R_{eq}} = \frac{I_g'' \cdot 16}{16 + \frac{4}{3}} \]
This leads to:
\[ 16I_g'' = I_G - I_g'' \]
Solving for \( I_g'' \):
\[ 12I_g'' = I_G \implies I_g'' = \frac{I_G}{12} \]

### Step 6: Relate the New Current to the New Deflection
The new deflection \( D'' \) can be expressed as:
\[ D'' = k \cdot I_g'' = k \cdot \frac{I_G}{12} \]
Where \( k \) is a constant of proportionality. Since the original deflection \( D = k \cdot I_G \), we can express the new deflection as:
\[ D'' = \frac{D}{12} \]

### Step 7: Final Calculation of Deflection
Since the original deflection was reduced to \( \frac{D}{5} \) when shunted with `4Ω`, the new deflection when shunted with `2Ω` will be:
\[ D'' = \frac{D}{12} \]
To express this in terms of the original deflection:
\[ D'' = \frac{5}{12}D \]

### Conclusion
Thus, the new deflection when the galvanometer is shunted with a `2Ω` wire will be \( \frac{5}{12} \) of the original deflection.