The deflection in galvanometer falls to `((1)/(4))^(th)` when it is shunted by `3Omega`. If additional shunt of `2Omega`is connected to earlier shunt in parallel, the deflection in galvanometer falls to

Answer

To solve the problem step-by-step, we will analyze the situation involving a galvanometer and shunt resistances.

### Step 1: Understand the Initial Condition
We know that when a shunt of \(3 \, \Omega\) is connected, the deflection in the galvanometer falls to \( \frac{1}{4} \) of its original value.

### Step 2: Set Up the Equation for the First Condition
Let \( I \) be the total current flowing through the circuit, \( I_g \) be the current through the galvanometer, and \( G \) be the resistance of the galvanometer. The current through the shunt \( S \) (which is \(3 \, \Omega\)) can be expressed as:
\[ I_g = \frac{S}{G + S} \cdot I \]
Given that the deflection falls to \( \frac{1}{4} \), we can write:
\[ \frac{I_g}{I} = \frac{1}{4} \]
Substituting \( S = 3 \, \Omega \):
\[ \frac{1}{4} = \frac{3}{G + 3} \]

### Step 3: Solve for \( G \)
Cross-multiplying gives:
\[ G + 3 = 12 \implies G = 9 \, \Omega \]

### Step 4: Introduce the Additional Shunt
Now, we add another shunt of \(2 \, \Omega\) in parallel with the existing \(3 \, \Omega\) shunt. We need to find the equivalent resistance \( S_{eq} \) of the two shunts in parallel:
\[ \frac{1}{S_{eq}} = \frac{1}{3} + \frac{1}{2} \]
Calculating the right-hand side:
\[ \frac{1}{S_{eq}} = \frac{2 + 3}{6} = \frac{5}{6} \implies S_{eq} = \frac{6}{5} \, \Omega \]

### Step 5: Set Up the Equation for the New Condition
Now we need to find the new deflection ratio with the equivalent shunt:
\[ \frac{I_g}{I} = \frac{S_{eq}}{G + S_{eq}} \]
Substituting \( S_{eq} = \frac{6}{5} \, \Omega \) and \( G = 9 \, \Omega \):
\[ \frac{I_g}{I} = \frac{\frac{6}{5}}{9 + \frac{6}{5}} = \frac{\frac{6}{5}}{\frac{45}{5} + \frac{6}{5}} = \frac{\frac{6}{5}}{\frac{51}{5}} = \frac{6}{51} = \frac{2}{17} \]

### Final Answer
Thus, the deflection in the galvanometer falls to:
\[ \frac{2}{17} \]