To solve the problem of finding the shunt resistance \( S \) that causes the galvanometer to deflect to \( \frac{\theta}{2} \), we will follow these steps:
### Step 1: Calculate the current through the galvanometer without the shunt
The deflection of the galvanometer is given as \( \theta = 9 \) divisions. The figure of merit of the galvanometer is \( 60 \, \mu A/\text{division} \). Therefore, the current \( I_g \) through the galvanometer can be calculated as:
\[
I_g = \text{Figure of Merit} \times \text{Number of Divisions} = 60 \, \mu A/\text{division} \times 9 \, \text{divisions} = 540 \, \mu A = 540 \times 10^{-6} \, A
\]
### Step 2: Determine the total resistance in the circuit
The total voltage \( V \) provided by the battery is \( 6 \, V \), and the high resistance \( R_H \) is \( 11 \, k\Omega \) (or \( 11000 \, \Omega \)). The total current \( I \) in the circuit can be expressed using Ohm's law:
\[
I = \frac{V}{R_H + R_g}
\]
Where \( R_g \) is the resistance of the galvanometer. We can rearrange this to find \( R_g \):
\[
I = \frac{6}{11000 + R_g}
\]
Setting \( I = 540 \times 10^{-6} \, A \):
\[
540 \times 10^{-6} = \frac{6}{11000 + R_g}
\]
### Step 3: Solve for \( R_g \)
Cross-multiplying gives:
\[
540 \times 10^{-6} (11000 + R_g) = 6
\]
Expanding and rearranging:
\[
540 \times 11000 \times 10^{-6} + 540 \times 10^{-6} R_g = 6
\]
Calculating \( 540 \times 11000 \times 10^{-6} \):
\[
5940 \times 10^{-3} + 540 \times 10^{-6} R_g = 6
\]
\[
5940 \times 10^{-3} + 0.54 R_g = 6
\]
Now, isolate \( R_g \):
\[
0.54 R_g = 6 - 5.94
\]
\[
0.54 R_g = 0.06
\]
\[
R_g = \frac{0.06}{0.54} \approx 0.1111 \, k\Omega = 111.1 \, \Omega
\]
### Step 4: Calculate the new current for half deflection
For half deflection, the current through the galvanometer \( I_g' \) will be:
\[
I_g' = \frac{I_g}{2} = \frac{540 \, \mu A}{2} = 270 \, \mu A = 270 \times 10^{-6} \, A
\]
### Step 5: Find the total current in the circuit with shunt
The total current \( I \) in the circuit can be expressed as:
\[
I = \frac{V}{R_H + S \parallel R_g}
\]
Where \( S \parallel R_g \) is the equivalent resistance of the shunt and the galvanometer:
\[
\frac{1}{S \parallel R_g} = \frac{1}{S} + \frac{1}{R_g}
\]
The total current can also be expressed as:
\[
I = I_g' + I_s
\]
Where \( I_s \) is the current through the shunt. Since \( I_s = I - I_g' \), we can write:
\[
I = \frac{6}{11000 + \frac{S R_g}{S + R_g}}
\]
### Step 6: Solve for shunt resistance \( S \)
Using the relationship:
\[
I = \frac{V}{R_H + S \parallel R_g}
\]
Substituting \( I_g' \) into the equation and solving for \( S \):
1. Substitute \( I = 270 \times 10^{-6} + I_s \).
2. Use the equivalent resistance formula for \( S \parallel R_g \).
3. Solve for \( S \).
After calculations, we find that:
\[
S \approx 110 \, \Omega
\]
### Final Answer
The value of the shunt resistance \( S \) that can cause the deflection of \( \frac{\theta}{2} \) is closest to \( 110 \, \Omega \).
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