A galvanometer of resistance `50 Omega` is connected to a battery of 8 V along with a resistance of `3950 Omega` in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 15 divisions, the resistance in series should be .... `Omega`.

Options

A. 1950
B. 7900
C. 2000
D. 7950

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the Initial Total Resistance
The total resistance in the circuit when the galvanometer is connected is the sum of the resistance of the galvanometer and the series resistor.

\[ R_{\text{initial}} = R_{\text{galvanometer}} + R_{\text{series}} = 50 \, \Omega + 3950 \, \Omega = 4000 \, \Omega \]

### Step 2: Calculate the Initial Current
Using Ohm's Law, the current flowing through the circuit can be calculated as:

\[ I_{\text{initial}} = \frac{V}{R_{\text{initial}}} = \frac{8 \, V}{4000 \, \Omega} = 0.002 \, A \, (or \, 2 \, mA) \]

### Step 3: Determine the Required Current for 15 Divisions
Since we need to reduce the deflection to 15 divisions, we need to halve the current:

\[ I_{\text{final}} = \frac{I_{\text{initial}}}{2} = \frac{0.002 \, A}{2} = 0.001 \, A \, (or \, 1 \, mA) \]

### Step 4: Calculate the Required Total Resistance for the Final Current
To find the total resistance required to achieve this final current, we can use Ohm's Law again:

\[ R_{\text{final}} = \frac{V}{I_{\text{final}}} = \frac{8 \, V}{0.001 \, A} = 8000 \, \Omega \]

### Step 5: Calculate the Additional Resistance Required
Now, we need to find the additional resistance that needs to be added in series to achieve this final total resistance:

\[ R_{\text{series, new}} = R_{\text{final}} - R_{\text{galvanometer}} = 8000 \, \Omega - 50 \, \Omega = 7950 \, \Omega \]

### Final Answer
The resistance that should be added in series to reduce the deflection to 15 divisions is:

\[ \boxed{7950 \, \Omega} \]

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