An unknown resistance `R_(1)` is connected is series with a resistance of `10 Omega`. This combination is connected to one gap of a meter bridge, while other gap is connected to another resistance `R_(2)`. The balance point is at `50 cm` Now , when the `10 Omega` resistance is removed, the balanced point shifts to `40 cm` Then the value of `R_(1)` is.

Options

A. `60 Omega`
B. `40 Omega`
C. `20 Omega`
D. `10 Omega`

Explanation

To solve for the unknown resistance \( R_1 \), we will follow these steps:

### Step 1: Set up the equations based on the meter bridge principle.
When the unknown resistance \( R_1 \) is connected in series with a \( 10 \, \Omega \) resistor, the balance point is at \( 50 \, \text{cm} \). According to the meter bridge principle:

\[ \frac{R_1 + 10}{R_2} = \frac{50}{100 - 50} \]

This simplifies to:

\[ \frac{R_1 + 10}{R_2} = \frac{50}{50} = 1 \]

Thus, we have:

\[ R_1 + 10 = R_2 \quad \text{(Equation 1)} \]

### Step 2: Set up the equation for the second scenario.
When the \( 10 \, \Omega \) resistor is removed, the balance point shifts to \( 40 \, \text{cm} \). Now we have:

\[ \frac{R_1}{R_2} = \frac{40}{100 - 40} \]

This simplifies to:

\[ \frac{R_1}{R_2} = \frac{40}{60} = \frac{2}{3} \]

Thus, we can express this as:

\[ R_1 = \frac{2}{3} R_2 \quad \text{(Equation 2)} \]

### Step 3: Substitute Equation 1 into Equation 2.
From Equation 1, we know \( R_2 = R_1 + 10 \). We can substitute this into Equation 2:

\[ R_1 = \frac{2}{3}(R_1 + 10) \]

### Step 4: Solve for \( R_1 \).
Now, we will solve the equation:

\[ R_1 = \frac{2}{3} R_1 + \frac{20}{3} \]

Multiplying through by \( 3 \) to eliminate the fraction:

\[ 3R_1 = 2R_1 + 20 \]

Rearranging gives:

\[ 3R_1 - 2R_1 = 20 \]

Thus:

\[ R_1 = 20 \, \Omega \]

### Conclusion
The value of the unknown resistance \( R_1 \) is \( 20 \, \Omega \).

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