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

Answer

To solve for the unknown resistance \( R_1 \) in the given problem, we can follow these steps:

### Step 1: Set up the initial condition
When the unknown resistance \( R_1 \) is connected in series with a \( 10 \, \Omega \) resistor, the balance point is at \( 50 \, \text{cm} \). This means that the ratio of the resistances is equal to the ratio of the lengths on the meter bridge.

### Step 2: Write the first equation
Using the balance point condition, we can write:
\[ \frac{R_1 + 10}{R_2} = \frac{50}{50} = 1 \]
From this, we can derive:
\[ R_1 + 10 = R_2 \quad \text{(Equation 1)} \]

### Step 3: Set up the second condition
When the \( 10 \, \Omega \) resistor is removed, the balance point shifts to \( 40 \, \text{cm} \). The remaining length on the other side of the meter bridge is \( 60 \, \text{cm} \).

### Step 4: Write the second equation
Using the new balance point, we can write:
\[ \frac{R_1}{R_2} = \frac{40}{60} = \frac{2}{3} \]
From this, we can derive:
\[ R_2 = \frac{3}{2} R_1 \quad \text{(Equation 2)} \]

### Step 5: Substitute Equation 2 into Equation 1
Now, we substitute \( R_2 \) from Equation 2 into Equation 1:
\[ R_1 + 10 = \frac{3}{2} R_1 \]

### Step 6: Solve for \( R_1 \)
Rearranging the equation gives:
\[ 10 = \frac{3}{2} R_1 - R_1 \]
\[ 10 = \frac{3}{2} R_1 - \frac{2}{2} R_1 \]
\[ 10 = \frac{1}{2} R_1 \]
Multiplying both sides by \( 2 \):
\[ R_1 = 20 \, \Omega \]

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

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