In a meter bridge, null point is `20 cm`, when the known resistance `R` is shunted by `10 Omega` resistance, null point is found to be shifted by 10 cm. Find the unknown resistance `X`.

Options

A. `10 cm`
B. `20 cm`
C. `30 cm`
D. `40cm`

Explanation

To solve the problem step by step, we will use the principles of a meter bridge and the concept of resistances in a balanced bridge circuit.

### Step 1: Understanding the Initial Setup
In the meter bridge, the null point is initially at 20 cm. This means that the lengths of the bridge wire on either side of the null point are:
- Length on one side (L1) = 20 cm
- Length on the other side (L2) = 100 cm - 20 cm = 80 cm

Let the known resistance be \( R \) and the unknown resistance be \( X \).

### Step 2: Setting Up the Balance Equation
Using the principle of the meter bridge, we can set up the equation based on the lengths:
\[ \frac{R}{X} = \frac{L1}{L2} = \frac{20}{80} \]
From this, we can simplify:
\[ \frac{R}{X} = \frac{1}{4} \]
This implies:
\[ X = 4R \]

### Step 3: Introducing the Shunt Resistance
Now, when a 10 Ω resistance is shunted across \( R \), the new effective resistance \( R' \) can be calculated using the formula for resistances in parallel:
\[ \frac{1}{R'} = \frac{1}{R} + \frac{1}{10} \]
This can be rearranged to:
\[ R' = \frac{R \cdot 10}{R + 10} \]

### Step 4: Analyzing the Shift in Null Point
After shunting the 10 Ω resistance, the null point shifts by 10 cm. The new null point is at:
- New length on one side (L1') = 20 cm - 10 cm = 10 cm
- New length on the other side (L2') = 100 cm - 10 cm = 90 cm

Now we can set up the new balance equation:
\[ \frac{R'}{X} = \frac{L1'}{L2'} = \frac{10}{90} = \frac{1}{9} \]
This implies:
\[ \frac{R'}{X} = \frac{1}{9} \]
Thus:
\[ X = 9R' \]

### Step 5: Substituting for \( R' \)
We already found that \( R' = \frac{R \cdot 10}{R + 10} \). Now substituting this into the equation for \( X \):
\[ X = 9 \left(\frac{R \cdot 10}{R + 10}\right) \]
This simplifies to:
\[ X = \frac{90R}{R + 10} \]

### Step 6: Equating the Two Expressions for \( X \)
From Step 2, we have \( X = 4R \). Now we equate the two expressions for \( X \):
\[ 4R = \frac{90R}{R + 10} \]

### Step 7: Solving for \( R \)
Cross-multiplying gives:
\[ 4R(R + 10) = 90R \]
Expanding this:
\[ 4R^2 + 40R = 90R \]
Rearranging gives:
\[ 4R^2 - 50R = 0 \]
Factoring out \( R \):
\[ R(4R - 50) = 0 \]
Thus, \( R = 0 \) or \( R = 12.5 \) Ω.

### Step 8: Finding the Unknown Resistance \( X \)
Substituting \( R = 12.5 \) Ω back into the equation for \( X \):
\[ X = 4R = 4 \times 12.5 = 50 \, \Omega \]

### Final Answer
The unknown resistance \( X \) is \( 50 \, \Omega \).

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