Two coils are connected in series in one gap of the Wheatstone's meter bridge and null point is obtained at the centre of the wire with a resistance of `100 Omega` in the other gap. When the two coils are connected in parallel in the same gap, the unknown resistance is to be changed by `84 Omega` to obtain the null point at the centre again. The resistance of the two coils will be

Options

A. `30Omega` and `80Omega`
B. `80Omega` and `30Omega`
C. `20Omega` and `80Omega`
D. `50Omega` and `50Omega`

Explanation

To solve the problem, we need to analyze the situation with the Wheatstone bridge and the two coils connected in series and parallel.

### Step-by-Step Solution:

1. **Understanding the Series Connection:**
When the two coils (let's denote their resistances as \( x_1 \) and \( x_2 \)) are connected in series, the total resistance is given by:
\[ R_{\text{total, series}} = x_1 + x_2 \]
Since the null point is obtained at the center of the bridge, the balancing condition can be expressed as:
\[ \frac{x_1 + x_2}{50} = \frac{100}{50} \]
This simplifies to:
\[ x_1 + x_2 = 100 \quad \text{(Equation 1)} \]

2. **Understanding the Parallel Connection:**
When the coils are connected in parallel, the equivalent resistance \( R' \) is given by:
\[ R' = \frac{x_1 x_2}{x_1 + x_2} \]
We know that when the coils are connected in parallel, the resistance in the other gap (which was 100 ohms) has to be changed by 84 ohms to maintain the null point at the center. This means:
\[ R' = 100 - 84 = 16 \quad \text{(Equation 2)} \]

3. **Substituting Equation 1 into Equation 2:**
From Equation 1, we know \( x_1 + x_2 = 100 \). We can substitute this into the equation for \( R' \):
\[ 16 = \frac{x_1 x_2}{100} \]
Rearranging gives:
\[ x_1 x_2 = 16 \times 100 = 1600 \quad \text{(Equation 3)} \]

4. **Solving the System of Equations:**
Now we have two equations:
- \( x_1 + x_2 = 100 \) (Equation 1)
- \( x_1 x_2 = 1600 \) (Equation 3)

We can express \( x_2 \) in terms of \( x_1 \):
\[ x_2 = 100 - x_1 \]
Substituting this into Equation 3:
\[ x_1(100 - x_1) = 1600 \]
Expanding and rearranging gives:
\[ 100x_1 - x_1^2 = 1600 \]
\[ x_1^2 - 100x_1 + 1600 = 0 \]

5. **Using the Quadratic Formula:**
We can solve this quadratic equation using the quadratic formula:
\[ x_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -100 \), and \( c = 1600 \):
\[ x_1 = \frac{100 \pm \sqrt{10000 - 6400}}{2} \]
\[ x_1 = \frac{100 \pm \sqrt{3600}}{2} \]
\[ x_1 = \frac{100 \pm 60}{2} \]
This gives us two possible values:
\[ x_1 = \frac{160}{2} = 80 \quad \text{or} \quad x_1 = \frac{40}{2} = 20 \]

6. **Finding \( x_2 \):**
Using \( x_1 + x_2 = 100 \):
- If \( x_1 = 80 \), then \( x_2 = 20 \).
- If \( x_1 = 20 \), then \( x_2 = 80 \).

Thus, the resistances of the two coils are \( 80 \Omega \) and \( 20 \Omega \).

### Final Answer:
The resistances of the two coils are \( 20 \Omega \) and \( 80 \Omega \).

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