Two resistances are connected in two gaps of Meter Bridge. The balance is 10 cm from the zero end. A resistance of 20 `Omega` is connected in series with the smaller of the two. The null point shifts to 20 cm. What is the value of the bigger resistance?

Options

A. `89 Omega`
B. `144Omega`
C. `100Omega`
D. None of the above

Explanation

To solve the problem step by step, we will use the principles of a meter bridge and the concept of resistance in series and parallel.

### Step 1: Understand the initial setup
In the meter bridge, we have two resistances, \( P \) and \( Q \). The initial balance point is at 10 cm from the zero end. This means that:

\[ \frac{P}{Q} = \frac{10}{90} = \frac{1}{9} \]

From this, we can express \( P \) in terms of \( Q \):

\[ P = \frac{Q}{9} \]

### Step 2: Introduce the series resistance
A resistance of \( 20 \, \Omega \) is connected in series with the smaller resistance \( P \). The new balance point shifts to 20 cm. Therefore, we can write the new ratio:

\[ \frac{P + 20}{Q} = \frac{20}{80} = \frac{1}{4} \]

### Step 3: Set up the equation
From the new balance point, we can express this as:

\[ P + 20 = \frac{Q}{4} \]

### Step 4: Substitute \( P \) from the first equation
We already have \( P = \frac{Q}{9} \). We can substitute this into the equation from Step 3:

\[ \frac{Q}{9} + 20 = \frac{Q}{4} \]

### Step 5: Clear the fractions
To eliminate the fractions, multiply through by 36 (the least common multiple of 9 and 4):

\[ 4Q + 720 = 9Q \]

### Step 6: Rearrange the equation
Now, rearranging gives:

\[ 720 = 9Q - 4Q \]
\[ 720 = 5Q \]

### Step 7: Solve for \( Q \)
Dividing both sides by 5:

\[ Q = \frac{720}{5} = 144 \, \Omega \]

### Step 8: Find \( P \)
Now, we can find \( P \) using the relation \( P = \frac{Q}{9} \):

\[ P = \frac{144}{9} = 16 \, \Omega \]

### Conclusion
The value of the bigger resistance \( Q \) is:

\[ \boxed{144 \, \Omega} \]