In a meter bridge `30 Omega ` resistance is connected in the left gap and a pair of resistance `20 Omega ` and Q in (Ohm) are in right gap and they are connected in series, then the balanceing length is 50 cm. When they were connected in parallel the balanceing length will be

Options

A. 20 cm
B. 40 cm
C. 60 cm
D. 81.81 cm

Answer

To solve the problem step by step, we will follow the process outlined in the video transcript while ensuring clarity and thoroughness.

### Step 1: Understand the Setup
We have a meter bridge with:
- A 30 Ω resistor in the left gap.
- A combination of a 20 Ω resistor and an unknown resistor Q in series in the right gap.
- The balancing length is given as 50 cm.

### Step 2: Use the Balancing Condition
The balancing condition for a meter bridge states that the ratio of the resistances is equal to the ratio of the lengths. Therefore, we can write:
\[ \frac{R_1}{L_1} = \frac{R_2}{L_2} \]
where \( R_1 = 30 \, \Omega \), \( L_1 = 50 \, \text{cm} \), \( R_2 = 20 + Q \, \Omega \), and \( L_2 = 100 - 50 = 50 \, \text{cm} \).

Substituting the values, we have:
\[ \frac{30}{50} = \frac{20 + Q}{50} \]

### Step 3: Simplify the Equation
Cross-multiplying gives:
\[ 30 \times 50 = (20 + Q) \times 50 \]
This simplifies to:
\[ 30 = 20 + Q \]

### Step 4: Solve for Q
Rearranging the equation gives:
\[ Q = 30 - 20 = 10 \, \Omega \]

### Step 5: Calculate the Equivalent Resistance in Parallel
Now, we need to find the equivalent resistance when the 20 Ω and 10 Ω resistors are connected in parallel. The formula for two resistors in parallel is:
\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \]
Substituting the values:
\[ \frac{1}{R_{eq}} = \frac{1}{20} + \frac{1}{10} \]
Finding a common denominator (which is 20):
\[ \frac{1}{R_{eq}} = \frac{1}{20} + \frac{2}{20} = \frac{3}{20} \]
Thus,
\[ R_{eq} = \frac{20}{3} \, \Omega \]

### Step 6: Set Up the New Balancing Condition
Now, we set up the balancing condition again with the new equivalent resistance:
\[ \frac{30}{x} = \frac{20/3}{100 - x} \]
where \( x \) is the new balancing length.

### Step 7: Cross-Multiply and Solve for x
Cross-multiplying gives:
\[ 30(100 - x) = \frac{20}{3}x \]
Expanding this:
\[ 3000 - 30x = \frac{20}{3}x \]
Multiplying through by 3 to eliminate the fraction:
\[ 9000 - 90x = 20x \]
Combining like terms:
\[ 9000 = 110x \]
Thus,
\[ x = \frac{9000}{110} = \frac{900}{11} \approx 81.81 \, \text{cm} \]

### Conclusion
The new balancing length when the resistors are connected in parallel is approximately **81.81 cm**.

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