When un known resistance and a resistance of `4Omega` are used in the left and right gaps of a meter bridge, the balance point is 50cm. the shift in the balance point if a `4Omega` resistance is now connected parallel to the resistor in right gap ______

Options

A. `(100)/(3)cm`
B. `(50)/(3)cm`
C. `(300)/(3)cm `
D. `(400)/(5)cm`

Answer

To solve the problem step by step, we will analyze the situation using the principles of a meter bridge and the concept of resistances in parallel.

### Step 1: Understand the initial setup
We have a meter bridge with an unknown resistance \( x \) on the left side and a known resistance of \( 4 \Omega \) on the right side. The balance point is at \( 50 \, \text{cm} \).

### Step 2: Use the meter bridge formula
The meter bridge operates on the principle that the ratio of the resistances is equal to the ratio of the lengths from the ends of the bridge to the balance point. The formula is given by:

\[ \frac{x}{y} = \frac{l_1}{l_2} \]

Where:
- \( x \) = unknown resistance
- \( y \) = known resistance (4 Ω)
- \( l_1 \) = length from one end to the balance point (50 cm)
- \( l_2 \) = length from the other end to the balance point (100 cm - 50 cm = 50 cm)

### Step 3: Substitute known values
Substituting the known values into the equation:

\[ \frac{x}{4} = \frac{50}{50} \]

This simplifies to:

\[ \frac{x}{4} = 1 \]

### Step 4: Solve for the unknown resistance
From the equation, we find:

\[ x = 4 \, \Omega \]

### Step 5: Introduce the new scenario
Now, we connect a \( 4 \Omega \) resistor in parallel with the \( 4 \Omega \) resistor already present in the right gap. The equivalent resistance \( y' \) of two \( 4 \Omega \) resistors in parallel can be calculated using the formula:

\[ \frac{1}{y'} = \frac{1}{4} + \frac{1}{4} \]

### Step 6: Calculate the equivalent resistance
Calculating the equivalent resistance:

\[ \frac{1}{y'} = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \]

Thus,

\[ y' = 2 \, \Omega \]

### Step 7: Set up the new balance point equation
Now, we set up the equation for the new balance point using the new equivalent resistance \( y' \):

\[ \frac{x}{y'} = \frac{l_1'}{100 - l_1'} \]

Substituting the known values:

\[ \frac{4}{2} = \frac{l_1'}{100 - l_1'} \]

### Step 8: Solve for the new balance point \( l_1' \)
Cross-multiplying gives:

\[ 4(100 - l_1') = 2l_1' \]

Expanding and rearranging:

\[ 400 - 4l_1' = 2l_1' \]
\[ 400 = 6l_1' \]
\[ l_1' = \frac{400}{6} = \frac{200}{3} \, \text{cm} \]

### Step 9: Calculate the shift in the balance point
The initial balance point was at \( 50 \, \text{cm} \). The new balance point is \( \frac{200}{3} \, \text{cm} \). The shift in the balance point \( \Delta l \) is:

\[ \Delta l = l_1' - l_1 = \frac{200}{3} - 50 \]

### Step 10: Solve for the shift
Converting \( 50 \) to a fraction with a common denominator:

\[ 50 = \frac{150}{3} \]

Now, substituting:

\[ \Delta l = \frac{200}{3} - \frac{150}{3} = \frac{50}{3} \, \text{cm} \]

### Final Answer
The shift in the balance point is \( \frac{50}{3} \, \text{cm} \).

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