Two resistances of values `20 Omega and 20 Omega` are introduced in the left and right gaps of a metre bridge. What is the shift in the null point if a resistance of `40 Omega` is connected in series with the resistance in the left gap ?

Options

A. 25 cm towards left of the centre
B. 25 cm towards right of the centre
C. 15 cm towards right of the centre
D. 15 cm towards left of the centre

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the initial setup
We have a meter bridge with two resistances of 20 Ω each placed in the left and right gaps. The initial null point is determined by the balance of the bridge.

### Step 2: Calculate the initial null point
Using the formula for the meter bridge, we have:

\[ \frac{L_1}{L_2} = \frac{R_1}{R_2} \]

Where:
- \(L_1\) is the length on the left side of the bridge,
- \(L_2\) is the length on the right side of the bridge,
- \(R_1 = 20 \, \Omega\) (left side),
- \(R_2 = 20 \, \Omega\) (right side).

Since both resistances are equal, we can set:

\[ \frac{L_1}{100 - L_1} = \frac{20}{20} \]

This simplifies to:

\[ L_1 = 100 - L_1 \]

Solving this gives:

\[ 2L_1 = 100 \implies L_1 = 50 \, \text{cm} \]

### Step 3: Introduce the new resistance
Now, we connect a resistance of 40 Ω in series with the 20 Ω resistance on the left side. The total resistance on the left side becomes:

\[ R_1' = 20 \, \Omega + 40 \, \Omega = 60 \, \Omega \]

### Step 4: Calculate the new null point
Now, we apply the meter bridge formula again with the new resistance:

\[ \frac{L_2}{100 - L_2} = \frac{R_1'}{R_2} \]

Substituting the values:

\[ \frac{L_2}{100 - L_2} = \frac{60}{20} \]

This simplifies to:

\[ \frac{L_2}{100 - L_2} = 3 \]

Cross-multiplying gives:

\[ L_2 = 3(100 - L_2) \]

Expanding and rearranging:

\[ L_2 + 3L_2 = 300 \implies 4L_2 = 300 \implies L_2 = 75 \, \text{cm} \]

### Step 5: Calculate the shift in the null point
The initial null point was at 50 cm, and the new null point is at 75 cm. The shift in the null point is:

\[ \text{Shift} = L_2 - L_1 = 75 \, \text{cm} - 50 \, \text{cm} = 25 \, \text{cm} \]

### Conclusion
The null point shifts 25 cm towards the right of the center.

---