The figure shows a meter bridge circuit with AB=100 cm, X=`12 Omega andR=18 Omega` and the jockey J in the position of balance. If R is now made `8Omega` through what distance will J have to be moved to obtain balance?

Options

A. 10 cm
B. 20 cm
C. 30 cm
D. 40 cm

Explanation

To solve the problem, we will use the principle of the meter bridge, which is based on the Wheatstone bridge principle. The balance condition can be expressed in terms of the lengths on the meter bridge and the resistances connected.

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
- Given:
- Length of the meter bridge, \( AB = 100 \, \text{cm} \)
- Resistance \( X = 12 \, \Omega \)
- Resistance \( R = 18 \, \Omega \)
- Let the length from point A to the jockey position be \( L \).

2. **Apply the Balance Condition:**
- According to the balance condition of the meter bridge:
\[ \frac{R}{X} = \frac{100 - L}{L} \]
- Substituting the values of \( R \) and \( X \):
\[ \frac{18}{12} = \frac{100 - L}{L} \]
- Simplifying the left side:
\[ \frac{3}{2} = \frac{100 - L}{L} \]

3. **Cross-Multiply to Solve for \( L \):**
- Cross-multiplying gives:
\[ 3L = 2(100 - L) \]
- Expanding the right side:
\[ 3L = 200 - 2L \]
- Rearranging the equation:
\[ 3L + 2L = 200 \]
\[ 5L = 200 \]
\[ L = \frac{200}{5} = 40 \, \text{cm} \]

4. **Change the Resistance \( R \):**
- Now, we change \( R \) to \( 8 \, \Omega \).
- We need to find the new length \( L' \) when \( R = 8 \, \Omega \):
\[ \frac{R}{X} = \frac{100 - L'}{L'} \]
- Substituting the new value of \( R \):
\[ \frac{8}{12} = \frac{100 - L'}{L'} \]
- Simplifying the left side:
\[ \frac{2}{3} = \frac{100 - L'}{L'} \]

5. **Cross-Multiply to Solve for \( L' \):**
- Cross-multiplying gives:
\[ 2L' = 3(100 - L') \]
- Expanding the right side:
\[ 2L' = 300 - 3L' \]
- Rearranging the equation:
\[ 2L' + 3L' = 300 \]
\[ 5L' = 300 \]
\[ L' = \frac{300}{5} = 60 \, \text{cm} \]

6. **Calculate the Distance the Jockey Needs to Move:**
- The distance \( \Delta X \) that the jockey has to move is:
\[ \Delta X = L' - L = 60 \, \text{cm} - 40 \, \text{cm} = 20 \, \text{cm} \]

### Final Answer:
The jockey has to be moved by **20 cm** to obtain balance when \( R \) is changed to \( 8 \, \Omega \).

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