1. Condition for balance of Wheatstone bridge

For a balanced Wheatstone bridge, $\frac{P}{Q} = \frac{R}{X}$ so, $X = \frac{Q}{P}R$

2. First balance condition

When $R = 400\,\Omega$, the bridge is balanced. Hence, \frac{P}{Q} = \frac{400}{X} \quad \Rightarrow \quad X = 400\frac{Q}{P} \tag{1}

3. After interchanging $P$ and $Q$

Now the ratio becomes $\frac{Q}{P} = \frac{405}{X}$ Thus, X = 405\frac{P}{Q} \tag{2}

4. Multiply equations (1) and (2)

From (1) and (2): $X^2 = 400 \times 405$ So, $X = \sqrt{400\cdot 405}$ $X = \sqrt{162000}$ $X = 402.49\,\Omega$

Hence, $X \approx 402.5\,\Omega$

5. Check options

• A: $402.5\,\Omega$ ✔
• B: $401.5\,\Omega$ ✘
• C: $403.5\,\Omega$ ✘
• D: $404.5\,\Omega$ ✘

Therefore, the correct option is: $\boxed{402.5\,\Omega}$

6. Comparison with stored answer

Stored correct answer: A

Our derived answer is also A.