In a Wheatstone bridge (see fig.), Resistances P and Q are approximately equal. When $$R = 400\Omega$$, the bridge is balanced. On interchanging P and Q, the value of R, for balance, is $$405\Omega$$. The value of Y is close to

First recall the Wheatstone-bridge balance condition. When no current flows through the galvanometer, the ratio of the resistances in one pair of opposite arms equals the ratio in the other pair, i.e.

$$\dfrac{P}{Q}= \dfrac{R}{Y}$$

Here $$P$$ and $$Q$$ are approximately equal resistances in the two upper arms, $$R$$ is the variable resistance in a lower arm, and $$Y$$ is the fixed resistance in the remaining lower arm that we have to find.

We are given two different balance situations:

First balance  -  the arms are $$P,Q,R,Y$$ in that order and balance is obtained with $$R_1=400\;\Omega$$. Using the balance condition, we write

$$\dfrac{P}{Q}= \dfrac{R_1}{Y}$$

Substituting $$R_1=400\;\Omega$$ gives

$$\dfrac{P}{Q}= \dfrac{400}{Y} \quad -(1)$$

For convenience introduce the ratio

$$k=\dfrac{P}{Q}$$

Equation (1) then becomes

$$k = \dfrac{400}{Y}\quad\Longrightarrow\quad Y = \dfrac{400}{k} \quad -(2)$$

Second balance  -  now $$P$$ and $$Q$$ are interchanged, so the arms are $$Q,P,R,Y$$. Balance is obtained with a new resistance value $$R_2 = 405\;\Omega$$. Again applying the balance condition, but with the new order, we obtain

$$\dfrac{Q}{P}= \dfrac{R_2}{Y}$$

Since $$\dfrac{Q}{P}=\dfrac{1}{k}$$ and $$R_2=405\;\Omega$$, this becomes

$$\dfrac{1}{k}= \dfrac{405}{Y} \quad -(3)$$

Cross-multiplying equation (3) gives

$$Y = 405\,k \quad -(4)$$

Now we have two separate expressions for $$Y$$, namely equation (2) and equation (4). Because they represent the same resistance, we equate them:

$$\dfrac{400}{k}= 405\,k$$

Multiplying both sides by $$k$$ to clear the denominator, we get

$$400 = 405\,k^2$$

Solving for $$k^2$$,

$$k^2 = \dfrac{400}{405}$$

Taking the (positive) square root,

$$k = \sqrt{\dfrac{400}{405}}$$

We next substitute this value of $$k$$ back into either equation (2) or equation (4) to find $$Y$$. Using equation (4):

$$Y = 405\,k = 405\,\sqrt{\dfrac{400}{405}}$$

To simplify, notice that

$$405\,\sqrt{\dfrac{400}{405}} = \sqrt{405^2}\,\sqrt{\dfrac{400}{405}} = \sqrt{405}\,\sqrt{405}\,\sqrt{\dfrac{400}{405}}$$ $$= \sqrt{405}\,\sqrt{400} = \sqrt{405\times400}$$

Therefore,

$$Y = \sqrt{400\times405}\; \Omega$$

Calculating the product inside the square root,

$$400\times405 = 162{,}000$$

and taking the square root:

$$Y = \sqrt{162{,}000}\; \Omega \approx 402.5\; \Omega$$

Hence, the correct answer is Option D.