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(B) The balancing condition for a Wheatstone's bridge is $\frac{P}{Q} = \frac{R_3}{R_4}$,where $R_3$ is the resistance in the third arm and $R_4$ is t

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$650$

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(B) The balancing condition for a Wheatstone's bridge is $\frac{P}{Q} = \frac{R_3}{R_4}$,where $R_3$ is the resistance in the third arm and $R_4$ is the resistance in the fourth arm $(S)$.Given $R_3 = 625\, \Omega$ and $R_4 = S$,we have:$\frac{P}{Q} = \frac{625}{S}$ ..................... $(1)$When $P$ and $Q$ are interchanged,the new resistance in the third arm becomes $625 + 51 = 676\, \Omega$. The bridge is balanced again,so:$\frac{Q}{P} = \frac{676}{S}$$\frac{P}{Q} = \frac{S}{676}$ ..................... $(2)$Equating $(1)$ and $(2)$:$\frac{625}{S} = \frac{S}{676}$$S^2 = 625 	imes 676$$S = \sqrt{625} 	imes \sqrt{676} = 25 	imes 26 = 650\, \Omega$.

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