b
(b) As bridge is in balanced condition, no current will flow through $BD.$
$R_{1}=R_{A B}+R_{B C}$
$=R+R=2 R$
$R_{2}=R_{A D}+R_{C D}=R+R=2 R$
Also, $R_{1}$ and $R_{2}$ are in parallel combination
Hence, equivalent resistance between $A$ and $C$ will be
$R_{e q}=\frac{R_{1} R_{2}}{R_{1}+R_{2}}=\frac{4 R^{2}}{4 R}=R$
