Explanation

(a) The balanced condition for Wheatstone bridge is
$\frac{P}{Q} = \frac{R}{S}$
as is obvious from the given values.
No, current flows through galvanometer is zero.
Now, P and R are in series, so
Resistance $R_{1} = P + R$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 10 + 15 = 25\Omega$
Similarly, Q and S are in series, so
Resistance $R_{2} = R + S = 20 + 30 = 50\Omega$
Net resistance of the network as $R_{1}$ and $R_{2}$ are in parallel.
$\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}}\ $
$\therefore R = \frac{25 \times 50}{25 + 50} = \frac{50}{3}\ \Omega$
Hence,
$I = \frac{V}{R} = \frac{6}{\frac{50}{3}} = 0.36A$