Explanation

(b)
Current $(i)\ = \ 1.5\ A$
Time $(t)\ 10\ min\ = \ 10\ \times \ 60\ = \ 600\ s$
Quantity of electricity passed $Q\ = \ i\ \times \ t\$
$= \ (1.5\ A)\ \times \ (600\ s)$
$= \ 900\ C$
Copper is deposited as
$Cu^{2 +} + 2e^{-} \longrightarrow Cu(s)$
2 moles of electrons or $2\ \times \ 96500\ C$ of current deposit copper = 63.56 g
900 C of current will deposit copper
$= \ \frac{63.56}{2\ \times 96500}\ \times \ 900$
= 0.296 g