Explanation

(a) For applying Faraday’s second law of electrolysis the same charge should be passed through copper and silver voltmeters.
Charge passed to liberate copper
$= 9 \times 20 \times 60 = 10800$ C
Charge passed to liberate silver
$\ = 10 \times 15 \times 50 = 9000$ C
As 9000 C charge liberates 10.8 g of Ag.
So 10800 C charge liberates
$\ = \frac{10.8 \times 10800}{9000} = 10.8 \times 1.2$ g of Ag.
Using Faraday’s second law of electrolysis
$\ \frac{m_{Cu}}{m_{Ag}} = \frac{E_{Cu}}{E_{Ag}} = \frac{63.5/2}{108/1}$
or $m_{Cu} = \frac{63.5 \times 10.8 \times 1.2}{2 \times 108} = 3.81$ g.