Correct Answer - A
Anode `{{:(2H_(2)SO_(4)toH_(2)S_(2)O_(8)+2H^(+)+2e^(-)),(2H_(2)OtoO_(2)+4H^(+)+4e^(-)):}}`
`underline(" Cathode")" "{2H_(2)OtoH_(2)+2OH^(-)-2e^(-)}3)`.
Net: `2H_(2)SO_(4)+8H_(2)OtoH_(2)S_(2)O_(8)+O_(2)+3H_(2)+6H^(+)+6OH^(-)`
Hence ratio of `n_(O_(2))` and `n_(H_(2))` is `1:3`
Correct Answer - (i) year ended 31st March, 2016: Loss on issue of Debentues - ₹ 1,00,000 to be written off from
Statement of Profit and Loss in the year...