When ${\rm{1 }}\,{\rm{A}}$ current flows for ${\rm{1}}\,\,{\rm{min}}$ through a silver voltmeter, it deposits ${\rm{0}}{\rm{.067}}\,\,{\rm{g}}$ of silver on the cathode, then how much charge will flow to deposit ${\rm{108}}\,\,{\rm{g}}$ of silver?

$9.67\,\, \times \,\,{10^4}\,\,{\rm{C}}\,{\rm{g}}_{\,{\rm{eq}}}^{ - 1}$

$10.6\,\, \times \,\,{10^4}\,\,{\rm{C}}\,{\rm{g}}_{\,{\rm{eq}}}^{ - 1}$

$8.7\,\, \times \,\,{10^4}\,\,{\rm{C}}\,{\rm{g}}_{\,{\rm{eq}}}^{ - 1}$

$4.3\,\, \times \,\,{10^4}\,\,{\rm{C}}\,{\rm{g}}_{\,{\rm{eq}}}^{ - 1}$

JEE PHYSICS MCQ: CURRENT ELECTRICITY

1 Medium

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Application +4 | -1

When

\mathrm{1 }\,\mathrm{A}

current flows for

\mathrm{1}\,\,\mathrm{min}

through a silver voltmeter, it deposits

\mathrm{0}\mathrm{.067}\,\,\mathrm{g}

of silver on the cathode, then how much charge will flow to deposit

\mathrm{108}\,\,\mathrm{g}

of silver?

View answer and explanation

Correct option: B

9.67\,\, \times \,\,{10^4}\,\,\mathrm{C}\,\mathrm{g}_{\,\mathrm{eq}}^{ - 1}

Option B (

9.67 \times {10^4}\mathrm{~C}\mathrm{~g}_{\mathrm{~eq}}^{ - 1}

) is correct because it is calculated using Faraday's law of electrolysis and the given mass of silver deposited.