**118.8cm**^{2} surface of the copper cathode of a voltameter is to be coated with 10^{-6}...

^{2}surface of the copper cathode of a voltameter is to be coated with 10

^{-6}...

### Question

118.8cm^{2} surface of the copper cathode of a voltameter is to be coated with 10^{-6}m thick copper of density 9 x 10^{3}kgm^{-3}. How long will the process run with 10A constant current?

[3.3 x 10^{-7}kgC^{-1}]

### Options

A) 5.4 min

B) 10.8 min

C) 15.0 min

D) 20.0 min

Related Lesson: Drift Velocity | Electric Current, Resistance, and Ohm's Law

The correct answer is A.

### Explanation:

From faradays law, M = itz

But mass, M = volume x density

and volume = cross sectional area x thickness

M = 118.8cm

∴ 118.8cm

∴ (118.8 x 9) x 10

But mass, M = volume x density

and volume = cross sectional area x thickness

M = 118.8cm

^{2}x 10^{-6}x 9 x 10^{3}∴ 118.8cm

^{2}x 10^{-6}x 9 x 10^{3}= 10 x t x 3.3 x 10^{-7}∴ (118.8 x 9) x 10

^{-7}= 10 x t x 3.3 x 10^{-7}∴ t = \(\frac{118.8 \times 9}{ 3.3}\)

t = 324 seconds

Convert to minutes by dividing by 60

t = \(\frac{324}{60}\)

t = 5.4 sec

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From faradays law, M = itz

But mass, M = volume x density

and volume = cross sectional area x thickness

M = 118.8cm

^{2}x 10^{-6}x 9 x 10^{3}∴ 118.8cm

^{2}x 10^{-6}x 9 x 10^{3}= 10 x t x 3.3 x 10^{-7}∴ (118.8 x 9) x 10

^{-7}= 10 x t x 3.3 x 10^{-7}∴ t = \(\frac{118.8 \times 9}{ 3.3}\)

t = 324 seconds

Convert to minutes by dividing by 60

t = \(\frac{324}{60}\)

t = 5.4 sec