A plate of area 10 cm 2 is to be electroplated with Cu density 9g-cc to a thickness of 10 - m on both sides using a 12 V battery- Calculate the energy spent by the battery in this process- ECE of Cu is 3 - 10-17 kg-C

The correct option is A 7.2 kJ #160;m=p(vol.)=9 × #160;10 × #160;20 #160;× #160;10^ 4=18 #160;× #160;10^ 2 #160;gram m=zQ Q=mz=18 ...

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Standard XII

Physics

Mixed Combination of Cells

A plate of ar...

Question

A plate of area 10 cm $^{2}$ is to be electroplated with Cu (density 9g/cc) to a thickness of 10 $μ m$ on both sides using a 12 V battery. Calculate the energy spent by the battery in this process. ECE of Cu is $3 \times 10^{- 17}$ kg/C

1.8 kJ

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3.6 kJ

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5.4 kJ

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7.2 kJ

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none of these

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Solution

The correct option is A 7.2 kJ
$m = p (v o l .) = 9 \times 10 \times 20 \times 10^{- 4} = 18 \times 10^{- 2}$ gram

$m = z Q$

$Q = m z = \frac{18 \times 10^{2}}{3 \times 10^{4}} = 600 C$
Energy spent $E = V Q$
$= 12 \times 600 = 7.2 kJ$

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A plate of area 10 cm2 is to be electroplated with Cu (density 9g/cc) to a thickness of 10 μm on both sides using a 12 V battery. Calculate the energy spent by the battery in this process. ECE of Cu is 3×10−17 kg/C

The correct option is A 7.2 kJm=p(vol.)=9×10×20×10−4=18×10−2 gramm=zQQ=mz=18×1023×104=600 CEnergy spent E=VQ=12×600=7.2 kJ

A plate of area 10 cm $^{2}$ is to be electroplated with Cu (density 9g/cc) to a thickness of 10 $μ m$ on both sides using a 12 V battery. Calculate the energy spent by the battery in this process. ECE of Cu is $3 \times 10^{- 17}$ kg/C

The correct option is A 7.2 kJ
$m = p (v o l .) = 9 \times 10 \times 20 \times 10^{- 4} = 18 \times 10^{- 2}$ gram

$m = z Q$

$Q = m z = \frac{18 \times 10^{2}}{3 \times 10^{4}} = 600 C$
Energy spent $E = V Q$
$= 12 \times 600 = 7.2 kJ$