KVPY SX SB Exam 2013 Question Paper

Question : 108 of 120

Marks: +1, -0

The density and eq. wt. of a metal are 10.5 g

\cm^{ -3}

and 100, respectively. The time required for a current of 3 amp to deposit a 0.005 mm thick layer of the same metal on an area of 80

\cm^{2}

is closest to

Solution:

\w={\E×i×t}/96500

\w={100×3×t}/96500

d=m/\V⇒m=\dV

\w = 10.5 × 80 ×

5 × 10^{-4}

= 42 × 10^{-2}

t = 135 sec

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