In the process of electroplating, m g of silver is deposited when 4 ampere of current flows for 2 minutes. The amount (in g) of silver deposited by 6 ampere of current flowing for 40 seconds will be
Question
In the process of electroplating, m g of silver is deposited when 4 ampere of current flows for 2 minutes. The amount (in g) of silver deposited by 6 ampere of current flowing for 40 seconds will be
Solution 1
To find the amount of silver deposited by 6 ampere of current flowing for 40 seconds, we can use the formula:
Amount of silver deposited = (Current × Time × Mass of silver deposited) / (Current × Time)
Given that 4 ampere of current flows for 2 minutes and deposits m grams of silver, we can write the equation as:
m = (4 × 2 × Mass of silver deposited) / (4 × 2)
Simplifying the equation, we get:
m = Mass of silver deposited
Therefore, the amount of silver deposited by 6 ampere of current flowing for 40 seconds will also be m grams.
Solution 2
To find the amount of silver deposited by 6 ampere of current flowing for 40 seconds, we can use the formula:
Amount of silver deposited = (Current × Time × Mass of silver deposited) / (Current × Time)
Given that 4 ampere of current flows for 2 minutes and deposits m grams of silver, we can write the equation as:
m = (4 × 2 × Mass of silver deposited) / (4 × 2)
Simplifying the equation, we get:
m = Mass of silver deposited
Therefore, the amount of silver deposited by 6 ampere of current flowing for 40 seconds will also be m grams.
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