When a current I was passed through an electrolyte solution for 40 minutes, a mass X g of a univalent metal was deposited at the cathode. What mass of the metal will be deposited when a current 2I is passed through the solution for 10 minutes?
X/4 g
X/2 g
2X g
4X g
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IN THIS CASES OF ELECTROLYSIS
AT THE FIRST ELECTROLYSIS PROCESS
CURRENT= I
TIME = 40MIN×60S=2400SECS
Q=IT
Q= I × 2400
Q= 2400I
M=Xg
AT SECOND ELECTROLYSIS PROCESS
THE CURRENT HAS BEEN DOUBLED AND THE TIME HAS BEEN REDUCED BY ¼
CURRENT= 2I
TIME= 10MIN×60=600SECS
Q=IT
Q= 2I × 600
Q= 1200I
TO GET THE MASS DEPOSITED WE USE FARADAY'S FIRST LAW OF ELECTROLYSIS
M~Q
M= ZQ
WE CONSIDER Z(ELECTROCHEMICAL EQUIVALENCE TO BE 1)
FOR THE FIRST CASE
M=1 × 2400I
M(i.e X) = 2400
Xg= 2400Ig
FOR THE SECOND CASE
M2= ZQ
M2= 1 × 1200I
M2= 1200Ig
MASS DEPOSITED= MASS OF FINAL/INITIAL
M2= 1200I Xg / 2400Ig
I cancels I
M2 = 1200Xg/2400
M2=X/2 
OPTION B
pls can someone clarify step by step me on this question i can't fully understand what my school solved 
I think d ans shud b c not b
from this, Y/(X g) = 1200 I/2400 I
Y will be =2x not x/2
