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If 10% of a substance decays in 10 days, then approximate percentage of substance left after 24 days

Answer»

78
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75
80

Solution :`N_(1)/N_(0)=(1/2)^((10)/(T)) or (90)/(100) =(1/2)^((10)/T)`
`(X)/(100)=(1/2)^(24//T)`
TAKING logs, we get
`log"" (9)/(100)=(10)/T log"" 1/2 & log ""(X)/(100)=(24)/(T) log""1/2`
Dividin, we get
`(log X)/(log 100) "/" log (90)/(100) =((24)/(T) log 1//2)/((10)/(T) log 1//2)=2.4`
Then, X=77.8


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