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Question
Radium decomposes at the rate proportional to the amount present at any time. If p percent of the amount disappears in one year, what percent of the amount of radium will be left after 2 years?
Solution
Let x be the amount of the radium at time t.
Then the rate of decomposition is `"dx"/"dt"` which is proportional to x.
∴ `"dx"/"dt" prop "x"`
∴ `"dx"/"dt"` = - kx, where k > 0
∴ `1/"x" "dx"` = - k dt
On integrating, we get
`int 1/"x" "dx" = - "k" int "dt"`
∴ log x = - kt + c
Let the original amount be x0, i.e. x = x0, when t = 0.
∴ log x0 = - k × 0 + c ∴ c = log x0
∴ log x = - kt + log x0
∴ log x - log x0 = - kt
∴ `log ("x"/"x"_0)` = - kt .....(1)
But p% of the amount disappears in one year,
∴ when t = 1, x= x0 - p % of x0 , i.e. x = x0 - `"px"_0/100`
∴ `log (("x"_0 - "px"_0/100)/"x"_0) = - "k" xx 1`
∴ k = `- log (1 - "p"/100) = - log ((100 - p)/100)`
∴ (1) becomes, `log ("x"/"x"_0) = "t" log ((100 - "p")/100)`
When t = 2, then
`log ("x"/"x"_0) = 2 log ((100 - "p")/100) = log ((100 - "p")/100)^2`
∴ `"x"/"x"_0 = ((100 - "p")/100)^2`
∴ x = `((100 - "p")/100)^2 "x"_0 = (1 - "p"/100)^2 "x"_0`
∴ % left after 2 years = `(100 xx (1 - "p"/100)^3 "x"_0)/"x"_0`
∴ `= 100 (1 - "p"/100)^2 = [10 (1 - "p"/100)]^2`
`= (10 - "p"/10)^2`
Hence, `= (10 - "p"/10)^2`% of the amount will be left after 2 years.
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Solution:
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∴ Bacteria are increased 8 times in 12 hours.
