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प्रश्न
The rate of disintegration of a radioactive element at any time t is proportional to its mass at that time. Find the time during which the original mass of 1.5 gm will disintegrate into its mass of 0.5 gm.
उत्तर
Let m be the mass of the radioactive element at time t.
Then the rate of disintegration is `"dm"/"dt"` which is proportional to m.
∴ `"dm"/"dt" prop "m"`
∴ `"dm"/"dt"` = - km, where k > 0
∴ `"dm"/"m"` = - k dt
On integrating, we get
`int 1/"m" "dm" = - "k" int "dt" + "c"`
∴ log m = - kt + c
Initially, i.e. when t = 0, m = 1.5
∴ log(1.5) = - k × 0 + c ∴ c = log`(3/2)`
∴ log m = - kt + log`(3/2)`
∴ log m - log`3/2` = - kt
∴ `log("2m"/3)` = - kt
When m = 0.5 = `1/2`, then
`log ((2 xx 1/2)/3) = - "kt"`
∴ `log (1/3)` = - kt
∴ log(3)-1 = - kt
∴ - log 3 = - kt
∴ t = `1/"k" log 3`
∴ the original mass will disintegrate to 0.5 gm when t = `1/"k" log 3`
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The rate of growth of population is proportional to the number present. If the population doubled in the last 25 years and the present population is 1,00,000, when will the city have population 4,00,000?
Let ‘p’ be the population at time ‘t’ years.
∴ `("dp")/"dt" prop "p"`
∴ Differential equation can be written as `("dp")/"dt" = "kp"`
where k is constant of proportionality.
∴ `("dp")/"p" = "k.dt"`
On integrating we get
`square` = kt + c ...(i)
(i) Where t = 0, p = 1,00,000
∴ from (i)
log 1,00,000 = k(0) + c
∴ c = `square`
∴ log `("p"/(1,00,000)) = "kt"` ...(ii)
(ii) When t = 25, p = 2,00,000
as population doubles in 25 years
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∴ t = `square ` years
