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प्रश्न
The rate of growth of bacteria is proportional to the number present. If initially, there were 1000 bacteria and the number doubles in 1 hour, find the number of bacteria after `2 1/2` hours.
[Take `sqrt2 = 1.414`]
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उत्तर
Let x be the number of bacteria at time t.
Then the rate of increase is `"dx"/"dt"` which is proportional to x.
∴ `"dx"/"dt" prop "x"`
∴ `"dx"/"dt"` = kx, where k is a constant
∴ `"dx"/"x" = "k dt"`
On integrating, we get
`int "dx"/"x" = "k" int "dt" + "c"`
∴ log x = kt + c
Initially, i.e. when t = 0, x = 1000
∴ log 1000 = k × 0 + c ∴ c = log 1000
∴ log x = ly + log 1000
∴ log x - log 1000 = kt
∴ log`("x"/1000) = "kt"` ...(1)
Now, when t = 1, x = 2 × 1000 = 2000
∴ `log (2000/1000)` = k ∴ k = log 2
∴ (1) becomes, log`("x"/1000) = t log 2
If t = `2 1/2 = 5/2,` then
log`("x"/1000) = 5/2 log 2 = log (2)^(5/2)`
∴ `("x"/1000) = (2)^(5/2) = 4sqrt2 = 4 xx 1.414 = 5.656`
∴ x = 5.656 × 1000 = 5656
∴ number of bacteria after `2 1/2`hours = 5656.
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Then the rate of increase of x is `"dx"/"dt"` which is proportional to x.
∴ `"dx"/"dt" ∝ "x"`
∴ `"dx"/"dt"` = kx, where k is a constant
∴ `square`
On integrating, we get
`int "dx"/"x" = "k" int "dt"`
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Initially, i.e. when t = 0, let x = x0
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