Advertisements
Advertisements
प्रश्न
The population of a town increases at a rate proportional to the population at that time. If the population increases from 40 thousands to 60 thousands in 40 years, what will be the population in another 20 years?
(Given: `sqrt(3/2)= 1.2247)`
उत्तर
Step 1: Use the Exponential Growth Model
P(t) = P0ekt
where:
- P = Population at time t
- P0 = Initial population
- k = Growth constant
- t = Time in years
Step 2: Find k Using Given Data
We are given:
- P0 = 40,000
- P(40) = 60,000
- t = 40 years
60,000 = 40,000e40k
`(60,000)/(40,000) = e^(40k)`
1.5 = e40k
Taking the natural logarithm:
ln (1.5) = 40k
`k = ln(1.5)/40`
Using ln (1.5) = 0.4055
`k = 0.4055/40 = 0.01014`
Step 3: Find Population After 60 Years
Now, we need to find P(60)
P(60) = 40,000e60k
P(60) = 40,000e60×0.01014
P(60) = 40,000e0.6084
P(60) = 40,000 × 1.837
P(60) ≈ 73,480
APPEARS IN
संबंधित 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.
A right circular cone has height 9 cm and radius of the base 5 cm. It is inverted and water is poured into it. If at any instant the water level rises at the rate of `(pi/"A")`cm/sec, where A is the area of the water surface A at that instant, show that the vessel will be full in 75 seconds.
The rate of growth of the population of a city at any time t is proportional to the size of the population. For a certain city, it is found that the constant of proportionality is 0.04. Find the population of the city after 25 years, if the initial population is 10,000. [Take e = 2.7182]
Choose the correct option from the given alternatives:
The decay rate of certain substances is directly proportional to the amount present at that instant. Initially there are 27 grams of substance and 3 hours later it is found that 8 grams left. The amount left after one more hour is
A person’s assets start reducing in such a way that the rate of reduction of assets is proportional to the square root of the assets existing at that moment. If the assets at the beginning ax ‘ 10 lakhs and they dwindle down to ‘ 10,000 after 2 years, show that the person will be bankrupt in `2 2/9` years from the start.
If the population of a town increases at a rate proportional to the population at that time. If the population increases from 40 thousand to 60 thousand in 40 years, what will be the population in another 20 years? `("Given" sqrt(3/2) = 1.2247)`
Choose the correct alternative:
Bacterial increases at the rate proportional to the number present. If original number M doubles in 3 hours, then number of bacteria will be 4M in
Choose the correct alternative:
The integrating factor of `("d"y)/("d"x) + y` = e–x is
Choose the correct alternative:
The integrating factor of `("d"^2y)/("d"x^2) - y` = ex, is e–x, then its solution is
Choose the correct alternative:
The solution of `dy/dx` = 1 is ______.
The population of city doubles in 80 years, in how many years will it be triple when the rate of increase is proportional to the number of inhabitants. `("Given" log3/log2 = 1.5894)`
Solution: Let p be the population at time t.
Then the rate of increase of p is `"dp"/"dt"` which is proportional to p.
∴ `"dp"/"dt" ∝ "p"`
∴ `"dp"/"dt"` = kp, where k is a constant
∴ `"dp"/"p"` = kdt
On integrating, we get
`int "dp"/"p" = "k" int "dt"`
∴ log p = kt + c
Initially, i.e., when t = 0, let p = N
∴ log N = k × 0 + c
∴ c = `square`
When t = 80, p = 2N
∴ log 2N = 80k + log N
∴ log 2N – log N = 80k
∴ `log ((2"N")/"N")` = 80k
∴ log (2) = 80k
∴ k = `square`
∴ p = 3N, then t = ?
∴ log p = `log2/80 "t" + log "N"`
∴ log 3N – log N = `square`
∴ t = `square` = `square` years
If the population grows at the rate of 8% per year, then the time taken for the population to be doubled, is (Given log 2 = 0.6912).
Let the population of rabbits surviving at a time t be governed by the differential equation `(dp(t))/dt = 1/2p(t) - 200`. If p(0) = 100, then p(t) equals ______
The rate of increase of bacteria in a certain culture is proportional to the number present. If it doubles in 7 hours, then in 35 hours its number would be ______.
In a certain culture of bacteria, the rate of increase is proportional to the number present. If it is found that the number doubles in 4 hours, complete the following activity to find the number of times the bacteria are increased in 12 hours.
The rate of disintegration of a radioactive element at time t is proportional to its mass at that time. The original mass of 800 gm will disintegrate into its mass of 400 gm after 5 days. Find the mass remaining after 30 days.
Solution: If x is the amount of material present at time t then `dx/dt = square`, where k is constant of proportionality.
`int dx/x = square + c`
∴ logx = `square`
x = `square` = `square`.ec
∴ x = `square`.a where a = ec
At t = 0, x = 800
∴ a = `square`
At t = 5, x = 400
∴ e–5k = `square`
Now when t = 30
x = `square` × `square` = 800 × (e–5k)6 = 800 × `square` = `square`.
The mass remaining after 30 days will be `square` mg.
If `(dy)/(dx)` = y + 3 > 0 and y = (0) = 2, then y (in 2) is equal to ______.
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
∴ from (ii) log2 = 25k
∴ k = `square`
∴ log`("p"/(1,00,000)) = (1/25log2).t`
(iii) ∴ when p = 4,00,000
`log ((4,00,000)/(1,00,000)) = (1/25log2).t`
∴ `log 4 = (1/25 log2).t`
∴ t = `square ` years