Advertisements
Advertisements
प्रश्न
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, find the number of times the bacteria are increased in 12 hours.
Solution: Let x be the number of bacteria in the culture at time t.
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"`
∴ log x = kt + c
Initially, i.e. when t = 0, let x = x0
∴ log x0 = k × 0 + c
∴ c = `square`
∴ log x = kt + log x0
∴ log x - log x0 = kt
∴ `log ("x"/"x"_0)`= kt ......(1)
Since the number doubles in 4 hours, i.e. when t = 4,
x = 2x0
∴ `log ((2"x"_0)/"x"_0)` = 4k
∴ k = `square`
∴ equation (1) becomes, `log ("x"/"x"_0) = "t"/4` log 2
When t = 12, we get
`log ("x"/"x"_0) = 12/4` log 2 = 3 log 2
∴ `log ("x"/"x"_0)` = log 23
∴ `"x"/"x"_0 = 8`
∴ x = `square`
∴ number of bacteria will be 8 times the original number in 12 hours.
उत्तर
Let x be the number of bacteria in the culture at time t.
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
∴ `"dx"/"x"` = k dt
On integrating, we get
`int "dx"/"x" = "k" int "dt"`
∴ log x = kt + c
Initially, i.e. when t = 0, let x = x0
∴ log x0 = k × 0 + c
∴ c = log x0
∴ log x = kt + log x0
∴ log x - log x0 = kt
∴ `log ("x"/"x"_0)`= kt .....(1)
Since the number doubles in 4 hours, i.e. when t = 4,
x = 2x0
∴ `log ((2"x"_0)/"x"_0)` = 4k
∴ log 2 = 4k
∴ k = `1/4 log 2`
∴ equation (1) becomes, `log ("x"/"x"_0) = "t"/4` log 2
When t = 12, we get
`log ("x"/"x"_0) = 12/4` log 2 = 3 log 2
∴ `log ("x"/"x"_0)` = log 23
∴ `"x"/"x"_0 = 8`
∴ x = 8x0
∴ number of bacteria will be 8 times the original number in 12 hours.
APPEARS IN
संबंधित प्रश्न
If the population of a country doubles in 60 years, in how many years will it be triple (treble) under the assumption that the rate of increase is proportional to the number of inhabitants?
(Given log 2 = 0.6912, log 3 = 1.0986)
If a body cools from 80°C to 50°C at room temperature of 25°C in 30 minutes, find the temperature of the body after 1 hour.
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.
Find the population of a city at any time t, given that the rate of increase of population is proportional to the population at that instant and that in a period of 40 years, the population increased from 30,000 to 40,000.
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?
Choose the correct option from the given alternatives:
If the surrounding air is kept at 20° C and a body cools from 80° C to 70° C in 5 minutes, the temperature of the body after 15 minutes will be
Show that the general solution of differential equation `"dy"/"dx" + ("y"^2 + "y" + 1)/("x"^2 + "x" + 1) = 0` is given by (x + y + 1) = (1 - x - y - 2xy).
The rate of depreciation `(dV)/ dt` of a machine is inversely proportional to the square of t + 1, where V is the value of the machine t years after it was purchased. The initial value of the machine was ₹ 8,00,000 and its value decreased ₹1,00,000 in the first year. Find its value after 6 years.
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 `5/2` hours `("Given" sqrt(2) = 1.414)`
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
Integrating factor of `("d"y)/("d"x) + y/x` = x3 – 3 is ______
Find the population of city at any time t given that rate of increase of population is proportional to the population at that instant and that in a period of 40 years the population increased from 30000 to 40000.
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" prop "p"`
∴ `"dp"/"dt"` = kp, where k is a constant.
∴ `"dp"/"p"` = k dt
On integrating, we get
`int "dp"/"p" = "k" int "dt"`
∴ log p = kt + c
Initially, i.e. when t = 0, let p = 30000
∴ log 30000 = k × 0 + c
∴ c = `square`
∴ log p = kt + log 30000
∴ log p - log 30000 = kt
∴ `log("p"/30000)` = kt .....(1)
when t = 40, p = 40000
∴ `log (40000/30000) = 40"k"`
∴ k = `square`
∴ equation (1) becomes, `log ("p"/30000)` = `square`
∴ `log ("p"/30000) = "t"/40 log (4/3)`
∴ p = `square`
Bacteria increases at the rate proportional to the number of bacteria present. If the original number N doubles in 4 hours, find in how many hours the number of bacteria will be 16N.
Solution: Let x be the number of bacteria in the culture at time t.
Then the rate of increase of x is `("d"x)/"dt"` which is proportional to x.
∴ `("d"x)/"dt" ∝ x`
∴ `("d"x)/"dt"` = kx, where k is a constant
∴ `("d"x)/x` = kdt
On integrating, we get
`int ("d"x)/x = "k" int "dt"`
∴ log x = kt + c .....(1)
∴ x = aekt where a = ec
Initially, i.e.,when t = 0, let x = N
∴ N = aek(0)
∴ a = `square`
∴ a = N, x = Nekt ......(2)
When t = 4, x = 2N
From equation (2), 2N = Ne4k
∴ e4k = 2
∴ ek = `square`
Now we have to find out t, when x = 16N
From equation (2),
16N = Nekt
∴ 16 = ekt
∴ `"t"/4 = square` hours
Hence, number of bacteria will be 16N in `square` hours
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
The bacteria increases at the rate proportional to the number of bacteria present. If the original number 'N' doubles in 4 h, then the number of bacteria in 12 h will be ____________.
If r is the radius of spherical balloon at time t and the surface area of balloon changes at a constant rate K, then ______.
If a curve y = f(x) passes through the point (1, - 1) and satisfies the differential equation, y (1 + xy) dx = x dy, then `f(-1/2)` 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
