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प्रश्न
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
विकल्प
`5 2/3` grams
`5 1/3` grams
5.1 grams
5 grams
उत्तर
`5 1/3` grams
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संबंधित 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.
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 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`]
The rate of decay of certain substances is directly proportional to the amount present at that instant. Initially, there is 25 gm of certain substance and two hours later it is found that 9 gm are left. Find the amount left after one more hour.
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.
A body cools according to Newton’s law from 100° C to 60° C in 20 minutes. The temperature of the surrounding being 20° C. How long will it take to cool down to 30° C?
Assume that a spherical raindrop evaporates at a rate proportional to its surface area. If its radius originally is 3 mm and 1 hour later has been reduced to 2 mm, find an expression for the radius of the raindrop at any time t.
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
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 normal lines to a given curve at each point (x, y) on the curve pass through (2, 0). The curve passes through (2, 3). Find the equation of the curve.
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.
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 solution of `("d"y)/("d"x) + x^2/y^2` = 0 is
The integrating factor of the differential equation `("d"y)/("d"x) - y` = x is ______
The rate of growth of population is proportional to the number present. If the population doubled in the last 25 years and present population is 1 lac., when will the city have population 4,00,000?
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 = 100000
∴ log 100000 = k × 0 + c
∴ c = `square`
∴ log p = kt + log 100000
∴ log p – log 100000 = kt
∴ `log ("P"/100000)` = kt ......(i)
Since the number doubled in 25 years, i.e., when t = 25, p = 200000
∴ `log (200000/100000)` = 25k
∴ k = `square`
∴ equation (i) becomes, `log("p"/100000) = square`
When p = 400000, then find t.
∴ `log(400000/100000) = "t"/25 log 2`
∴ `log 4 = "t"/25 log 2`
∴ t = `25 (log 4)/(log 2)`
∴ t = `square` years
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 rate of decay of certain substance is directly proportional to the amount present at that instant. Initially, there are 27 gm of certain substance and 3 h later it is found that 8 gm are left, then the amount left after one more hour is ______.
The equation of tangent at P(- 4, - 4) on the curve x2 = - 4y is ______.
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 the lengths of the transverse axis and the latus rectum of a hyperbola are 6 and `8/3` respectively, then the equation of the hyperbola is ______.
If r is the radius of spherical balloon at time t and the surface area of balloon changes at a constant rate K, then ______.
The population of a town increases at a rate proportional to the population at that time. If the population increases from 26,000 to 39,000 in 50 years, then the population in another 25 years will be ______ `(sqrt(3/2) = 1.225)`
The length of the perimeter of a sector of a circle is 24 cm, the maximum area of the sector is ______.
The rate of growth of bacteria is proportional to the number present. If initially, there are 1000 bacteria and the number doubles in 1 hour, the number of bacteria after `21/2` hours will be ______. `(sqrt(2) = 1.414)`
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.
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 N be the number of bacteria present at time ‘t’.
Since the rate of increase of N is proportional to N, the differential equation can be written as –
`(dN)/dt αN`
∴ `(dN)/dt` = KN, where K is constant of proportionality
∴ `(dN)/N` = k . dt
∴ `int 1/N dN = K int 1 . dt`
∴ log N = `square` + C ...(1)
When t = 0, N = N0 where N0 is initial number of bacteria.
∴ log N0 = K × 0 + C
∴ C = log N0
Also when t = 4, N = 2N0
∴ log (2 N0) = K . 4 + `square` ...[From (1)]
∴ `log((2N_0)/N_0)` = 4K,
∴ log 2 = 4K
∴ K = `square` ...(2)
Now N = ? when t = 12
From (1) and (2)
log N = `1/4 log 2 . (12) + log N_0`
log N – log N0 = 3 log 2
∴ `log(N_0/N_0)` = `square`
∴ N = 8 N0
∴ Bacteria are increased 8 times in 12 hours.
