RD Sharma Solutions for Class 12 Maths Exercise 22.11 Chapter 22 Differential Equations are given here. Each question in these materials is related to the topics included in Class 12 CBSE syllabus and is designed by our experts to provide the best solutions to the problems in which students face difficulties. Download RD Sharma Solutions for Class 12 Maths Chapter 22 in their respective links.
RD Sharma Solutions for Class 12 Maths Chapter 22 Exercise 11
Access RD Sharma Solutions for Class 12 Maths Chapter 22 Exercise 11
EXERCISE 22.11
Question. 1
Solution:
From the question, it is given that,
Initial radius of balloon = 1 unit
After 3 seconds radius of the balloon = 2 units
So, let us assume A be the surface area of the balloon,
(dA/dt) ∝ t
Then,
dA/dt = λt
d(4πr2)/dt = λt
8πr (dr/dt) = λt
Integrating on both sides we get,
8π ∫rdr = λ ∫t dt
8π (r2/2) = (λt2/2) + c
4πr2 = (λt2/2) + c …. … [equation (i)]
From question, r = 1 unit when t = 0,
4π (1)2 = 0 + c
4π = c
From equation (i) c = 4πr2 – (λt2/2)
Then, 4πr2 = (λt2/2) + 4π … [equation (ii)]
And also form question, given r = 2 units when t = 3 sec
4π(2)2 = (λ(3)2/2) + 4π
16π = (9/2) λ + 4π
(9/2) λ = 12π
By cross multiplication, we get,
λ = (24/9) π
λ = (8/3) π
So, now equation (ii) becomes,
4πr2 = (8π/6) t2 + 4π
4π (r2 – 1) = (4/3) πt2
By cross multiplication,
r2 – 1 = (1/3) t2
r2 = 1 + (1/3)t2
r = √(1 + (1/3)t2)
Question. 2
Solution:
From the question, it is given that,
Population grows at the rate of 5% per year.
So, let us assume the population after time t is p and the initial population is Po,
(dp/dt) = 5% × P
Then,
dP/dt = P/20
By cross multiplication, we get,
20 (dP/P) = dt
Integrating on both sides we get,
20 ∫(dP/P) = ∫ dt
20 log P = t + c …. … [equation (i)]
From question, P = Po unit when t = 0,
20 log (Po) = 0 + c
20log(P/Po) = c
Then, equation (i) becomes,
20 log (P) = t + 20 log (Po)
By cross multiplication, we get,
20 log (P/Po) = t
Let time is t, when P = 2Po,
Then, 20 log (2P/2Po) = t1
t1 = 20 log 2
Therefore, the time period required is 20 log 2 years.
Question. 3
Solution:
From the question, it is given that,
The present population is 1,00,000
Then, the population of the city doubled in the past 25 years,
So, let us assume P be the surface area of the balloon,
(dP/dt) ∝ P
Then,
dP/dt = λP
dP/dt = λ dt
Integrating on both sides, we get,
∫ dP/dt = λ ∫ dt
Log P = λt + c … [equation (i)]
From question, P = Po t when t = 0,
log (Po) = 0 + c
c = log (Po)
Then, equation (i) becomes,
log (P) = λt + log (Po)
log (P/Po) = λt … [equation (ii)]
And also form question, given P = 2Po when t = 25
log (2Po/Po) = 25λ
log 2 = 25λ
By cross multiplication, we get,
λ = log2/25
So, now equation (ii) becomes,
log (P/Po) = (log2/25)t
let us assume that t1 be the time to become population 5,00,000 from 1,00,000,
Then, log (5,00,000/1,00,000) = (log2/25) t1
By cross multiplication, we get,
t1 = 25 log 5/log 2
= 25(1.609)/(0.6931)
= 58
Therefore, the required time is 58 years.
Question. 4
Solution:
From the question, it is given that,
The bacteria count is 1,00,000
The number is increased by 10% in 2 hours.
So, let us assume C be the surface area of the balloon,
(dC/dt) ∝ C
Then,
dC/dt = λC
dC/dt = λ dt
Integrating on both sides we get,
∫ dC/dt = λ ∫ dt
Log C = λt + log k … [equation (i)]
From question, t = 0 when c = 1,00,000,
log (1,00,000) = λ × 0 + log k … [equation (ii)]
log (1,00,000) = log k … [equation (iii)]
And also form question, given t = 2, c = 1,00,000 + 1,00,000 × (10/100) = 110000
So, from equation (i) we have,
log 110000 = λ × 2 + log K … [equation (iv)]
Now, subtracting equation (ii) from equation (iv), we have,
Log 110000 – log 100000 = 2 λ
Then, log 11 × 10000 – log 10 × 10000 = 2 λ
Log ((11 × 10000)/(10 × 10000)) = 2λ
Log (11/10) = 2λ
So, λ = ½ log (11/10) … [equation (v)]
Now we need to find the time ‘t’ in which the count reaches 200000.
Then, substituting the values of λ and k from equations (iii) and (v) in equation (i),
Log 200000 = ½ log (11/10)t + log 100000
½ log (11/10)t = log 200000 – log 100000
½ log (11/10)t = log (200000/100000)
½ log(11/10)t = log 2
Therefore, the required time t = 2log2/log(11/10) hours.
Question. 5
Solution:
From the question, it is given that,
The interest is compounded continuously at 6% per annum
So, let us assume P is the principal,
(dP/dt) = Pr/100
Then,
dP/dt = (r/100) dt
Integrating on both sides, we get,
∫ dP/P = ∫ r/100 dt
Log P = (rt/100) + c … [equation (i)]
Let us assume Po is the initial principal at t = 0
log (Po) = 0 + C
C = log (Po)
Now, substitute the value of C in equation (i)
log (P) = rt/100 + log (Po)
log (P/Po) = rt/100
Now in case 1:
Po = 1000, t = 10 years and r = 6
log (P/1000) = (6 × 10)/100
log P – log 1000 = 0.6
log P = loge0.6 + log 1000
Taking log common in both terms,
log P = log (e0.6 + 1000)
log P = log (1.822 + 1000)
log P = log 1822
Then,
P = ₹ 1822
₹ 1000 will be ₹ 1822 after 10 years,
Now in case 2:
Let us assume t1 is the time to double ₹ 1000,
P = 2000, Po = 1000, r = 6%
log (P/Po) = rt/100
log (2000/1000) = 6t1/100
(100 log2)/6 = t1
(100 × 0.6931)/6 = t1
t1 = 11.55 years
Therefore, it will take approximately 12 years to double
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