6. ⇒ (MHT CET 2021 20th September Evening Shift )

The population of a city increases at a rate proportional to the population at that time. If the population of the city increase from 20 lakhs to 40 lakhs in 30 years, then after another 15 years the population is

A. $10 \sqrt{2}$ lakhs

B. $40 \sqrt{2}$ lakh

C. $30 \sqrt{2}$ lakhs

D. None of these

Correct Option is (B)

We have $\frac{dP}{dt} \propto P \Rightarrow \frac{dP}{dt} = kP$

$∴ \int \frac{dP}{P} = \int kdt \Rightarrow \log P = kt + c$

From given data, we write

$\begin{aligned} \log 20 = k (0) + c \Rightarrow c = \log 20 \\ ∴ \log P = kt + \log 20 \end{aligned}$

Also $\log 40 = 30 k + \log 20$

$\begin{aligned} ∴ \log 40 - \log 20 = 30 k \Rightarrow k = \frac{1}{30} \log 2 \\ ∴ \log P = (\frac{\log 2}{30}) t + \log 20 \end{aligned}$

When $t = 30 + 15 = 45$ ,

$\begin{aligned} ∴ \log P & = (\frac{\log 2}{30}) (45) + \log 20 = (\log 2) (\frac{3}{2}) + \log 20 \\ = \log (2)^{\frac{3}{2}} + \log 20 = \log (2 \sqrt{2} \times 20) \\ ∴ P & = 40 \sqrt{2} lakhs \end{aligned}$

7. ⇒ (MHT CET 2021 20th September Morning Shift )

A differential equation for the temperature 'T' of a hot body as a function of time, when it is placed in a bath which is held at a constant temperature of 32 $^{\circ}$ F, is given by (where k is a constant of proportionality)

A. $\frac{dT}{dt} = kT - 32$

B. $\frac{dT}{dt} = kT + 32$

C. $\frac{dT}{dt} = - k (T - 32)$

D. $\frac{dT}{dt} = 32 kT$

Correct Option is (C)

The temperature T of the body will decrease with time.

The body is kept in a bath of temperature 32 $^{\circ}$ F.

$∴ \frac{dT}{dt} α - (T - 32) \Rightarrow \frac{dT}{dt} = - k (T - 32)$

8. ⇒ (MHT CET 2021 20th September Morning Shift )

An ice ball melts at the rate which is proportional to the amount of ice at that instant. Half the quantity of ice melts in 20 minutes, $x_{0}$ is the initial quantity of ice. If after 40 minutes the amount of ice left is ${Kx}_{0}$ , then $K =$

A. $\frac{1}{2}$

B. $\frac{1}{8}$

C. $\frac{1}{4}$

D. $\frac{1}{3}$

Correct Option is (C)

Given that the rate of melting ( $\frac{d x}{d t}$ ) is proportional to the amount of ice ( $x$ ) present at that moment, we can write the differential equation as :

$\frac{d x}{d t} = - k x$

(Here the negative sign indicates melting or decrease in quantity)

Let's solve this ordinary differential equation :

Separate the variables :

$\frac{d x}{x} = - k d t$

Now, integrate both sides :

$\int \frac{d x}{x} = - k \int d t$

This gives :

$\ln | x | = - k t + C$

Where $C$ is the constant of integration.

Given that half the quantity of ice melts in 20 minutes, we have :

$x = \frac{x_{0}}{2}$ when $t = 20$

Plug this into our equation to find $C$ :

$\ln (\frac{x_{0}}{2}) = - 20 k + C$

Now, using the initial condition that $x = x_{0}$ when $t = 0$ :

$\ln (x_{0}) = C$

From these two, we get :

$\ln (\frac{x_{0}}{2}) = \ln (x_{0}) - 20 k$

Which gives :

$k = \frac{\ln 2}{20}$

After 40 minutes, let $x = K x_{0}$ , plug into the equation :

$\ln (K x_{0}) = - \frac{\ln 2}{20} \times 40 + \ln (x_{0})$

$\ln (K x_{0}) = - 2 \ln 2 + \ln (x_{0})$

$\ln (K) = - 2 \ln 2$

$K = e^{- 2 \ln 2}$

$K = e^{\ln (2^{- 2})}$

$K = 2^{- 2}$

$K = \frac{1}{4}$

So, the correct answer is :

Option C :

$K = \frac{1}{4}$

⇒Defferential Equation