1. ⇒ (MHT CET 2023 12th May Evening Shift )

Water flows from the base of rectangular tank, of depth 16 meters. The rate of flow of the water is proportional to the square root of depth at any time $t$ . If depth is $4 m$ when $t = 2$ hours, then after 3.5 hours the depth (in meters) is

A. 0

B. 0.25

C. 0.5

D. 3

Correct Option is (B)

MHT CET 2023 12th May Evening Shift Mathematics - Differential Equations Question 1 English Explanation

Given that $\frac{d x}{dt} \propto \sqrt{x}$

$∴ \frac{d x}{dt} = a \sqrt{x}$ , for real number a

$∴ \int \frac{d x}{\sqrt{x}} = \int a d t$

$∴ 2 \sqrt{x} = at + c$ .... (i)

When $t = 0, x = 16$

$∴$ (i) $\Rightarrow c = 8$

$∴$ (i) becomes $2 \sqrt{x} =$ at +8 .... (ii)

When $t = 2, x = 4$

$∴$ (ii) $\Rightarrow a = - 2$

$∴$ (ii) becomes $2 \sqrt{x} = - 2 t + 8$ .... (iii)

$∴$ when $t = 3.5$

(iii) $\Rightarrow x = 0.25 m$

2. ⇒ (MHT CET 2023 12th May Morning Shift )

The decay rate of radio active material at any time $t$ is proportional to its mass at that time. The mass is 27 grams when $t = 0$ . After three hours it was found that 8 grams are left. Then the substance left after one more hour is

A. $\frac{27}{8}$ grams

B. $\frac{81}{4}$ grams

C. $\frac{16}{3}$ grams

D. $\frac{16}{9}$ grams

Correct Option is (C)

Let ' $x$ ' be the mass of the material at time ' $t$ '.

$∴ \frac{d x}{dt} = - k x, (- ve$ sign indicates decay. $)$

$∴ \int \frac{d x}{x} = - k \int dt$

$∴ \log | x | = - kt + c$

When $t = 0, x = 27$

$∴ c = \log 27$

$∴ \log | x | = - k t + \log 27$

When $t = 3, x = 8$

$∴ k = \log (\frac{3}{2})$

When $t = 4$ , we get

$\begin{array}{ll} \log | x | = - 4 \log (\frac{3}{2}) + \log 27 \\ ∴ & \log | x | = \log (\frac{16}{3}) \\ ∴ & x = \frac{16}{3} grams \end{array}$

3. ⇒ (MHT CET 2023 10th May Evening Shift )

The differential equation of all circles, passing through the origin and having their centres on the $X$ -axis, is

A. $y^{2} = x^{2} + x y \frac{d y}{d x}$

B. $x^{2} = y^{2} + 2 x y \frac{d y}{d x}$

C. $y^{2} = x^{2} + 2 x y \frac{d y}{d x}$

D. $x^{2} = y^{2} - x y \frac{d y}{d x}$

Correct Option is (C)

The system of circles which passes through origin and whose centre lies on $X$ -axis is

$x^{2} + y^{2} - 2 b x = 0$ .... (i)

Differentiating w.r.t $x$ , we get

$2 x + 2 y \frac{d y}{d x} = 2 b$ ..... (ii)

Substituting (ii) in (i), we get

$\begin{aligned} x^{2} + y^{2} - 2 x^{2} - 2 x y \frac{d y}{d x} = 0 \\ \Rightarrow y^{2} - x^{2} - 2 x y \frac{d y}{d x} = 0 \\ \Rightarrow y^{2} = x^{2} + 2 x y \frac{d y}{d x} \end{aligned}$

4. ⇒ (MHT CET 2023 10th May Morning Shift )

The population $P = P (t)$ at time $t$ of certain species follows the differential equation $\frac{d P}{d t} = 0.5 P - 450$ . If $P (0) = 850$ , then the time at which population becomes zero is

A. $2 \log 18$

B. $\log 9$

C. $\frac{1}{2} \log 18$

D. $\log 18$

Correct Option is (A)

$\begin{aligned} Given differential equation is \\ \frac{dP}{dt} = 0.5 P - 450 \\ = \frac{P}{2} - \frac{900}{2} \\ ∴ \frac{dP}{dt} = \frac{P - 900}{2} \\ ∴ \frac{2 dP}{P - 900} = dt \\ Integrating on both sides, we get \\ 2 \log | P - 900 | = t + c \\ P (0) = 850 i.e., P = 850 when t = 0 \\ ∴ c = 2 \log 50 \\ ∴ 2 \log | P - 900 | = t + 2 \log 50 \\ When P = 0, \\ 2 \log 900 = t + 2 \log 50 \\ \Rightarrow t = 2 (\log 900 - \log 50) \\ = 2 \log \frac{900}{50} = 2 \log 18 \end{aligned}$

5. ⇒ (MHT CET 2021 21th September Evening Shift )

If the half life period of a substance is 5 years, then the total amount of the substance left after 15 years, when initial amount is 64 gms is

A. 8 gms

B. 16 gms

C. 2 gms

D. 32 gms

Correct Option is (A)

Half life period $= 5$ years.

Initial quantity of substance $= 64$ grams

$∴$ Quantity left after 5 years $= \frac{64}{2} = 32$ grams

$∴$ Quantity left after 10 years $= \frac{32}{2} = 16$ grams

$∴$ Quantity left after 15 years $= \frac{16}{2} = 8$ grams

⇒Defferential Equation