11. ⇒ (MHT CET 2023 9th May Morning Shift )

A spherical raindrop evaporates at a rate proportional to its surface area. If originally its radius is $3 mm$ and 1 hour later it reduces to $2 mm$ , then the expression for the radius $R$ of the raindrop at any time $t$ is

A. $6 R = t + 2$

B. $R (t + 2) = 6$

C. $R = 6 (t + 2)$

D. $6 R = t$

Correct Option is (B)

According to the given conditions, when $t = 0, R = 3$ and when $t = 1, R = 2$

This condition is satisfied by only option (B)

12. ⇒ (MHT CET 2021 21th September Morning Shift )

A body at an unknown temperature is placed in a room which is held at a constant temperature of $30^{\circ} F$ . If after 10 minutes the temperature of the body is $0^{\circ} F$ and after 20 minutes the temperature of the body is $15^{\circ} F$ , then the expression for the temperature of the body at any time $t$ is

A. $T = - 60 e^{- 0.069 t} - 30$

B. $T = - 60 e^{- 0.03010 t} + 30$

C. $T = 60 e^{- 0.069 t} + 30$

D. $T = 60 e^{- 0.069 t} - 30$

Correct Option is (B)

$\begin{aligned} We have \frac{dT}{dt} \propto (30 - T) \\ ∴ \frac{dT}{dt} = - K (30 - T) \Rightarrow \int \frac{dT}{30 - T} = \int - Kt \\ ∴ \log | 30 - T | = - kt + c .....(1) \end{aligned}$

From given data, we write

$\log | 30 - 0 | = - 10 K + c$ ..... (2)

$\log | 30 - 0 | = - 20 K + c$ ..... (3)

Solving (2) and (3), we get

$\log (\frac{30}{15}) = 10 K \Rightarrow K = \frac{1}{10} \log 2$

Substituting value of $K$ in eq. (2), we get

$\log 30 = (- 10) (\frac{\log 2}{10}) + c \Rightarrow c = \log 60$

Thus eq. (1) becomes

$\begin{aligned} \log | 30 - T | = \frac{- \log 2}{10} t + \log 60 \\ ∴ \log | \frac{30 - T}{60} | = \frac{- 0.3010}{10} t = - 0.03010 t \\ ∴ \frac{30 - T}{60} = e^{- 0.03010 t} ∴ T = - 60 e^{- 0.03010 t} + 30 \end{aligned}$

13. ⇒ (MHT CET 2021 21th September Morning Shift )

A stone is thrown into a quite lake and the waves formed move in circles. If the radius of a circular wave increases at the rate of 4 cm/sec, then the rate of increase in its area, at the instant when its radius is 10 cm, is _________ cm $^{2}$ /sec.

A. 80 $π$

B. 10 $π$

C. 8 $π$

D. 40 $π$

Correct Option is (A)

We have $\frac{dr}{dt} = 4 cm / \sec$ and $A = π r^{2}$

$\begin{aligned} ∴ \frac{d A}{d t} = 2 π r \frac{d r}{d t} \\ ∴ \frac{d A}{d t} = 2 π (10) (4) = 80 π \end{aligned}$

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

The surface area of a spherical balloon is increasing at the rate $2 {cm}^{2} / \sec$ . Then rate of increase in the volume of the balloon is , when the radius of the balloon is $6 cm$ .

A. $4 {cm}^{3} / \sec$ .

B. $16 {cm}^{3} / \sec$ .

C. $36 {cm}^{3} / \sec$ .

D. $6 {cm}^{3} / \sec$ .

Correct Option is (D)

We have $\frac{dA}{dt} = 2$ and $A = 4 π r^{2}$

$\begin{aligned} ∴ \frac{dA}{dt} = 8 π r \frac{dr}{dt} \Rightarrow 2 = 8 π (6) \frac{dr}{dt} \\ ∴ \frac{dr}{dt} = \frac{1}{24 π} \end{aligned}$

Here $V = \frac{4}{3} π r^{3}$

$\begin{aligned} ∴ \frac{dV}{dt} & = 4 π r^{2} \frac{dr}{dt} \\ = 4 π (6)^{2} (\frac{1}{24 π}) = 6 {cm}^{3} / \sec \end{aligned}$

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

Water is being poured at the rate of 36 m $^{3}$ /min. into a cylindrical vessel, whose circular base is of radius 3 m. Then the wate level in the cylinder is rising at the rate of

A. $\frac{4}{π}$ m/min

B. $4 π$ m/min

C. $\frac{1}{4 π}$ m/min

D. $\frac{2}{π}$ m/min

Correct Option is (A)

$\begin{aligned} Volume of cylinder = π r^{2} h \\ ∴ \frac{dv}{dt} = π r^{2} \frac{dh}{dt} \\ ∴ 36 = π (3)^{2} \frac{dh}{dt} \Rightarrow \frac{dh}{dt} = \frac{4}{π} m / \min \end{aligned}$

⇒Application Of Derivative