A spherical rain drop evaporates at a rate pr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $V$ be the volume and $r$ be the radius of the spherical rain drop. The volume is given by $V = \frac{4}{3}\pi r^3$. The rate of change of vol

Vedclass · Accepted Answer

$\frac{dr}{dt} + K = 0$

Vedclass · Answer

(A) Let $V$ be the volume and $r$ be the radius of the spherical rain drop. The volume is given by $V = \frac{4}{3}\pi r^3$. The rate of change of volume is $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$. According to the problem,the rate of evaporation is proportional to the surface area $S = 4\pi r^2$. Since it is evaporating,the rate of change of volume is negative: $\frac{dV}{dt} = -K(4\pi r^2)$,where $K > 0$. Equating the two expressions for $\frac{dV}{dt}$: $4\pi r^2 \frac{dr}{dt} = -K(4\pi r^2)$. Dividing both sides by $4\pi r^2$ (assuming $r 
eq 0$): $\frac{dr}{dt} = -K$,which can be written as $\frac{dr}{dt} + K = 0$.

$A$ spherical rain drop evaporates at a rate proportional to its surface area. The differential equation corresponding to the rate of change of the radius of the rain drop,if the constant of proportionality is $K > 0$,is

Similar Questions

If the population grows at the rate of $8 \%$ per year,then the time taken for the population to be doubled,is (Given $\log 2 = 0 \cdot 6912$)

The rate of disintegration of a radioactive element at time $t$ is proportional to its mass at that time. Then the time during which the original mass of $6 \text{ gm}$ will disintegrate into its mass of $3 \text{ gm}$ is proportional to

The slope of the tangent at any point $(x, y)$ on the curve is equal to the product of the coordinates of that point. If the equation of the normal to the curve at the point $(\sqrt{2}, e)$ is $ax + by = 1$,then $\frac{b}{a} =$

Any curve passes through the point $(3, -4)$. If the slope of the tangent to the curve at any point $(x, y)$ is $\frac{2y}{x}$,then the equation of the curve is . . . . . . .

If surrounding air is kept at $20^{\circ} C$ and a body cools from $80^{\circ} C$ to $70^{\circ} C$ in $5$ minutes,then the temperature of the body after $15$ minutes will be: (in $^{\circ} C$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module