The rate of decay of mass of a certain substa | vedclass

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

Question

(C) Given that the rate of decay is proportional to the mass: $\frac{dm}{dt} = -Km$. Integrating both sides: $\int \frac{dm}{m} = \int -K dt$. This gi

Vedclass · Accepted Answer

$\left(\frac{1}{K}\right) \log \left(\frac{m_{0}}{m_{1}}\right)$

Vedclass · Answer

(C) Given that the rate of decay is proportional to the mass: $\frac{dm}{dt} = -Km$. Integrating both sides: $\int \frac{dm}{m} = \int -K dt$. This gives $\log m = -Kt + c$. At $t = 0$,$m = m_{0}$,so $\log m_{0} = c$. Substituting $c$ back: $\log m = -Kt + \log m_{0}$. Rearranging: $\log m - \log m_{0} = -Kt$,which implies $\log \left(\frac{m}{m_{0}}\right) = -Kt$. When $m = m_{1}$,we have $\log \left(\frac{m_{1}}{m_{0}}\right) = -Kt$. Solving for $t$: $t = -\frac{1}{K} \log \left(\frac{m_{1}}{m_{0}}\right) = \frac{1}{K} \log \left(\frac{m_{0}}{m_{1}}\right)$.

The rate of decay of mass of a certain substance at time $t$ is proportional to the mass at that instant. The time during which the original mass of $m_{0} \text{ gm}$ will be reduced to $m_{1} \text{ gm}$ is (where $K$ is the constant of proportionality).

Similar Questions

The solution of the differential equation $\frac{dy}{dx} + \frac{x}{y} \cdot \frac{x^2+y^2-1}{2(x^2+y^2)+1} = 0$ is

If the tangent at a point $P(x, y)$ of a curve is perpendicular to the line that joins the origin with the point $P$,then the curve is

The equation to the orthogonal trajectories of the system of parabolas $y = ax^2$ is

If the slope of the tangent to a curve at any point $(x, y)$ is given by $3x^2 + 2x + 5$ and the curve passes through the point $(0, 1)$,then the equation of the curve is:

The solution of the differential equation $x \frac{d^2y}{dx^2} = 1$,given that $y = 1$ and $\frac{dy}{dx} = 0$ when $x = 1$,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module