An ice ball melts at a rate proportional to t | vedclass

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

Question

(C) Let $x(t)$ be the amount of ice at time $t$. The rate of melting is proportional to the amount of ice,so $\frac{dx}{dt} = -cx$,where $c > 0$. Inte

Vedclass · Accepted Answer

$1/4$

Vedclass · Answer

(C) Let $x(t)$ be the amount of ice at time $t$. The rate of melting is proportional to the amount of ice,so $\frac{dx}{dt} = -cx$,where $c > 0$. Integrating this,we get $x(t) = x_0 e^{-ct}$. Given that half the ice melts in $15 	ext{ minutes}$,at $t = 15$,$x(15) = \frac{1}{2} x_0$. So,$\frac{1}{2} x_0 = x_0 e^{-15c}$,which implies $e^{-15c} = \frac{1}{2}$. We need to find the amount of ice left after $30 	ext{ minutes}$,which is $x(30) = x_0 e^{-30c}$. $x(30) = x_0 (e^{-15c})^2 = x_0 \left(\frac{1}{2}ight)^2 = \frac{1}{4} x_0$. Comparing this with $k x_0$,we get $k = \frac{1}{4}$.

An ice ball melts at a rate proportional to the amount of ice present at that instant. Half of the initial quantity of ice melts in $15 \text{ minutes}$. Let $x_0$ be the initial quantity of ice. If after $30 \text{ minutes}$ the amount of ice left is $k x_0$,then the value of $k$ is:

Similar Questions

The differential equation $y \frac{dy}{dx} + x = c$ represents

Let the population of rabbits surviving at time $t$ be governed by the differential equation $\frac{dp(t)}{dt} = \frac{1}{2}p(t) - 200$. If $p(0) = 100$,then $p(t)$ equals:

The rate at which the population of a city increases varies as the population. In a period of $20$ years,the population increased from $4$ lakhs to $6$ lakhs. In another $20$ years,the population will be (in $lakhs$)

$A$ curve passes through the point $(3,2)$ for which the segment of the tangent line contained between the coordinate axes is bisected at the point of contact. The equation of the curve is

If $y = At^2 + \frac{B}{t}$ ($A, B$ are parameters) is the general solution of the differential equation $f(t) y''(t) + g(t) y'(t) + h(t) y = 0$,then $2 f(t) + t^2 h(t) =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module