A wet substance in the open air loses its moi | vedclass

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

Question

(A) Let $y$ be the amount of moisture at time $t$.The rate of change of moisture is proportional to the moisture content:$\frac{dy}{dt} = -ky$ (where

Vedclass · Accepted Answer

$\frac{2 \log 10}{\log 2}$

Vedclass · Answer

(A) Let $y$ be the amount of moisture at time $t$.The rate of change of moisture is proportional to the moisture content:$\frac{dy}{dt} = -ky$ (where $k > 0$ is a constant).Separating variables and integrating:$\int \frac{dy}{y} = -\int k dt \Rightarrow \ln y = -kt + C$.At $t = 0$,let the initial moisture be $y_0 = 1$ (representing $100 \%$).Then $\ln(1) = -k(0) + C \Rightarrow C = 0$.So,$\ln y = -kt$.Given that at $t = 1$ hour,the sheet loses half its moisture,so $y = 0.5$ (or $1/2$):$\ln(0.5) = -k(1) \Rightarrow k = -\ln(0.5) = \ln(2)$.Now,we need to find $t$ when $99 \%$ of the moisture is lost,meaning $1 \%$ remains:$y = 0.01 = \frac{1}{100} = 10^{-2}$.Substituting into the equation $\ln y = -kt$:$\ln(10^{-2}) = -(\ln 2)t$.$-2 \ln(10) = -(\ln 2)t$.$t = \frac{2 \ln 10}{\ln 2} = \frac{2 \log 10}{\log 2}$.

$A$ wet substance in the open air loses its moisture at a rate proportional to the moisture content. If a sheet hung in the open air loses half its moisture during the first hour,then the time $t$,in which $99 \%$ of the moisture will be lost,is

Similar Questions

$A$ spherical balloon expands at a rate proportional to its surface area. Initially, its radius is $2 \text{ cm}$ and $5$ minutes later it increases to $7 \text{ cm}$. What will be the surface area of the spherical balloon after $12$ minutes (in $\text{ cm}^2$)?

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

Let $f$ be a twice differentiable non-negative function such that $(f(x))^2 = 25 + \int_{0}^{x} ((f(t))^2 + (f'(t))^2) dt$. Then the mean of $f(\log_e(1)), f(\log_e(2)), \ldots, f(\log_e(625))$ is equal to:

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).

$A$ population grows at the rate of $10 \%$ of the population per year. How long does it take for the population to double?

Vedclass Test Series

Exam Paper Generator

Online Exam Module