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

The general solution of the differential equation $x \cos y \mathrm{~d} y=\left(x \mathrm{e}^{\mathrm{x}} \log x+\mathrm{e}^x\right) \mathrm{d} x$ is given by

If order and degree of the differential equation $\left(\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}\right)^5+4 \frac{\left(\frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}\right)^5}{\left(\frac{\mathrm{~d}^3 y}{\mathrm{~d} x^3}\right)}+\frac{\mathrm{d}^3 y}{\mathrm{~d} x^3}=\sin x$, are $m$ and $n$ respectively, then the value of $\left(\mathrm{m}^2+\mathrm{n}^2\right)$ is equal to

If a body cools from $80^{\circ} \mathrm{C}$ to $60^{\circ} \mathrm{C}$ in the room temperature of $30^{\circ} \mathrm{C}$ in 30 min , then the temperature of a body after one hour is