$A$ body is heated to $110^{\circ} C$ and placed in air at $10^{\circ} C$. After $1$ hour,its temperature is $60^{\circ} C$. The additional time required for it to cool to $30^{\circ} C$ is

Verify that the given function $y = \sqrt{a^{2} - x^{2}}$,where $x \in (-a, a)$,is a solution of the differential equation $x + y \frac{dy}{dx} = 0$ (where $y \neq 0$).

Given $\frac{d^2 y}{d x^2}+\cot x \frac{d y}{d x}+4 y \operatorname{cosec}^2 x=0$. Changing the independent variable $x$ to $z$ by the substitution $z=\log \tan \frac{x}{2}$,the equation is changed to

$A$ spherical rain drop evaporates at a rate proportional to its surface area. If initially its radius is $3 \ mm$ and after $1 \ second$ it is reduced to $2 \ mm$,then at any time $t$ its radius is (where $0 \leq t < 3$)

Water at $100^{\circ} C$ cools in $15 \text{ minutes}$ to $75^{\circ} C$ in a room temperature of $25^{\circ} C$. Then the temperature of water after $30 \text{ minutes}$ is

Let $\Gamma$ denote a curve $y = y(x)$ which is in the first quadrant and let the point $(1,0)$ lie on it. Let the tangent to $\Gamma$ at a point $P$ intersect the $y$-axis at $Y_p$. If $PY_p$ has length $1$ for each point $P$ on $\Gamma$,then which of the following options is/are correct?
$(1)$ $y=\ln\left(\frac{1+\sqrt{1-x^2}}{x}\right)-\sqrt{1-x^2}$
$(2)$ $xy^{\prime}+\sqrt{1-x^2}=0$
$(3)$ $y=-\ln\left(\frac{1+\sqrt{1-x^2}}{x}\right)+\sqrt{1-x^2}$
$(4)$ $xy^{\prime}-\sqrt{1-x^2}=0$