The differential equation $\frac{dy}{dx} = \frac{\sqrt{1-y^2}}{y}$ determines a family of circles with

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 $lakhs$)

The temperature $T(t)$ of a body at time $t=0$ is $160^{\circ} F$ and it decreases continuously as per the differential equation $\frac{dT}{dt}=-K(T-80)$,where $K$ is a positive constant. If $T(15)=120^{\circ} F$,then $T(45)$ is equal to . . . . . . . . (in $^{\circ} F$)

Let $S$ be the set of real numbers $p$ such that there is no non-zero continuous function $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying $\int_0^x f(t) dt = p f(x)$ for all $x \in \mathbb{R}$. Then,$S$ is

The equation of the curve passing through $\left(2, \frac{9}{2}\right)$ and having the slope $\left(1-\frac{1}{x^2}\right)$ at $(x, y)$ is

Water is drained from a vertical cylindrical tank by opening a valve at the base of the tank. It is known that the rate at which the water level drops is proportional to the square root of water depth $y$,where the constant of proportionality $k > 0$ depends on the acceleration due to gravity and the geometry of the hole. If $t$ is measured in minutes and $k = \frac{1}{15}$,then the time required to drain the tank if the water is $4 \text{ m}$ deep to start with is .......... $\text{min}$.