$A$ container is in the shape of an inverted cone. Its height is $6 \ m$ and the radius at the top is $4 \ m$. If it is filled with water at the rate of $3 \ m^3/min$,then the rate of change of the height of the water (in $m/min$) when the water level is $3 \ m$,is

$A$ point is in motion along a hyperbola $y = \frac{10}{x}$ such that its abscissa $x$ increases uniformly at a rate of $1 \text{ unit/s}$. Find the rate of change of its ordinate when the point passes through $(5, 2)$.

The point on the curve $y^2 = 18x$,at which the rate of change of $Y$-coordinate is twice the rate of change of $X$-coordinate is . . . . . . . $\left(\frac{dx}{dt} \neq 0\right)$

$A$ man of $5 \text{ feet}$ height is walking away from a light fixed at a height of $15 \text{ feet}$ at the rate of $K \text{ miles/hour}$. If the rate of increase of his shadow is $\frac{11}{5} \text{ feet/sec}$, then $K=$ (Take $1 \text{ mile} = 5280 \text{ feet}$)

$A$ vessel in the shape of an inverted cone of height $10 \ ft$ and semi-vertical angle $30^{\circ}$ is full of water. Due to a hole at the vertex, the slant height of the water in the vessel is decreasing at a constant rate of $\frac{1}{\sqrt{3}} \ ft/min$. The rate (in $cu. \ ft/min$) at which the volume of water in the vessel is decreasing, when the volume of water is $\frac{8 \pi}{\sqrt{3}} \ cu. \ ft$, is

The surface area of a spherical ball is increasing at the rate of $4 \pi \,cm^2/s$. The rate at which the radius is increasing when the surface area is $16 \pi \,cm^2$ is: (in $\,cm/s$)