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 $\frac{1}{\sqrt{3}}$ feet per minute. The rate (in cu. feet/min) at which the volume of water in the vessel is decreasing, when the volume of water is $\frac{8 \pi}{\sqrt{3}}$ cubic feet, is

The area (in sq. units) of the triangle formed by the tangent and normal drawn to the curve $\left(\frac{x}{3}\right)^n+\left(\frac{y}{4}\right)^n=2$ at $(3,4)$ and $x$-axis is

If the curves $a x^2+b y^2=1$ and $c x^2+d y^2=1$ intersect orthogonally, then $\frac{b-a}{d-c}=$

The radius of a sphere is changing. At an instant of time the rate of change in its volume and its surface area are equal. Then the value of radius at that instant is?