The diameter of one of the bases of a truncated cone is $100 \, mm$. If the diameter of this base is increased by $21 \%$ such that it still remains a truncated cone with the height and the other base unchanged,the volume also increases by $21 \%$. The radius of the other base (in $mm$) is

The displacement of a particle at time $t$ is given by $x = At^2 + Bt + C$,where $A, B$ and $C$ are constants. If $v$ is the velocity,then $4Ax - v^2 = ....$

The rate of change of the volume of a sphere with respect to its surface area,when its radius is $2 \text{ cm}$,is . . . . . . $\text{cm}^3 / \text{cm}^2$.

The radius of a sphere increases at the rate of $0.04 \text{ cm/sec}$. The rate of increase in the volume of that sphere with respect to its surface area,when its radius is $10 \text{ cm}$ is

$A$ stone is dropped into a still pond,creating ripples that move at a speed of $3.5 \text{ cm/sec}$. At the instant when the radius of the ripples is $7.5 \text{ cm}$,how fast is the enclosed area increasing?

$A$ water tank has the shape of an inverted right circular cone with its axis vertical and vertex lowermost. Its semi-vertical angle is $\tan ^{-1}(0.5)$. Water is poured into it at a constant rate of $5 \ m^3/h$. Find the rate at which the level of the water is rising at the instant when the depth of the water in the tank is $4 \ m$.