$A$ spherical balloon is being inflated at the rate of $35 \, cc/min$. The rate of increase of the surface area of the balloon when its diameter is $14 \, cm$ is ....... $sq \, cm/min$.

If the volume of a sphere increases at the rate of $2 \pi \text{ cm}^3/\text{s}$,then the rate of increase of its radius (in $\text{cm}/\text{s}$),when the volume is $288 \pi \text{ cm}^3$,is

The radius of the base of a cone is increasing at the rate $3 \text{ cm/min}$ and the altitude is decreasing at the rate $4 \text{ cm/min}$. The rate at which the lateral surface area is changing,when the radius is $7 \text{ cm}$ and altitude is $24 \text{ cm}$,is:

$A$ particle moves along the curve $y=x^2+2x$. Then,the point on the curve such that the $x$ and $y$ coordinates of the particle change with the same rate is

Two particles $A$ and $B$ move from rest along a straight line with constant accelerations $f$ and $f'$ respectively. If $A$ takes $m$ seconds more than that of $B$ and describes $n$ units more than that of $B$ in acquiring the same velocity,then:

The volume of a sphere is increasing at the rate of $4 \pi \text{ cm}^3/\text{sec}$. When its volume is $288 \pi \text{ cm}^3$,the rate of increase (in $\text{cm/sec}$) in its radius is