The radius of a sphere is measured to be $20 \, cm$ with a possible error of $0.02 \, cm$. The consequent error in the surface area of the sphere is ....... $sq \, cm$.

$A$ particle moves along a straight line according to the law $s=16-2t+3t^{3}$,where $s$ metres is the distance of the particle from a fixed point at the end of $t$ seconds. The acceleration of the particle at the end of $2 \ s$ is

The displacement of a particle at the time $t$ is given by $s = \sqrt{1+t}$. Then its acceleration $a$ is proportional to:

$A$ particle moves along the curve $\frac{x^2}{16} + \frac{y^2}{4} = 1$. When the rate of change of abscissa is $4$ times that of its ordinate,then the quadrant in which the particle lies is

$A$ point $P$ is moving on the curve $x^3 y^4 = 2^7$. The $x$-coordinate of $P$ is decreasing at the rate of $8 \text{ units per second}$. When the point $P$ is at $(2, 2)$,the $y$-coordinate of $P$:

The volume of a spherical balloon is increasing at the rate of $2 \ cm^3/sec$. When its radius is $4 \ cm$,the rate of change of its surface area (in $cm^2/sec$) is