The rate of change of the radius of a sphere is $0.1 \ cm/s$. Find the rate of change of its surface area when its radius is $200 \ cm$.

$A$ man $2 \text{ m}$ tall walks at the rate of $1 \frac{2}{3} \text{ m/s}$ towards a street light which is $5 \frac{1}{3} \text{ m}$ above the ground. The rate at which the length of his shadow is changing when he is $3 \frac{1}{3} \text{ m}$ away from the base of the light is . . . . . . (in $\text{ m/s}$)

$A$ 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

If a particle moves such that the displacement $s$ is proportional to the square of the velocity $v$ acquired,then its acceleration is

$A$ stone is dropped in a quiet lake and it is observed that waves move in circles. If the radius of a circular wave increases at the rate of $2 \text{ cm/sec}$, then the rate of increase in its area at the instant when its radius is $10 \text{ cm}$, is in $\text{cm}^2\text{/sec}$: (in $\pi$)

$A$ bullet is shot horizontally and its distance $S$ cm at time $t$ second is given by $S=1200t-15t^2$. Then, the distance covered by the bullet when it comes to rest is: (in $\text{ cm}$)