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

The surface area of a spherical balloon being inflated increases at a constant rate. If initially,the radius of the balloon is $3$ units and after $5$ seconds,it becomes $7$ units,then its radius after $9$ seconds is

$A$ circular plate of radius $5 \text{ cm}$ is heated. Due to expansion,its radius increases at the rate of $0.05 \text{ cm/s}$. The rate at which its area is increasing when the radius is $5.2 \text{ cm}$ is:

If the surface area of a spherical balloon of radius $6 \text{ cm}$ is increasing at the rate $2 \text{ cm}^2/\text{sec}$,then the rate of increase in its volume in $\text{cm}^3/\text{sec}$ is

$A$ is a point on the circle with radius $8$ and centre at $O$. $A$ particle $P$ is moving on the circumference of the circle starting from $A$. $M$ is the foot of the perpendicular from $P$ on $OA$ and $\angle POM = \theta$. When $OM = 4$ and $\frac{d\theta}{dt} = 6 \text{ radians/sec}$,then the rate of change of $PM$ is (in units/sec)

What will be the rate of change of the volume of a sphere with radius $r$,with respect to its diameter?