Question

A solid sphere moves at a terminal velocity of $20{\text{ms}}^{-1}$ in air at a place where $g=9.8{\text{ms}}^{-2}$. The sphere is taken in a gravity free hall having air at the same pressure and pushed downwards at a speed of $20{\text{ms}}^{-1}$. Then, identify the correct statement(s):

[Assume density of air to be very low.]

• A. Intial acceleration of sphere will be $9.8{\text{ms}}^{-2}$ downwards.
• B. Initial acceleration of sphere will be $9.8{\text{ms}}^{-2}$ upwards.
• C. The magnitude of acceleration of the sphere will decrease as time passes.
• D. Sphere will eventually stop.

Answer 1

Answer: (B), (C), (D)

The correct options are
B Initial acceleration of sphere will be $9.8{\text{ms}}^{-2}$ upwards.
C The magnitude of acceleration of the sphere will decrease as time passes.
D Sphere will eventually stop.

Since there is no gravity, there will be no initial acceleration in downward direction. Hence the option $\left(a\right)$ is incorrect.
$20\text{m/s}$ is the terminal velocity. At this velocity, the viscous force is equal to the weight of the sphere (assuming the buoyant force to be negligible because air has a very low density).
i.e $F_v=mg$ at $v=20\text{m/s}$

When sphere is thrown downwards at a speed of $20\text{m/s}$, once again, viscous force $F_v=mg$ will act vertically upwards.
$W=mg=0$ [since gravity free space]
$\Rightarrow a=\frac{F_v}{m}=\frac{mg}{m}=g$
The sphere gets an initial acceleration $a=9.8{\text{m/s}}^2$ in upward direction.
$\therefore$ Hence option $\left(b\right)$ is correct.

$F_v$ acting in opposite direction of velocity will cause retardation in the sphere. Hence, its velocity will decrease continously.
$F_v\propto v$
$\Rightarrow F_v\downarrow$ as $v\downarrow$
Hence, magnitude of acceleration (retardation) also reduces as time passes.
$\therefore$Option $\left(c\right)$ is correct

Since the viscous force $\left(F_v\right)$ continously opposes the motion of the sphere, it will eventually stop i.e at some instant $v=0$ will be achieved.
Hence, the option $\left(d\right)$ is also correct.