An air bubble is released from rest at the bottom of a tall glass...

Let the mass of the air bubble be $m$ and its radius be $r$.

The forces acting on the bubble as it rises are:
1. Upward buoyant force, $F_B$ (given as constant)
2. Downward weight, $W = mg$
3. Downward viscous drag force, $F_v$

According to Stokes' law, the viscous drag force is proportional to the velocity $v$ of the bubble:
$F_v = 6\pi\eta r v = kv$
where $k = 6\pi\eta r$ is a positive constant.

The net upward force $F_{\text{net}}$ on the bubble is:
$F_{\text{net}} = F_B - W - F_v$
$ma = F_B - mg - kv$

Dividing by $m$, we get the upward acceleration $a$ as a function of upward velocity $v$:
$a = \dfrac{F_B - mg}{m} - \left(\dfrac{k}{m}\right)v$

Let $a_0 = \dfrac{F_B - mg}{m}$ and $c = \dfrac{k}{m}$. Since $F_B$, $m$, $g$, and $k$ are constants, $a_0$ and $c$ are also positive constants. The equation simplifies to:
$a = a_0 - cv$

This is the equation of a straight line with a positive y-intercept ($a_0$) and a negative slope ($-c$).

Looking at the given graphs:
- Graph (P) represents a straight line with a negative slope and a positive intercept, which perfectly matches our derived equation.
- The acceleration becomes zero when the bubble reaches its terminal velocity $v_1 = \dfrac{a_0}{c}$.

Therefore, Graph (P) best represents the variation of the bubble's upward acceleration as a function of its upward velocity.