Physics Q95 – AP-EAMCET 2026

Home Physics Mechanical Properties of FluidsQuestion #253

A bubble rises in a viscous liquid. If radius of bubble doubles, terminal velocity becomes

Clue 1 of 3: Forces on a Rising Bubble

Identify the forces acting on a bubble rising in a viscous liquid: buoyant force, gravitational force (weight), and viscous drag force.

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Step 1: Identify Forces and Condition for Terminal Velocity✦ Active

When a bubble rises in a viscous liquid, three main forces act on it:

1. Gravitational force (weight) acting downwards: $W = m g = \frac{4}{3} π r^{3} ρ_{b} g$ , where $ρ_{b}$ is the density of the bubble.

2. Buoyant force acting upwards: $F_{B} = \frac{4}{3} π r^{3} ρ_{l} g$ , where $ρ_{l}$ is the density of the liquid.

3. Viscous drag force acting downwards (opposite to the direction of motion): $F_{D} = 6 π η r v_{t}$ , where $η$ is the coefficient of viscosity of the liquid, $r$ is the radius of the bubble, and $v_{t}$ is the terminal velocity.

At terminal velocity, the net force on the bubble is zero. Since the bubble is rising, the upward buoyant force balances the sum of the downward gravitational and drag forces.

F_{B} = W + F_{D}

💡 Teacher's Secret Hint

Remember that for a rising bubble, the buoyant force is typically greater than the weight, so the drag force acts downwards, opposing the upward motion.

Step 2: Derive the Expression for Terminal Velocity○ Expand

Substitute the force expressions into the balance equation:

\frac{4}{3} π r^{3} ρ_{l} g = \frac{4}{3} π r^{3} ρ_{b} g + 6 π η r v_{t}

Rearrange to solve for $v_{t}$ :

6 π η r v_{t} = \frac{4}{3} π r^{3} g (ρ_{l} - ρ_{b})

v_{t} = \frac{4 π r^{3} g (ρ_{l} - ρ_{b})}{18 π η r}

v_{t} = \frac{2 r^{2} g (ρ_{l} - ρ_{b})}{9 η}

💡 Teacher's Secret Hint

Ensure careful algebraic manipulation, especially when canceling terms and isolating $v_{t}$ . The factor of $r$ in the denominator cancels one $r$ from $r^{3}$ in the numerator.

Step 3: Analyze the Proportionality of Terminal Velocity with Radius○ Expand

From the derived formula, $v_{t} = \frac{2 r^{2} g (ρ_{l} - ρ_{b})}{9 η}$ , we can see that the terminal velocity $v_{t}$ is directly proportional to the square of the radius $r$ , assuming all other parameters ( $g$ , $ρ_{l}$ , $ρ_{b}$ , $η$ ) remain constant.

v_{t} \propto r^{2}

💡 Teacher's Secret Hint

Recognizing proportionality is key to quickly solving problems involving changes in parameters without re-calculating the full formula.

Step 4: Calculate the Change in Terminal Velocity○ Expand

Let the initial radius be $r_{1} = r$ and the initial terminal velocity be $v_{t 1}$ .

If the radius doubles, the new radius is $r_{2} = 2 r$ .

Using the proportionality:

\frac{v_{t 2}}{v_{t 1}} = \frac{r_{2}^{2}}{r_{1}^{2}}

\frac{v_{t 2}}{v_{t 1}} = \frac{(2 r)^{2}}{r^{2}}

\frac{v_{t 2}}{v_{t 1}} = \frac{4 r^{2}}{r^{2}}

\frac{v_{t 2}}{v_{t 1}} = 4

Therefore, $v_{t 2} = 4 v_{t 1}$ . The terminal velocity becomes four times its original value.

💡 Teacher's Secret Hint

Always clearly define your initial and final states for variables to avoid confusion during substitution.

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Physics Question 95 – AP-EAMCET 2026

Clue 1 of 3: Forces on a Rising Bubble

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