Physics Q35 – JEE-MAIN 2026

Home Physics Mechanical Properties of FluidsQuestion #257

If an air bubble of diameter

2 mm

rises steadily through a liquid of density

2000 {kg/m}^{3}

at a rate of

0.5 cm/s

, then the coefficient of viscosity of liquid is _______ Poise. (Take

g = 10 {m/s}^{2}

)

Clue 1 of 3: Terminal Velocity Concept

When an object rises or falls steadily through a fluid, it reaches a terminal velocity where the net force on it is zero. This means the upward forces balance the downward forces.

🧠 Full Solution Path

Step 1: Identify Forces and Principle✦ Active

When an air bubble rises steadily through a liquid, it reaches terminal velocity. At this point, the net force on the bubble is zero. The upward buoyant force is balanced by the downward viscous drag force. The weight of the air bubble is negligible compared to the buoyant force due to the liquid.

F_{B} = F_{D}

\frac{4}{3} π r^{3} ρ_{L} g = 6 π η r v

Step 2: Rearrange and Substitute Values○ Expand

Solve the equation for the coefficient of viscosity $η$ and substitute the given values after converting them to SI units.

η = \frac{4 π r^{3} ρ_{L} g}{3 \times 6 π r v} = \frac{2 r^{2} ρ_{L} g}{9 v}

\text{Given:} \quad D = 2 \text{ mm} \implies r = 1 \text{ mm} = 1 \times 10^{-3} \text{ m} \ \rho_L = 2000 \text{ kg/m}^3 \ v = 0.5 \text{ cm/s} = 0.5 \times 10^{-2} \text{ m/s} = 5 \times 10^{-3} \text{ m/s} \ g = 10 \text{ m/s}^2 \ \eta = \frac{2 \times (1 \times 10^{-3})^2 \times 2000 \times 10}{9 \times (5 \times 10^{-3})} \ \eta = \frac{2 \times 10^{-6} \times 20000}{45 \times 10^{-3}} = \frac{40000 \times 10^{-6}}{45 \times 10^{-3}} = \frac{0.04}{0.045} = \frac{40}{45} = \frac{8}{9} \text{ Pa} \cdot \text{s}$$

💡 Teacher's Secret Hint

Ensure all units are consistent (SI units) before performing calculations to avoid errors.

Step 3: Convert to Poise○ Expand

The calculated viscosity is in Pascal-seconds (Pa $\cdot$ s). Convert this to Poise using the conversion factor $1 Pa \cdot s = 10 Poise$ .

η = \frac{8}{9} Pa \cdot s \times 10 \frac{Poise}{Pa \cdot s} = \frac{80}{9} Poise \approx 8.88 Poise

\text{The closest option is } 8.8 \text{ Poise.}

💡 Teacher's Secret Hint

Pay attention to the required units for the final answer, as unit conversions are a common source of error.

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Physics Question 35 – JEE-MAIN 2026

Clue 1 of 3: Terminal Velocity Concept

🧠 Full Solution Path