Physics Q32 – JEE-MAIN 2025

Home Physics Mechanical Properties of FluidsQuestion #273

A solid steel ball of diameter

3.6

mm acquired terminal velocity

2.45 \times 10^{- 2}

m/s while falling under gravity through an oil of density

925

kg m

^{- 3}

. Take density of steel as

7825

kg m

^{- 3}

and g as

9.8

m/s

^{2}

. The viscosity of the oil in SI unit is

Clue 1 of 3: Forces in Fluid Dynamics

When an object falls through a viscous fluid and reaches terminal velocity, the net force on it is zero. This means the downward gravitational force is balanced by the upward buoyant force and the upward viscous drag force.

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Step 1: Identify the governing principle and formula✦ Active

When a sphere falls through a viscous fluid and reaches terminal velocity, the net force on it is zero. The forces acting are gravitational force ( $F_{g}$ ), buoyant force ( $F_{b}$ ), and viscous drag force ( $F_{v}$ ). At terminal velocity, $F_{g} = F_{b} + F_{v}$ .

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

This simplifies to the formula for terminal velocity:

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

💡 Teacher's Secret Hint

Remember to convert all units to SI before calculation, especially diameter to radius in meters.

Step 2: Rearrange the formula to solve for viscosity○ Expand

Rearrange the terminal velocity formula to solve for the viscosity $η$ :

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

Step 3: Substitute the given values and calculate○ Expand

Given values are: diameter $D = 3.6$ mm $⟹ r = 1.8 \times 10^{- 3}$ m, $v_{t} = 2.45 \times 10^{- 2}$ m/s, $ρ_{f} = 925$ kg m $^{- 3}$ , $ρ_{s} = 7825$ kg m $^{- 3}$ , $g = 9.8$ m/s $^{2}$ . Substitute these into the formula:

η = \frac{2 \times (1.8 \times 10^{- 3})^{2} \times (7825 - 925) \times 9.8}{9 \times (2.45 \times 10^{- 2})}

η = \frac{2 \times (3.24 \times 10^{- 6}) \times (6900) \times 9.8}{9 \times (2.45 \times 10^{- 2})}

η = \frac{437683.2 \times 10^{- 6}}{22.05 \times 10^{- 2}} = \frac{437683.2}{22.05} \times 10^{- 4} \approx 19849.57 \times 10^{- 4} \approx 1.985 Pa s

Rounding to two decimal places, the viscosity of the oil is approximately $1.99$ Pa s.

💡 Teacher's Secret Hint

Pay close attention to powers of 10 and unit conversions to avoid common calculation errors.

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Physics Question 32 – JEE-MAIN 2025

Clue 1 of 3: Forces in Fluid Dynamics

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