A spherical ball of radius 1 × 10⁻⁴ m and density 10⁵ kg/m³ falls freely under gravity through a distance h before entering a tank of water. If after entering in water the velocity of the ball does not change, then the value of h is approximately : (The coefficient of viscosity of water is 9.8 × 10⁻⁶ N s/m²)
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Question

A spherical ball of radius 1 × 10⁻⁴ m and density 10⁵ kg/m³ falls freely under gravity through a distance h before entering a tank of water. If after entering in water the velocity of the ball does not change, then the value of h is approximately : (The coefficient of viscosity of water is 9.8 × 10⁻⁶ N s/m²)

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G1 Physics · Accepted Answer

For the velocity to remain unchanged after entering water, the entry speed must equal the terminal speed in water. Terminal speed by Stokes' law: v(t) = [2gR²(ρ(ball) - ρ(water))]/(9η) Using: g = 10 m/s² R = 1 × 10⁻⁴ m ρ(ball) = 10⁵ kg/m³ ρ(water) = 10³ kg/m³ η = 9.8 × 10⁻⁶ N s/m² we get approximately: v(t) = 224.5 m/s For free fall from height h: v = √(2gh) Therefore: h = v²/(2g) ≈ (224.5)²/20 ≈ 2518 m