A spherical ball of radius 1 × 10⁻⁴ m and density 10⁵ kg/m³ falls… | JEE Main 2024 (9 Apr, Shift 2)

Q28JEE Main 2024 · 9 APR · SHIFT 2Medium

Terminal Velocity

A spherical ball of radius $1 \times 10^{-4}\text{ m}$ and density $10^{5}\text{ kg/m}^{3}$ 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 $9.8 \times 10^{-6}\text{ N s/m}^{2}$ )

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Answer: (a)

Velocity unchanged in water means it enters at terminal velocity: $\sqrt{2gh} = v_t$ .
$v_t = \dfrac{2r^2\rho g}{9\eta} = \dfrac{2 \times 10^{-8} \times 10^5 \times 9.8}{9 \times 9.8 \times 10^{-6}} \approx 222.2$ m/s (water density neglected, $\rho \gg \sigma$ )
$h = \dfrac{v_t^2}{2g} = \dfrac{(222.2)^2}{19.6} \approx 2518$ m
So the answer is (a) 2518 m.

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