Q 11-09-135JEE MainJEE Main 2020 (4 Sep, Shift 1)Medium
An air bubble of radius $1\ \text{cm}$ in water has an upward acceleration of $9.8\ \text{cm s}^{-2}$. The density of water is $1\ \text{g cm}^{-3}$ and water offers negligible drag force on the bubble. The mass of the bubble is $(g = 980\ \text{cm s}^{-2})$
Answer: (C) $4.15\ \text{g}$
Buoyant force $= V\rho g$ with $V = \tfrac43\pi(1)^{3} = 4.19\ \text{cm}^{3}$.
$$V\rho g - mg = ma \Rightarrow m = \frac{V\rho g}{g + a} = \frac{4.19\times980}{989.8} \approx 4.15\ \text{g}$$
Solution by Sreeraj P, M.Sc Physics