A air bubble of radius 1 cm in water has an upward acceleration $9.8 \mathrm{~cm} \mathrm{~s}^{-2}$. The density of water is $1 \mathrm{gm} \mathrm{cm}^{-3}$ and water offers negligible drag force on the bubble. The mass of the bubble is: $\left(\mathrm{g}=980 \mathrm{~cm} / \mathrm{s}^{2}\right)$.

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Fluid Mechanics

A air bubble of radius 1 cm in water has an upward acceleration

9.8 cm s^{- 2}

. The density of water is

1 gm cm^{- 3}

and water offers negligible drag force on the bubble. The mass of the bubble is:

(g = 980 cm / s^{2})

(A)4.51 gm
(B)1.52 gm
(C)3.15 gm
(D)4.15 gm✓ Correct

Answer: (D)

Solution

The bubble accelerates upward because the buoyant force exceeds its weight:

ρ_{w} V g - m g = ma

, so

m = \frac{ρ _{w} V g}{g + a}

. Its volume is

V = \frac{4}{3} π r^{3} = \frac{4}{3} π (1)^{3} = \frac{4 π}{3}

^{3}

. With

ρ_{w} = 1

g/cm

^{3}

g = 980

cm/s

^{2}

and

a = 9.8

cm/s

^{2}

m = \frac{1 \times \frac{4 π}{3} \times 980}{980 + 9.8} = \frac{\frac{4 π}{3} \times 980}{989.8} \approx 4.15

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Single correct option

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