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)

Question

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) 3.15 gm (B) 4.51 gm (C) 4.15 gm (D) 1.52 gm

Tardigrade Admin · Accepted Answer

Volume V =(4 π/3) r 3=(4 π/3) ×(1)3=4.19 cm 3 a =9.8 cm / s 2 B - mg = ma m =( B / g + a ) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/fa5d69d6b809196d0b98541dd7182a2a-.png" /> m =(( V ρω g )/ g + a )=( V ρω/1+( a ) g ) =((4.19) × 1/1+(9.8)980)=(4.19/1.01)=4.15 gm

Q. A air bubble of radius $1 c m$ in water has an upward acceleration $9.8 c m s^{- 2} .$ The density of water is $1 g m c m^{- 3}$ and water offers negligible drag force on the bubble. The mass of the bubble is $(g = 980 c m / s^{2})$

$3.15 g m$

$4.51 g m$

$4.15 g m$

$1.52 g m$

Volume $V = \frac{4 π}{3} r^{3} = \frac{4 π}{3} \times (1)^{3} = 4.19 c m^{3}$
$a = 9.8 c m / s^{2}$
$B - m g = ma$
$m = \frac{B}{g + a}$
$m = \frac{( V ρ _{ω} g )}{g + a} = \frac{V ρ _{ω}}{1 + \frac{a}{g}}$
$= \frac{( 4.19 ) \times 1}{1 + \frac{9.8}{980}} = \frac{4.19}{1.01} = 4.15 g m$

Q. A air bubble of radius 1cm in water has an upward acceleration 9.8cms−2. The density of water is 1gmcm−3 and water offers negligible drag force on the bubble. The mass of the bubble is (g=980cm/s2)

3.15gm

4.51gm

4.15gm

1.52gm