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Step-by-Step Solution
Step 1: Identify the forces acting on the bubble
When the air bubble is submerged in water, it experiences two main forces:
Buoyant Force (upward), which equals the weight of the displaced water.
Weight of the bubble (downward), which is the mass of the bubble multiplied by the acceleration due to gravity.
Step 2: Write the expression for buoyant force
The buoyant force, $F_{\text{buoyant}}$, is given by the weight of the fluid displaced. Since the bubble has radius $r = 1 \text{ cm}$, its volume $V$ is:
$V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi \times (1)^3 = \frac{4}{3}\pi \text{ cm}^3$
For water with density $\rho_{\text{water}} = 1 \text{ gm/cm}^3$, the mass of displaced water is
$m_{\text{displaced}} = \rho_{\text{water}} \times V = 1 \times \frac{4}{3}\pi = \frac{4}{3}\pi \text{ gm}.$
Hence, the buoyant force (in cgs units) is
$F_{\text{buoyant}} = m_{\text{displaced}} \cdot g = \left(\frac{4}{3}\pi\right) \cdot 980 \text{ dynes}.$
Step 3: Set up the equation using Newton’s Second Law
Let the mass of the bubble be $m$. The net upward force on the bubble equals the bubble’s mass times its upward acceleration $a = 9.8 \text{ cm/s}^2$. Thus:
$F_{\text{net}} = (m)(a).$
But $F_{\text{net}}$ is also the difference between the buoyant force and the weight of the bubble:
$F_{\text{net}} = F_{\text{buoyant}} - (m \cdot g).$
So,
$(m)(a) = F_{\text{buoyant}} - m g.$
Step 4: Substitute the values and solve for m
From Step 2, $F_{\text{buoyant}} = \left(\frac{4}{3}\pi\right) \times 980.$ Plugging into the net force equation:
$m \cdot a = \left(\frac{4}{3}\pi \times 980\right) - m \cdot 980.$
Rearrange to isolate $m$:
$\frac{4}{3}\pi \times 980 = m \cdot 980 + m \cdot a,$
$\frac{4}{3}\pi \times 980 = m \left(980 + 9.8\right).$
Thus,
$m = \frac{\frac{4}{3}\pi \times 980}{980 + 9.8} = \frac{\frac{4}{3}\pi \times 980}{989.8}.$
Calculating this numerically gives approximately $4.15 \text{ gm}.$
Step 5: Final Answer
Therefore, the mass of the bubble is $4.15 \text{ gm}.$
