1. Given data

• Radius of air bubble: $r = 1\ \text{cm}$
• Density of water: $\rho = 1\ \text{g cm}^{-3}$
• Upward acceleration of bubble: $a = 9.8\ \text{cm s}^{-2}$
• Acceleration due to gravity: $g = 980\ \text{cm s}^{-2}$
• Drag force is negligible.

2. Forces acting on the bubble

Since the bubble is moving upward, the forces are:

• Upthrust (buoyant force) upward: $F_b = \rho V g$
• Weight of bubble downward: $W = mg$

Net upward force is: $F_b - mg = ma$

So, $\rho V g = m(g+a)$

Hence, $m = \frac{\rho V g}{g+a}$

3. Volume of the bubble

Since the bubble is spherical, $V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (1)^3 = \frac{4\pi}{3}\ \text{cm}^3$

Using $\pi \approx 3.14$, $V \approx \frac{4 \times 3.14}{3} = 4.1867\ \text{cm}^3$

4. Substitute in the formula

$m = \frac{(1)(4.1867)(980)}{980+9.8}$

$m = \frac{4102.966}{989.8}$

$m \approx 4.15\ \text{g}$

5. Match with options

• A: $1.52\ \text{g}$
• B: $4.51\ \text{g}$
• C: $3.15\ \text{g}$
• D: $4.15\ \text{g}$

Therefore, the correct option is: $\boxed{\text{D: } 4.15\ \text{gm}}$

6. Comparison with stored correct answer

Stored correct answer: D

Our derived answer is also D, so they agree.