An air bubble of radius 1 cm in water has an upward acceleration of 9.8 cms$$^{-2}$$. 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$$^2)$$.

We have an air bubble in water. The radius of the bubble is given as $$r = 1\ \text{cm}$$, the upward (net) acceleration is $$a = 9.8\ \text{cm s}^{-2}$$, the density of water is $$\rho_w = 1\ \text{g cm}^{-3}$$ and the acceleration due to gravity in the CGS system is $$g = 980\ \text{cm s}^{-2}$$. Water drag is said to be negligible.

First we calculate the volume of the spherical bubble. The formula for the volume of a sphere is stated as

$$V \;=\; \frac{4}{3}\,\pi\,r^{3}.$$

Substituting $$r = 1\ \text{cm}$$, we get

$$V \;=\; \frac{4}{3}\,\pi\,(1\ \text{cm})^{3} \;=\; \frac{4\pi}{3}\ \text{cm}^{3}.$$

Evaluating the numerical value with $$\pi \approx 3.1416$$,

$$V \;=\; \frac{4 \times 3.1416}{3} \;=\; 4.18879\ \text{cm}^{3}.$$

Next we write the forces acting on the bubble. The upward buoyant force is equal to the weight of the displaced water. Hence,

$$F_{\text{buoyant}} \;=\; \rho_w\,g\,V.$$

The downward weight of the bubble itself is

$$W \;=\; m\,g,$$

where $$m$$ is the (unknown) mass of the bubble. Because drag is negligible, the net upward force on the bubble must provide the upward acceleration $$a$$. Therefore, using Newton’s second law,

$$F_{\text{net}} \;=\; F_{\text{buoyant}} \;-\; W \;=\; m\,a.$$

Substituting the expressions for the buoyant force and the weight, we have

$$\rho_w\,g\,V \;-\; m\,g \;=\; m\,a.$$

Now we collect the terms containing $$m$$ on one side:

$$\rho_w\,g\,V \;=\; m\,g \;+\; m\,a \;=\; m\,(g + a).$$

Solving for $$m$$ gives

$$m \;=\; \frac{\rho_w\,g\,V}{g + a}.$$

We now substitute the numerical values:

$$m \;=\; \frac{(1\ \text{g cm}^{-3})\,(980\ \text{cm s}^{-2})\,(4.18879\ \text{cm}^{3})}{980\ \text{cm s}^{-2} + 9.8\ \text{cm s}^{-2}}.$$

First compute the numerator:

$$980 \times 4.18879 \;=\; 4105.0162\ \text{g cm s}^{-2}.$$

Next compute the denominator:

$$980 + 9.8 \;=\; 989.8\ \text{cm s}^{-2}.$$

Now divide:

$$m \;=\; \frac{4105.0162}{989.8}\ \text{g}.$$

Carrying out the division,

$$m \;\approx\; 4.147\ \text{g}.$$

Rounding to three significant figures,

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

Hence, the correct answer is Option C.