Question

The volume of a small ball is calculated to be $25 \text{ cm}^3$ and it weighs $30 \text{ g}$ in air. Will this ball float or sink in water?

• A. Sink
• B. Partially sink
• C. Float
• D. Can not be predicted from the given information

Hint:

Always remember the standard density of water in CGS units is exactly $1 \text{ g/cm}^3$ (or $1000 \text{ kg/m}^3$ in SI units). If mass $>$ volume numerically in CGS, it sinks in water.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
To determine whether an object floats or sinks in a fluid, we must compare the density of the object to the density of the fluid.
Step 2: Key Formula or Approach:
Density ($\rho$) = $\frac{\text{Mass } (m)}{\text{Volume } (V)}$
If $\rho_{\text{object}}>\rho_{\text{fluid}}$, the object will sink.
If $\rho_{\text{object}}<\rho_{\text{fluid}}$, the object will float.
If $\rho_{\text{object}} = \rho_{\text{fluid}}$, it will float suspended within the fluid.
Step 3: Detailed Explanation:
Given values for the ball:
Mass ($m$) = $30 \text{ g}$
Volume ($V$) = $25 \text{ cm}^3$
Calculate the density of the ball: \[ \rho_{\text{ball}} = \frac{30 \text{ g}}{25 \text{ cm}^3} = 1.2 \text{ g/cm}^3 \] The density of water is a standard known value: $\rho_{\text{water}} = 1.0 \text{ g/cm}^3$.
Comparing the two densities, we see that $1.2 \text{ g/cm}^3>1.0 \text{ g/cm}^3$.
Since the density of the ball is greater than that of water, the buoyant force cannot support its weight, and the ball will sink.
Step 4: Final Answer:
The ball will sink.