An ice-cube of edge is 10 cm floating in water. Time period of small vertical oscillations of the cube is: (Specific gravity of ice is 0.9)

Options

A. 0.6s
B. 1.6s
C. 2.4s
D. 3.6s

Explanation

To solve the problem of finding the time period of small vertical oscillations of an ice cube floating in water, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Edge length of the ice cube, \( a = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Specific gravity of ice, \( SG = 0.9 \)

2. **Convert Specific Gravity to Density:**
- The specific gravity is defined as the ratio of the density of the object to the density of the fluid (water in this case).
- Let \( \sigma \) be the density of ice and \( \rho \) be the density of water.
- Since the specific gravity of ice is given as 0.9, we can express this as:
\[ SG = \frac{\sigma}{\rho} \implies \sigma = 0.9 \rho \]
- The density of water at \( 4^\circ C \) is approximately \( 1000 \, \text{kg/m}^3 \), hence:
\[ \sigma = 0.9 \times 1000 = 900 \, \text{kg/m}^3 \]

3. **Determine the Height of the Ice Cube:**
- The height of the ice cube \( h \) is equal to its edge length:
\[ h = 0.1 \, \text{m} \]

4. **Use the Formula for Time Period of Vertical Oscillation:**
- The formula for the time period \( T \) of small oscillations of a floating object is given by:
\[ T = 2\pi \sqrt{\frac{\sigma h}{\rho g}} \]
- Substituting the values we have:
- \( \sigma = 900 \, \text{kg/m}^3 \)
- \( h = 0.1 \, \text{m} \)
- \( \rho = 1000 \, \text{kg/m}^3 \)
- \( g \approx 10 \, \text{m/s}^2 \)

5. **Calculate the Time Period:**
- Plugging in the values:
\[ T = 2\pi \sqrt{\frac{900 \times 0.1}{1000 \times 10}} \]
- Simplifying the expression:
\[ T = 2\pi \sqrt{\frac{90}{10000}} = 2\pi \sqrt{0.009} = 2\pi \times 0.09487 \]
- Calculating the numerical value:
\[ T \approx 2\pi \times 0.09487 \approx 0.596 \, \text{s} \]
- Rounding to one decimal place gives:
\[ T \approx 0.6 \, \text{s} \]

### Final Answer:
The time period of small vertical oscillations of the ice cube is approximately **0.6 seconds**.