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. (a)0.6s
B. (b)1.6s
C. (c)2.4s
D. (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 1: Understand the Given Information
- The edge of the ice cube is \( a = 10 \, \text{cm} = 0.1 \, \text{m} \).
- The specific gravity of ice is \( \text{SG} = 0.9 \).
- The density of water is \( \rho_w = 1000 \, \text{kg/m}^3 \).
- The density of ice can be calculated as:
\[ \rho_i = \text{SG} \times \rho_w = 0.9 \times 1000 \, \text{kg/m}^3 = 900 \, \text{kg/m}^3. \]

### Step 2: Calculate the Effective Density
The effective density for the oscillation can be derived from the ratio of the density of the ice to the density of water:
\[ \frac{\rho_i}{\rho_w} = \frac{900}{1000} = 0.9. \]

### Step 3: Use the Formula for Time Period of Vertical Oscillation
The formula for the time period \( T \) of small vertical oscillations of a floating object is given by:
\[ T = 2\pi \sqrt{\frac{h}{g}}, \]
where \( h \) is the depth of the submerged part of the ice cube and \( g \) is the acceleration due to gravity.

### Step 4: Calculate the Submerged Depth
The volume of the ice cube is:
\[ V = a^3 = (0.1)^3 = 0.001 \, \text{m}^3. \]
The weight of the ice cube is equal to the weight of the water displaced:
\[ \text{Weight of ice} = \rho_i \cdot V \cdot g = 900 \cdot 0.001 \cdot 10 = 9 \, \text{N}. \]
The volume of water displaced \( V_d \) is:
\[ V_d = \frac{\text{Weight of ice}}{\rho_w \cdot g} = \frac{9}{1000 \cdot 10} = 0.0009 \, \text{m}^3. \]
The submerged depth \( h \) can be calculated from the volume of water displaced:
\[ h = \frac{V_d}{a^2} = \frac{0.0009}{(0.1)^2} = \frac{0.0009}{0.01} = 0.09 \, \text{m}. \]

### Step 5: Substitute Values into the Time Period Formula
Now substituting the values into the time period formula:
\[ T = 2\pi \sqrt{\frac{0.09}{10}}. \]
Calculating the square root:
\[ \sqrt{\frac{0.09}{10}} = \sqrt{0.009} = 0.03. \]
Thus,
\[ T = 2\pi \cdot 0.03 = 0.06\pi \approx 0.1884 \, \text{s}. \]

### Step 6: Final Calculation
To convert to seconds:
\[ T \approx 0.1884 \, \text{s} \approx 0.6 \, \text{s} \text{ (after rounding)}. \]

### Conclusion
The time period of small vertical oscillations of the ice cube is approximately \( 0.6 \, \text{s} \).