A cylindrical piece of cork of height 4.9 cm floats in water with its axis vertical and is made to execute slın. Calculate its time period of oscillation if its density is 0.2 gm/cc ?

Explanation

To solve the problem of finding the time period of oscillation for a cylindrical piece of cork floating in water, we can follow these steps:

### Step 1: Identify Given Data
- Height of the cork (h) = 4.9 cm = 0.049 m (convert to meters for SI units)
- Density of cork (ρ_cork) = 0.2 g/cm³ = 200 kg/m³ (convert to kg/m³)
- Density of water (ρ_water) = 1 g/cm³ = 1000 kg/m³ (convert to kg/m³)
- Acceleration due to gravity (g) = 9.8 m/s²

### Step 2: Use the Formula for Time Period of Oscillation
The time period (T) of oscillation for a floating body is given by the formula:
\[ T = 2\pi \sqrt{\frac{h \cdot \rho_{cork}}{\rho_{water} \cdot g}} \]

### Step 3: Substitute the Values into the Formula
Substituting the values we have:
\[ T = 2\pi \sqrt{\frac{0.049 \cdot 200}{1000 \cdot 9.8}} \]

### Step 4: Calculate the Values Inside the Square Root
First, calculate the numerator:
\[ 0.049 \cdot 200 = 9.8 \]

Now calculate the denominator:
\[ 1000 \cdot 9.8 = 9800 \]

Now substitute these values back into the formula:
\[ T = 2\pi \sqrt{\frac{9.8}{9800}} \]

### Step 5: Simplify the Square Root
Calculate the fraction:
\[ \frac{9.8}{9800} = 0.001 \]

Now take the square root:
\[ \sqrt{0.001} = 0.03162 \]

### Step 6: Calculate the Time Period
Now substitute this back into the equation for T:
\[ T = 2\pi \cdot 0.03162 \]
\[ T \approx 0.198 \text{ seconds} \]

### Step 7: Final Calculation
Using \( \pi \approx 3.14 \):
\[ T \approx 2 \cdot 3.14 \cdot 0.03162 \approx 0.198 \text{ seconds} \]

### Conclusion
The time period of oscillation for the cylindrical piece of cork is approximately **0.198 seconds**.

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