The amplitude of a simple pendulum, oscillating in air with a small spherical bob, decreases from 10 cm to 8 cm in 40 s. Assuming that Stokes law is valid, and ratio of the coefficient of viscosity of air to that of carbon dioxide is 1.3, the time in which amplitude of this pendulum will reduce from 10 cm to 5 cm in carbon dioxide will be close to (In 5=1.601, In 2=0.693).

Options

A. 231s
B. 208s
C. 161s
D. 142s

Explanation

To solve the problem step by step, we will use the concept of damped oscillations and the relationship between amplitude and time.

### Step-by-Step Solution:

1. **Understanding the Damping Formula**:
The amplitude \( A(t) \) of a damped oscillator can be expressed as:
\[ A(t) = A_0 e^{-\frac{b}{2m} t} \]
where \( A_0 \) is the initial amplitude, \( b \) is the damping coefficient, \( m \) is the mass of the bob, and \( t \) is time.

2. **Given Data for Air**:
- Initial amplitude \( A_0 = 10 \, \text{cm} \)
- Amplitude after 40 seconds \( A(40) = 8 \, \text{cm} \)
- Time \( t = 40 \, \text{s} \)

3. **Setting Up the Equation for Air**:
Using the formula:
\[ 8 = 10 e^{-\frac{b_{\text{air}}}{2m} \cdot 40} \]
Dividing both sides by 10:
\[ \frac{8}{10} = e^{-\frac{b_{\text{air}}}{2m} \cdot 40} \]
Simplifying gives:
\[ \frac{4}{5} = e^{-\frac{b_{\text{air}}}{2m} \cdot 40} \]

4. **Taking the Natural Logarithm**:
Taking the natural logarithm of both sides:
\[ \ln\left(\frac{4}{5}\right) = -\frac{b_{\text{air}}}{2m} \cdot 40 \]
Rearranging gives:
\[ \frac{b_{\text{air}}}{2m} = -\frac{40 \ln\left(\frac{4}{5}\right)}{1} \]

5. **Using the Ratio of Viscosities**:
Given that the ratio of the coefficients of viscosity is:
\[ \frac{b_{\text{air}}}{b_{\text{CO2}}} = 1.3 \]
Therefore, we can express \( b_{\text{CO2}} \) as:
\[ b_{\text{CO2}} = \frac{b_{\text{air}}}{1.3} \]

6. **Setting Up the Equation for CO2**:
For the second case where the amplitude decreases from \( 10 \, \text{cm} \) to \( 5 \, \text{cm} \):
\[ 5 = 10 e^{-\frac{b_{\text{CO2}}}{2m} t} \]
Dividing both sides by 10:
\[ \frac{1}{2} = e^{-\frac{b_{\text{CO2}}}{2m} t} \]

7. **Taking the Natural Logarithm Again**:
Taking the natural logarithm:
\[ \ln\left(\frac{1}{2}\right) = -\frac{b_{\text{CO2}}}{2m} t \]
Rearranging gives:
\[ t = -\frac{2m}{b_{\text{CO2}}} \ln\left(\frac{1}{2}\right) \]

8. **Substituting for \( b_{\text{CO2}} \)**:
Substituting \( b_{\text{CO2}} = \frac{b_{\text{air}}}{1.3} \):
\[ t = -\frac{2m \cdot 1.3}{b_{\text{air}}} \ln\left(\frac{1}{2}\right) \]

9. **Relating Times for Air and CO2**:
From the previous steps, we can relate the times:
\[ t_{\text{CO2}} = \frac{1.3 \cdot 40}{\ln\left(\frac{4}{5}\right) / \ln\left(\frac{1}{2}\right)} \]

10. **Calculating the Final Time**:
Using \( \ln\left(\frac{4}{5}\right) = \ln(4) - \ln(5) = 1.386 - 1.601 \) and \( \ln(2) = 0.693 \):
\[ t_{\text{CO2}} \approx \frac{1.3 \cdot 40}{\ln(4) - 2 \ln(2)} \]
After calculating, we find that \( t_{\text{CO2}} \) is approximately \( 161 \, \text{s} \).

### Final Answer:
The time in which the amplitude of this pendulum will reduce from 10 cm to 5 cm in carbon dioxide will be close to **161 seconds**.