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. (a)231s
B. (b)208s
C. (c)161s
D. (d)142s

Explanation

To solve the problem step by step, we will follow the reasoning based on the information provided in the question and the video transcript.

### Step 1: Understand the Problem
We need to find the time it takes for the amplitude of a simple pendulum to decrease from 10 cm to 5 cm in carbon dioxide, given that it decreases from 10 cm to 8 cm in 40 seconds in air.

### Step 2: Use the Information Provided
We know:
- Amplitude in air decreases from \( A_1 = 10 \, \text{cm} \) to \( A_2 = 8 \, \text{cm} \) in \( t_1 = 40 \, \text{s} \).
- The ratio of the coefficients of viscosity of air to carbon dioxide is \( \frac{\eta_{\text{air}}}{\eta_{\text{CO2}}} = 1.3 \).

### Step 3: Relate the Amplitude Decrease to Time
The amplitude \( A \) of a damped harmonic oscillator decreases exponentially with time, which can be expressed as:
\[ A(t) = A_0 e^{-\lambda t} \]
where \( \lambda \) is the damping coefficient.

### Step 4: Set Up the Equations
For air, we can write:
\[ \frac{A_2}{A_1} = \frac{8}{10} = e^{-\lambda_{\text{air}} t_1} \]
Taking the natural logarithm of both sides:
\[ \ln\left(\frac{8}{10}\right) = -\lambda_{\text{air}} \cdot 40 \]

For carbon dioxide, we want to find the time \( t_2 \) for the amplitude to decrease from 10 cm to 5 cm:
\[ \frac{A_3}{A_1} = \frac{5}{10} = e^{-\lambda_{\text{CO2}} t_2} \]
Taking the natural logarithm:
\[ \ln\left(\frac{5}{10}\right) = -\lambda_{\text{CO2}} t_2 \]

### Step 5: Relate the Damping Coefficients
Using the viscosity ratio:
\[ \lambda_{\text{CO2}} = \frac{\lambda_{\text{air}}}{1.3} \]

### Step 6: Substitute and Solve
Substituting \( \lambda_{\text{CO2}} \) into the equation for carbon dioxide:
\[ \ln\left(\frac{5}{10}\right) = -\frac{\lambda_{\text{air}}}{1.3} t_2 \]

Now we can express \( t_2 \) in terms of \( t_1 \):
\[ t_2 = -\frac{1.3 \ln\left(\frac{5}{10}\right)}{\lambda_{\text{air}}} \]

From our earlier equation for air:
\[ \lambda_{\text{air}} = -\frac{\ln\left(\frac{8}{10}\right)}{40} \]

### Step 7: Substitute \( \lambda_{\text{air}} \) into \( t_2 \)
Now substitute \( \lambda_{\text{air}} \) into the equation for \( t_2 \):
\[ t_2 = -\frac{1.3 \ln\left(\frac{5}{10}\right)}{-\frac{\ln\left(\frac{8}{10}\right)}{40}} \]
This simplifies to:
\[ t_2 = 1.3 \cdot 40 \cdot \frac{\ln\left(\frac{5}{10}\right)}{\ln\left(\frac{8}{10}\right)} \]

### Step 8: Calculate the Values
Using the logarithm values provided:
- \( \ln\left(\frac{5}{10}\right) = \ln(0.5) = -0.693 \)
- \( \ln\left(\frac{8}{10}\right) = \ln(0.8) = -0.223 \) (calculated as \( \ln(8) - \ln(10) \))

Now substituting these values:
\[ t_2 = 1.3 \cdot 40 \cdot \frac{-0.693}{-0.223} \]
Calculating this gives:
\[ t_2 \approx 1.3 \cdot 40 \cdot 3.106 \approx 162.2 \, \text{s} \]

### Step 9: Final Answer
The time in which the amplitude of the pendulum will reduce from 10 cm to 5 cm in carbon dioxide is approximately 162 seconds.