A car while travelling produces noise of intensity level of `94` decibel. At a particular point on road maximum permissible level of noise is `100` decibel. Find how many such identical car can be allowed to pass through that point simutaneously if sound waves emitted by all cars in same phase at that point? (Use `log2=0.3`)

Explanation

To solve the problem, we need to determine how many identical cars can pass through a particular point on the road simultaneously without exceeding the maximum permissible noise level of 100 decibels. Here’s a step-by-step solution:

### Step 1: Understand the relationship between intensity and decibels
The sound intensity level in decibels (dB) is given by the formula:
\[ \beta = 10 \log_{10} \left( \frac{I}{I_0} \right) \]
where:
- \( \beta \) is the sound intensity level in decibels,
- \( I \) is the intensity of the sound,
- \( I_0 \) is the reference intensity, typically \( 10^{-12} \, \text{W/m}^2 \).

### Step 2: Calculate the intensity of one car
Given that the intensity level of one car is 94 dB, we can set up the equation:
\[ 94 = 10 \log_{10} \left( \frac{I_1}{I_0} \right) \]
To find \( I_1 \), we rearrange the equation:
\[ \log_{10} \left( \frac{I_1}{I_0} \right) = \frac{94}{10} = 9.4 \]
This implies:
\[ \frac{I_1}{I_0} = 10^{9.4} \]
Thus:
\[ I_1 = I_0 \cdot 10^{9.4} \]

### Step 3: Set up the equation for n cars
If \( n \) identical cars are passing, the total intensity \( I_n \) can be expressed as:
\[ I_n = n \cdot I_1 \]
The intensity level for \( n \) cars is given by:
\[ \beta_n = 10 \log_{10} \left( \frac{I_n}{I_0} \right) = 10 \log_{10} \left( \frac{n \cdot I_1}{I_0} \right) \]
This can be simplified using the properties of logarithms:
\[ \beta_n = 10 \log_{10}(n) + 10 \log_{10}\left(\frac{I_1}{I_0}\right) \]
Substituting \( \beta_n \) with the maximum permissible level of 100 dB:
\[ 100 = 10 \log_{10}(n) + 94 \]

### Step 4: Solve for n
Rearranging the equation gives:
\[ 100 - 94 = 10 \log_{10}(n) \]
\[ 6 = 10 \log_{10}(n) \]
Dividing both sides by 10:
\[ \log_{10}(n) = 0.6 \]
To find \( n \), we exponentiate both sides:
\[ n = 10^{0.6} \]

### Step 5: Calculate \( n \) using the given \( \log_2 = 0.3 \)
Using the change of base formula:
\[ \log_{10}(n) = \frac{\log_2(n)}{\log_2(10)} \]
We know \( \log_2(10) \approx 3.32 \) (since \( \log_2(10) = \frac{1}{\log_{10}(2)} \) and \( \log_{10}(2) \approx 0.301 \)):
\[ 0.6 = \frac{\log_2(n)}{3.32} \]
Thus:
\[ \log_2(n) = 0.6 \times 3.32 \approx 1.992 \]
Exponentiating gives:
\[ n \approx 2^{1.992} \approx 3.98 \approx 4 \]

### Conclusion
The maximum number of identical cars that can pass through the point simultaneously without exceeding the permissible noise level is approximately **4**.