The clocktower ("ghantaghar") of Dehradun is famous for the sound of its bell, which can be heard, albeit faintly, upto the outskirts of the city 8 km away. Let the intensity of this faint sound be 30 dB. The clock is situated 80 m high. The intensity at the base of the tower is :-

Options

A. 60 dB.
B. 70 dB.
C. 80 dB
D. 90 dB

Explanation

To solve the problem, we need to find the intensity of sound at the base of the clock tower given that the intensity of sound at a distance of 8 km (8000 m) is 30 dB.

### Step-by-Step Solution:

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

2. **Convert the given loudness to intensity**:
For the faint sound at 8 km:
\[ L_1 = 30 \, \text{dB} \]
Using the formula:
\[ 30 = 10 \log_{10} \left( \frac{I_1}{I_0} \right) \]
Rearranging gives:
\[ \log_{10} \left( \frac{I_1}{I_0} \right) = 3 \implies \frac{I_1}{I_0} = 10^3 \implies I_1 = 1000 \times I_0 \]
Substituting \( I_0 = 10^{-12} \, \text{W/m}^2 \):
\[ I_1 = 1000 \times 10^{-12} = 10^{-9} \, \text{W/m}^2 \]

3. **Use the inverse square law for sound intensity**:
The intensity of sound decreases with the square of the distance from the source. If \( R_1 = 8000 \, \text{m} \) (distance to the outskirts) and \( R_2 = 80 \, \text{m} \) (height of the tower), we can express the relationship as:
\[ \frac{I_2}{I_1} = \left( \frac{R_1}{R_2} \right)^2 \]
Therefore:
\[ I_2 = I_1 \left( \frac{R_1}{R_2} \right)^2 \]

4. **Calculate \( I_2 \)**:
Substituting the values:
\[ I_2 = 10^{-9} \left( \frac{8000}{80} \right)^2 \]
Simplifying:
\[ I_2 = 10^{-9} \left( 100 \right)^2 = 10^{-9} \times 10000 = 10^{-5} \, \text{W/m}^2 \]

5. **Convert \( I_2 \) back to decibels**:
Now, we need to find the loudness \( L_2 \) at the base of the tower:
\[ L_2 = 10 \log_{10} \left( \frac{I_2}{I_0} \right) \]
Substituting \( I_2 \) and \( I_0 \):
\[ L_2 = 10 \log_{10} \left( \frac{10^{-5}}{10^{-12}} \right) = 10 \log_{10} (10^7) = 10 \times 7 = 70 \, \text{dB} \]

### Final Answer:
The intensity at the base of the tower is **70 dB**.