The clocktower ("ghantaghar") of Dehradun is | vedclass

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(B) Given: Distance $r_{1} = 8 \,km = 8000 \,m$, Intensity level $L_{1} = 30 \,dB$.Distance at the base $r_{2} = 80 \,m$, Intensity level $L_{2} = ?$.

Vedclass · Accepted Answer

$70$

Vedclass · Answer

(B) Given: Distance $r_{1} = 8 \,km = 8000 \,m$, Intensity level $L_{1} = 30 \,dB$.Distance at the base $r_{2} = 80 \,m$, Intensity level $L_{2} = ?$.The intensity level $L$ is given by $L = 10 \log_{10} \left( \frac{I}{I_{0}} ight)$.Since the sound source acts as a point source, intensity $I = \frac{P}{4 \pi r^{2}}$, which implies $I \propto \frac{1}{r^{2}}$.The difference in intensity levels is given by:$L_{2} - L_{1} = 10 \log_{10} \left( \frac{I_{2}}{I_{1}} ight) = 10 \log_{10} \left( \frac{r_{1}}{r_{2}} ight)^{2} = 20 \log_{10} \left( \frac{r_{1}}{r_{2}} ight)$.Substituting the values:$L_{2} - 30 = 20 \log_{10} \left( \frac{8000}{80} ight)$$L_{2} - 30 = 20 \log_{10} (100)$$L_{2} - 30 = 20 	imes 2 = 40$$L_{2} = 40 + 30 = 70 \,dB$.Thus, the intensity at the base of the tower is $70 \,dB$.

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