If the amplitude of sound is doubled and the frequency reduced to one-fourth, the intensity of sound at the same point will

Solution :
Factors on which intensity depends are
(i) ) Amplitude $$\left( a \right)$$ of vibration of the source, $$I \propto {a^2}$$
(ii) Surface area $$\left( A \right)$$ of the vibrating body, $$I \propto A$$
(iii) Density $$\left( \rho \right)$$ of the medium, $$I \propto \rho $$
(iv) Frequency $$\left( \nu \right)$$ of the source, $$I \propto {\nu ^2}$$
(v) Motion of the medium which changes effective velocity $$v$$ of sound, $$I \propto v$$
$${\text{As}}\,\,I \propto {a^2}\,\,{\text{and}}\,\,I \propto {\nu ^2}$$
Therefore, intensity becomes $$\frac{{{2^2}}}{{{4^2}}} = \frac{1}{4}th.$$