c
(c)
Given situation is as shown below.
Speed of approach of source
$=v_s=r \omega=30 \,ms ^{-1}$
Using formula for Doppler's effect,
Maximum frequency,
$f_{\max }=\left(\frac{v}{v-v_s}\right) \cdot f$
Minimum frequency,
$f_{\min }=\left(\frac{v}{v+v_s}\right) \cdot f$
So, ratio of
$\frac{f_{\max }}{f_{\min }}=\frac{v+v_s}{v-v_s}=\frac{330+30 }{330-30}$
$\Rightarrow f_{\max }=1.2$
