Answer:
Option B
Explanation:
As the observer is moving towards the source, so frequency of waves emitted by the source will be given by the formula.
$f_{observed}= f_{actual}.(\frac{1+\frac{v}{c}}{1-\frac{v}{c}})^{\frac{1}{2}}$
Here, frequency $\frac{v}{c}=\frac{1}{2}$
So, $f_{observed}= f_{actual} (\frac{3/2}{1/2})^{\frac{1}{2}}$
$\therefore$ $f_{observed}= 10\times\sqrt{3}=17.3GHz$
