Answer:
Option C
Explanation:
As $\frac{1}{2}m_Av_A^2 = \frac{1}{2}m_{B}v_B^2$
$\frac{v_{A}}{v_{B}} = \sqrt{\frac{m_{B}}{m_{A}}}$
$\frac{P_{B}}{P_{A}} = \frac{m_{B}v_{B}}{m_{A}v_{A}}$
$\frac{m_{B}}{m_{A}}\sqrt{\frac{m_{A}}{m_{B}}} =\sqrt{\frac{m_{B}}{m_{A}}}$ = $\frac{1}{\sqrt{3}}$
