⇒Kinetic Theory of Gases

Correct Option is (B)

We know,

$\begin{array}{rl}& {\mathrm{v}}_{\mathrm{ms}}^2=\frac{3\mathrm{PV}}{\mathrm{Nm}}\\ & \Rightarrow \mathrm{P}=\frac{1}{3}\frac{\mathrm{mN}}{\mathrm{V}}{\mathrm{v}}_{\mathrm{mms}}^2\\ & \Rightarrow \mathrm{P}\propto {\mathrm{v}}_{\mathrm{ms}}^2\\ & \therefore \frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}=\frac{{\mathrm{m}}_2}{{\mathrm{m}}_1}\times {\left[\frac{{\mathrm{v}}_2}{{\mathrm{v}}_1}\right]}^2\\ & =\frac{\left(\frac{{\mathrm{m}}_1}{2}\right)}{{\mathrm{m}}_1}{\left[\frac{2{\mathrm{v}}_1}{{\mathrm{v}}_1}\right]}^2\\ & \dots ..(\because \text{ given }{\mathrm{m}}_2=\frac{{\mathrm{m}}_1}{2}\text{ and }{\mathrm{v}}_2=2{\mathrm{v}}_1)\end{array}$

$\begin{array}{rl}\frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}& =2\\ \therefore {\mathrm{P}}_2& =2{\mathrm{P}}_1\\ & =2\mathrm{P}....(\text{given}{\mathrm{P}}_1=\mathrm{P})\end{array}$