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Q.
The temperature of an ideal gas is increased from $120\, K$ to $480\, K$. If at $120\, K$, the root mean square speed of gas molecules is $v$, then at $480 \,K$ it will be
The root mean square velocity is given by
$v_{ rms }=\sqrt{\frac{3 R T}{M}}$
So, $\frac{v_{1}}{v_{2}}=\sqrt{\frac{T_{1}}{T_{2}}}$
Now, $T_{1}=120\, K ,$
$ T_{2}=480\, K ,$
$ v_{1}=v$
So, $\frac{v_{1}}{v_{2}}=\sqrt{\frac{120}{480}}$
$=\sqrt{\frac{1}{4}}=\frac{1}{2}$
$\Rightarrow \frac{v}{v_{2}}=\frac{1}{2}$
$\Rightarrow v_{2}=2 \,v$