$(a)\;1.3m/s\qquad(b)\;0.6m/s\qquad(c)\;2.1m/s\qquad(d)\;1.8m/s$

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$U_{AV} = \sqrt{(\large\frac{8RT}{\pi m})}$

Given

$U_{AV} = 0.3m/s$ at 300K

$\therefore U_1 = 0.3 = \sqrt{(\large\frac{8R\times300}{\pi m})}$-------(1)

At T = 273+927 = 1200K

$\therefore U_2 = \sqrt{(\large\frac{8R\times1200}{\pi m})}$-------(2)

$\therefore \large\frac{U_2}{0.3} = \sqrt{(\large\frac{1200}{300})}$

$\Rightarrow U_2 = 0.6m/s$

Hence answer is (b)

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