$\begin {array}{1,1}(a)\;7\times10^4cm\;sec^{-1}\\(b)\;8.028\times10^4cm\;sec^{-1}\\(c)\;6\times10^5cm\;sec^{-1}\\(d)\;8.028\times10^6cm\;sec^{-1} \end{array}$

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Given

Pressure = 3atm

Volume = 12.5 litre

Weight = 15 g

For gas We have PV = $\large\frac{w}{m}RT$

$3\times12.5 = \large\frac{15}{m}\times 0.0821\times T$

$\large\frac{T}{m} = 30.45$

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

$ = \sqrt{\large\frac{8\times 8.314\times 10^7\times 30.4\times 7}{22}}$

$ = 8.028\times10^4 cm\;Sec^{-1}$

Hence answer is (b)

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