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(D) The $r.m.s.$ velocity of gas molecules is given by $V_{rms} = \sqrt{\frac{3RT}{M}}$,which implies $V_{rms} \propto \sqrt{T}$.Initial temperature $

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$10$

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(D) The $r.m.s.$ velocity of gas molecules is given by $V_{rms} = \sqrt{\frac{3RT}{M}}$,which implies $V_{rms} \propto \sqrt{T}$.Initial temperature $T_1 = 30 + 273 = 303 \ K$.Final temperature $T_2 = 90 + 273 = 363 \ K$.The ratio of velocities is $\frac{V_2}{V_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{363}{303}} \approx \sqrt{1.198} \approx 1.0945$.The percentage increase is given by $\left( \frac{V_2 - V_1}{V_1} ight) 	imes 100 = (1.0945 - 1) 	imes 100 \approx 9.45 \%$.Rounding to the nearest provided option,the percentage increase is approximately $10 \%$.

When the temperature of a gas is raised from $30^o C$ to $90^o C$,the percentage increase in the $r.m.s.$ velocity of the molecules will be (in $\%$)

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