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Two molecules of a gas have speeds of 9 \times 10^{6} \mathrm{~m} / \mathrm{s}$ and $1 \times 10^{6} \mathrm{~m} / \mathrm{s} , respectively. What is the root mean square speed of these molecules?

Option: 1

\sqrt{39 \times 10^{6}} \mathrm{~m} / \mathrm{s}


Option: 2

6 \times 10^{6} \mathrm{~m} / \mathrm{s}


Option: 3

\sqrt{41 \times 10^{6} \mathrm{~m} / \mathrm{s}}


Option: 4

8 \times 10^{6} \mathrm{~m} / \mathrm{s}


Answers (1)

best_answer

By definition,

v_{\mathrm{rms}}=\sqrt{\frac{v_{1}^{2}+v_{2}^{2}}{2}}$, where $v_{1}$ and $v_{2} are the individual velocities of the two particles.

\text{Given,} \quad v_{1}=9 \times 10^{6} \mathrm{~m} / \mathrm{s}
and             \mathrm{v_{2} =1 \times 10^{6} \mathrm{~m} / \mathrm{s}}

\mathrm{\therefore v_{\text {rms }} =\sqrt{\frac{\left(9 \times 10^{6}\right)^{2}+\left(1 \times 10^{6}\right)^{2}}{2}}}
\mathrm{=\sqrt{\frac{81 \times 10^{12}+1 \times 10^{12}}{2}}}

\mathrm{ =\sqrt{\frac{(81+1) \times 10^{12}}{2}}=\sqrt{\frac{82 \times 10^{12}}{2}} }
\mathrm{=\sqrt{41} \times 10^{6} \mathrm{~m} / \mathrm{s}}

Posted by

sudhir.kumar

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