Answer
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Hint: Mean speed $(v_{avg})$ is the sum of the speeds of individual particles in a gas divided by the total number of particles ‘n’ present in the gas. The root mean square speed $(v_{rms})$ is defined as the square root of the mean of the squares of the speeds of all particles of the same gas.
Complete step by step answer:
Speed of three particles given to us is $2u,10u{\text{ and 11u}}$
We can calculate the mean speed $({v_{avg}})$ of three particles as follows:
${v_{avg}} = \dfrac{{{u_1} + u{}_2 + {u_3}}}{n}$
Where,
$\Rightarrow {v_{avg}}$ = mean speed
$\Rightarrow$ n = number of particles =3
$\Rightarrow {u_1}$= speed of particle of 1 = $2u$
$\Rightarrow {u_2}$= speed of particle of 2 = $10u$
$\Rightarrow {u_3}$= speed of particle of 3 = $11u$
Now, substitute the given values to us and calculate the mean speed $({v_{avg}})$
$\Rightarrow {v_{avg}} = \dfrac{{2u + 10u + 11u}}{3}$
$\Rightarrow {v_{avg}} = \dfrac{{23}}{3} = 7.67u$
Thus, the mean speed $({v_{avg}})$ of three particles is $7.66u$.
Now, we have to calculate the root mean square speed $({v_{rms}})$ of three particles is as follows:
\[\Rightarrow {v_{rms}} = \sqrt {\dfrac{{{u_1}^2 + u{{{}_2}^2} + {u_3}^2}}{n}} \]
\[\Rightarrow {v_{rms}} = \sqrt {\dfrac{{{{(2u)}^2} + {{(10u)}^2} + {{(11u)}^2}}}{3}} \]
\[\Rightarrow {v_{rms}} = \sqrt {\dfrac{{(4 + 100 + 121){u^2}}}{3}} \]
\[\Rightarrow {v_{rms}} = 8.66u\]
Thus, the root mean square speed $({v_{rms}})$ of three particles is \[8.66u\]
By comparing the value of root mean square speed $(v_{rms})$ and mean speed $(v_{avg})$ we can say that root mean speed is greater than the mean speed.
Now, calculate the difference between root mean square speed $(v_{rms})$ and mean speed.
$\Rightarrow {v_{rms}} - {v_{avg}} = 8.66u - 7.66u = 1u$
Thus, the root mean square speed exceeds the mean speed by about $u$.
Thus, Option A is correct.
Note: Gas particles move randomly in all directions with different speeds. Root mean square speed velocity is the single value of the velocity of the particles. The root mean square speed is always equal or greater than the mean speed.
Complete step by step answer:
Speed of three particles given to us is $2u,10u{\text{ and 11u}}$
We can calculate the mean speed $({v_{avg}})$ of three particles as follows:
${v_{avg}} = \dfrac{{{u_1} + u{}_2 + {u_3}}}{n}$
Where,
$\Rightarrow {v_{avg}}$ = mean speed
$\Rightarrow$ n = number of particles =3
$\Rightarrow {u_1}$= speed of particle of 1 = $2u$
$\Rightarrow {u_2}$= speed of particle of 2 = $10u$
$\Rightarrow {u_3}$= speed of particle of 3 = $11u$
Now, substitute the given values to us and calculate the mean speed $({v_{avg}})$
$\Rightarrow {v_{avg}} = \dfrac{{2u + 10u + 11u}}{3}$
$\Rightarrow {v_{avg}} = \dfrac{{23}}{3} = 7.67u$
Thus, the mean speed $({v_{avg}})$ of three particles is $7.66u$.
Now, we have to calculate the root mean square speed $({v_{rms}})$ of three particles is as follows:
\[\Rightarrow {v_{rms}} = \sqrt {\dfrac{{{u_1}^2 + u{{{}_2}^2} + {u_3}^2}}{n}} \]
\[\Rightarrow {v_{rms}} = \sqrt {\dfrac{{{{(2u)}^2} + {{(10u)}^2} + {{(11u)}^2}}}{3}} \]
\[\Rightarrow {v_{rms}} = \sqrt {\dfrac{{(4 + 100 + 121){u^2}}}{3}} \]
\[\Rightarrow {v_{rms}} = 8.66u\]
Thus, the root mean square speed $({v_{rms}})$ of three particles is \[8.66u\]
By comparing the value of root mean square speed $(v_{rms})$ and mean speed $(v_{avg})$ we can say that root mean speed is greater than the mean speed.
Now, calculate the difference between root mean square speed $(v_{rms})$ and mean speed.
$\Rightarrow {v_{rms}} - {v_{avg}} = 8.66u - 7.66u = 1u$
Thus, the root mean square speed exceeds the mean speed by about $u$.
Thus, Option A is correct.
Note: Gas particles move randomly in all directions with different speeds. Root mean square speed velocity is the single value of the velocity of the particles. The root mean square speed is always equal or greater than the mean speed.
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