
Three particles have speeds of $2u, 3u$ and $4u$. Which of the following statements is correct?
A. The r.m.s speed exceeds the mean speed.
B. The mean speed exceeds the r.m.s speed.
C. The r.m.s speed equals the mean speed.
D. The r.m.s speed exceeds the mean speed by less than u.
Answer
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Hint: The mean speed of a group of objects is the average value of the speed of all the given particles whereas the root mean square velocity is the square root of the mean of the squares of all the speed values for all particles given. Now we simply need to find out the values of each of them and compare them for this question
Complete step by step answer:
Let’s calculate the root mean square speed for the particles. Sum of square of the speed of the particles ${\left( {2u} \right)^2} + {\left( {3u} \right)^2} + {\left( {4u} \right)^2} = 29{u^2}$
Now the root mean of the sum of square of the speed $ = \sqrt {\dfrac{{29{u^2}}}{3}} = u\sqrt {\dfrac{{29}}{3}} $
Let’s calculate the mean speed for the particles.
Mean of all the speed $ = \dfrac{{2u + 3u + 4u}}{3} = \dfrac{{9u}}{3} = 3u$
Let’s compare $\sqrt {\dfrac{{29}}{3}} $ and $3$. We know that
$\dfrac{{29}}{3} > \dfrac{{27}}{3}$
So,
$\Rightarrow \sqrt {\dfrac{{29}}{3}} > \sqrt {\dfrac{{27}}{3}} $
$\Rightarrow \sqrt {\dfrac{{29}}{3}} > 3$
Now we have to make another comparison whether the value $\sqrt {\dfrac{{29}}{3}} $ is how much greater than 3
We know that $\sqrt {16} = 4$
Now let’s compare $\sqrt {\dfrac{{29}}{3}} $ and 4
$\dfrac{{48}}{3} > \dfrac{{29}}{3}$
So,
$\sqrt {\dfrac{{48}}{3}} > \sqrt {\dfrac{{29}}{3}} $
$\Rightarrow 4 > \sqrt {\dfrac{{29}}{3}} $
Now we are sure that the r.m.s speed exceeds the mean speed by less than u from the above made calculations as the value of $\sqrt {\dfrac{{29}}{3}} < 4$.
In actual value $\sqrt {\dfrac{{29}}{3}} = 3.109$
Thus, the r.m.s speed exceeds the mean speed for the particles.
Thus, the option D is correct.
Note: The major mistakes a student makes is in the area regarding the confusion that arises in option (A) and (D) in single choice questions, if the question is multiple correct then both the options are correct. In case of single choice, we need to make a rough approximation related to the values of the mean and r.m.s speed as we did in the solution part to come up with the correct answer.
Complete step by step answer:
Let’s calculate the root mean square speed for the particles. Sum of square of the speed of the particles ${\left( {2u} \right)^2} + {\left( {3u} \right)^2} + {\left( {4u} \right)^2} = 29{u^2}$
Now the root mean of the sum of square of the speed $ = \sqrt {\dfrac{{29{u^2}}}{3}} = u\sqrt {\dfrac{{29}}{3}} $
Let’s calculate the mean speed for the particles.
Mean of all the speed $ = \dfrac{{2u + 3u + 4u}}{3} = \dfrac{{9u}}{3} = 3u$
Let’s compare $\sqrt {\dfrac{{29}}{3}} $ and $3$. We know that
$\dfrac{{29}}{3} > \dfrac{{27}}{3}$
So,
$\Rightarrow \sqrt {\dfrac{{29}}{3}} > \sqrt {\dfrac{{27}}{3}} $
$\Rightarrow \sqrt {\dfrac{{29}}{3}} > 3$
Now we have to make another comparison whether the value $\sqrt {\dfrac{{29}}{3}} $ is how much greater than 3
We know that $\sqrt {16} = 4$
Now let’s compare $\sqrt {\dfrac{{29}}{3}} $ and 4
$\dfrac{{48}}{3} > \dfrac{{29}}{3}$
So,
$\sqrt {\dfrac{{48}}{3}} > \sqrt {\dfrac{{29}}{3}} $
$\Rightarrow 4 > \sqrt {\dfrac{{29}}{3}} $
Now we are sure that the r.m.s speed exceeds the mean speed by less than u from the above made calculations as the value of $\sqrt {\dfrac{{29}}{3}} < 4$.
In actual value $\sqrt {\dfrac{{29}}{3}} = 3.109$
Thus, the r.m.s speed exceeds the mean speed for the particles.
Thus, the option D is correct.
Note: The major mistakes a student makes is in the area regarding the confusion that arises in option (A) and (D) in single choice questions, if the question is multiple correct then both the options are correct. In case of single choice, we need to make a rough approximation related to the values of the mean and r.m.s speed as we did in the solution part to come up with the correct answer.
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