Remember that the relationship between rms speed and average speed for molecules can involve constants such as \( \pi \), and care should be taken when substituting numerical values.
To find the value of \( x \) such that the root mean square (rms) speed of an oxygen molecule in a vessel at a particular temperature is given by \((1+\frac{5}{x})^{\frac{1}{2}} v\), where \( v \) is the average speed, let's follow these steps:
We know the relation between the rms speed \((v_{\text{rms}})\), the average speed \((v_{\text{avg}})\), and the most probable speed \((v_{\text{mp}})\) for any gas:
where \(k\) is the Boltzmann constant, \(T\) is the temperature, and \(m\) is the mass of a molecule.
Given the rms speed of an oxygen molecule is \((1+\frac{5}{x})^{\frac{1}{2}} v\).
Express the given relation in terms of known speed ratios:
\(v_{\text{rms}} = \sqrt{\frac{3}{8/\pi}} v_{\text{avg}}\) because the ratio \(\frac{v_{\text{rms}}}{v_{\text{avg}}} = \sqrt{\frac{3\pi}{8}}\).
This implies:
\((1+\frac{5}{x})^{\frac{1}{2}} v = \sqrt{\frac{3\pi}{8}} v\)
Since \(v\) cancels from both sides, equate and solve:
\((1+\frac{5}{x})^{\frac{1}{2}} = \sqrt{\frac{3\pi}{8}}\)
By squaring both sides, we get:
\(1+\frac{5}{x} = \frac{3\pi}{8}\)
Using \(\pi = \frac{22}{7}\), substitute into the equation:
\(1+\frac{5}{x} = \frac{3 \times \frac{22}{7}}{8}\)
\(1+\frac{5}{x} = \frac{66}{56}\)
\(1+\frac{5}{x} = \frac{33}{28}\)
Rearrange the equation for \(x\):
\(\frac{5}{x} = \frac{33}{28} - 1\)
\(\frac{5}{x} = \frac{33}{28} - \frac{28}{28}\)
\(\frac{5}{x} = \frac{5}{28}\)
Solving for \(x\), we get:
\(x = 28\)
Therefore, the value of \( x \) is 28, which is the correct option.
A body of mass 1000 kg is moving horizontally with a velocity of 6 m/s. If 200 kg extra mass is added, the final velocity (in m/s) is: