The correct answer is C:15bar
Partial pressure of \(N_2 =\) mole fraction \(\times P_{Total}\) of \(N_2\)
\(X_{N_2} = \frac{\text{moles of }N_2}{\text{ total moles}}\)
moles of \(N_2 = \frac{7}{28} = \frac{1}{4};\)
moles of \(Ar = \frac{8}{40} = \frac{1}{5}\)
\(X_{N_2} = \frac{(\frac{1}{4})}{\frac{1}{4} + \frac{1}{5}} = \frac{5}{9}\)
\(P_{N_2} = \frac{5}{9} \times 27 = 15\) bar
A bob of heavy mass \(m\) is suspended by a light string of length \(l\). The bob is given a horizontal velocity \(v_0\) as shown in figure. If the string gets slack at some point P making an angle \( \theta \) from the horizontal, the ratio of the speed \(v\) of the bob at point P to its initial speed \(v_0\) is :
The matter is made up of very tiny particles and these particles are so small that we cannot see them with naked eyes.
The three states of matter are as follows: