Analyzing the lone pairs of electrons in the given molecules.
\(H_2O\): Oxygen has 2 lone pairs, and each hydrogen atom has 0 lone pairs. Total = 2 lone pairs.
\(N_2\): Nitrogen in \(N_2\) has no lone pairs in the molecular structure. Total = 0 lone pairs.
\(CO\): Carbon in CO has no lone pairs, but oxygen has 2 lone pairs. Total = 2 lone pairs.
\(XeF_4\): Xenon in \(XeF_4\) has 2 lone pairs, and each fluorine atom has 3 lone pairs. Total = 2 lone pairs.
\(NH\_3\): Nitrogen in \(NH_3\) has 1 lone pair, and each hydrogen atom has 0 lone pairs. Total = 1 lone pair.
\(NO\): Nitrogen in NO has 1 lone pair, and oxygen has 2 lone pairs. Total = 3 lone pairs.
\(CO_2\): Carbon in \(CO_2\) has no lone pairs, and oxygen has 2 lone pairs on each oxygen atom. Total = 4 lone pairs.
\(F_2\): Each fluorine atom in \(F_2\) has 3 lone pairs. Total = 3 lone pairs. From the analysis, the molecules that contain only 2 lone pairs are \(H_2O\), \(CO\), and \(XeF_4\).
Therefore, the correct answer is 4 molecules.
In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 
Two circular discs of radius \(10\) cm each are joined at their centres by a rod, as shown in the figure. The length of the rod is \(30\) cm and its mass is \(600\) g. The mass of each disc is also \(600\) g. If the applied torque between the two discs is \(43\times10^{-7}\) dyne·cm, then the angular acceleration of the system about the given axis \(AB\) is ________ rad s\(^{-2}\).
