The number of moles (n) is given by: $n = \frac{mass (g)}{molar mass (g/mol)}$. The number of molecules is given by $N = nN_A$, where $N_A$ is Avogadro's number.
For 1.0 g of H$_2$: Molar mass of H$_2$ = 2.016 g/mol Moles of H$_2 = \frac{1.0 g}{2.016 g/mol} \approx 0.496$ mol
Number of molecules = 0.496 $\times$ $N_A$
Let's check the options:
(1) 14 g of N$_2$: Molar mass of N$_2$ = 28 g/mol Moles of N$_2 = \frac{14 g}{28 g/mol} = 0.5$ mol Number of molecules = 0.5 $\times$ $N_A$
Thus, 14 g of N$_2$ has approximately the same number of molecules as 1 g of H$_2$.
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 :