The comparison of boiling points among organic compounds requires an understanding of intermolecular forces. In the case of aldehydes, ketones, hydrocarbons, and alcohols, these forces play a crucial role in determining their boiling points.
Statement I Explanation: Aldehydes and ketones possess a carbonyl group (C=O) that induces polarity within the molecule, leading to dipole-dipole interactions. Hydrocarbons, on the other hand, are generally non-polar and exhibit weaker London dispersion forces. As a result, aldehydes and ketones have higher boiling points compared to analogous hydrocarbons due to these dipole-dipole interactions facilitating a modest degree of molecular association.
Statement II Explanation: Alcohols contain an -OH group that can engage in hydrogen bonding, a significant intermolecular force that raises boiling points. Although aldehydes and ketones engage in dipole-dipole interactions, these are weaker than hydrogen bonds. Consequently, aldehydes and ketones show lower boiling points than alcohols of comparable molecular mass since they lack the ability to form hydrogen bonds.
Based on these explanations, both Statement I and Statement II correctly describe the reasons for the observed boiling point trends among these compounds. Therefore, the correct answer is: Both Statement I and Statement II are correct.
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 :
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