
For the AC circuit shown in the figure, $ R = 100 \, \text{k}\Omega $ and $ C = 100 \, \text{pF} $, and the phase difference between $ V_{\text{in}} $ and $ (V_B - V_A) $ is 90°. The input signal frequency is $ 10^x $ rad/sec, where $ x $ is:
Two parabolas have the same focus $(4, 3)$ and their directrices are the $x$-axis and the $y$-axis, respectively. If these parabolas intersect at the points $A$ and $B$, then $(AB)^2$ is equal to:
A point particle of charge \( Q \) is located at \( P \) along the axis of an electric dipole 1 at a distance \( r \) as shown in the figure. The point \( P \) is also on the equatorial plane of a second electric dipole 2 at a distance \( r \). The dipoles are made of opposite charge \( q \) separated by a distance \( 2a \). For the charge particle at \( P \) not to experience any net force, which of the following correctly describes the situation?

Hydrogen bonding implies the formation of hydrogen bonds which are an attractive intermolecular force. An example of hydrogen bonding is the bond between the H atom and the O atom in water.
A special type of intermolecular attractive force arises only in the compounds having Hydrogen atoms bonded to an electronegative atom. This force is known as the Hydrogen bond. For instance, in water molecules, the hydrogen atom is bonded to a highly electronegative Oxygen.
Association: The molecules of carboxylic acids exist as dimer because of the hydrogen bonding. The molecular masses of such compounds are found to be double than those calculated from their simple formula.
Dissociation: In aqueous solution, HF dissociates and gives the difluoride ion instead of fluoride ion. This is due to hydrogen bonding in HF. The molecules of HCl, HBr, HI do not form a hydrogen bond. This explains the non-existence of compounds like KHCl2, KHBr2, KHI2.