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?

The group two or alkaline earth metals are s-block elements with two electrons in their s-orbital. They are alkaline earth metals. They are named so because of the alkaline nature of the hydroxides and oxides.
Alkaline earth metals are characterized by two s-electrons. This group of elements includes:
Elements whose atoms have their s-subshell filled with their two valence electrons are called alkaline earth metals. Their general electronic configuration is [Noble gas] ns2. They occupy the second column of the periodic table and so-called as group two metals also.