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:
Since both branches are identical, the phase difference between \( V_A \) and \( V_B \) and \( V_{in} \) are the same, but in opposite directions. Thus, the phase difference between \( V_{in} \) and \( V_A \) must be \( 45^\circ \), as \( V_{in} \) and \( |V_A - V_D| \) have a difference of \( 90^\circ \). Now, clearly: \[ |R| = (xc) \] Given: \[ 100 \times 10^3 = \frac{10^{12}}{w \times 100} \] Solving for \(w\): \[ w = 10^5 \, \text{rad/s} \] \[ \boxed{w = 10^5 \, \text{rad/s}} \]
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,

An alternating current is represented by the equation, $\mathrm{i}=100 \sqrt{2} \sin (100 \pi \mathrm{t})$ ampere. The RMS value of current and the frequency of the given alternating current are

What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)