The correct answer is: 2
Power gain
\(=[\frac{\triangle{ic}}{\triangle{iB}}]×\frac{R_0}{R_i}\)
\(=[\frac{10^{-2}}{10^{-4}}]×\frac{2}{1}\)
= 2 × 104
\(⇒ x = 2\)
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,

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 : \(A_2 \rightleftharpoons 2A\)
According to Ampere’s law, magnetic fields are related to the electric current that is produced in them. This law specifies that the magnetic field is associated with a given current or vice-versa, provided that the electric field doesn’t change with time.
Ampere’s circuital law can be written as the line integral of the magnetic field surrounding the closed loop which is equal to the number of times the algebraic sum of currents passing through the loop.

According to the second equation, if the magnetic field is integrated along the blue path, then it is equal to the current enclosed, I.
The magnetic field doesn’t vary at a distance r because of symmetry. The path length (in blue) in figure 1 has to be equal to the circumference of a circle,2πr.
