The problem asks to match the laws of electromagnetism (List-I) with their corresponding equations (List-II). These laws are derived from Maxwell’s equations.
Gauss’s Law of Electrostatics states that the total electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space. The corresponding equation is:
\[ \oint \mathbf{E} \cdot d\mathbf{a} = \frac{1}{\epsilon_0} \int \rho dV. \]
This matches D – I.
Faraday’s Law of Electromagnetic Induction states that the electromotive force induced in a closed loop is equal to the negative rate of change of the magnetic flux through the loop. The corresponding equation is:
\[ \oint \mathbf{E} \cdot d\mathbf{l} = - \frac{d}{dt} \int \mathbf{B} \cdot d\mathbf{a}. \]
This matches B – III.
Ampere’s Law (with Maxwell’s correction) relates the line integral of the magnetic field around a closed loop to the current passing through the loop and the displacement current. For steady currents, the corresponding equation is:
\[ \oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I. \]
This matches C – IV.
Gauss’s Law for Magnetostatics states that the net magnetic flux through a closed surface is zero, indicating there are no magnetic monopoles. The corresponding equation is:
\[ \oint \mathbf{B} \cdot d\mathbf{a} = 0. \]
This matches A – II.
Final Matching:
A – II, B – III, C – IV, D – I
A battery of emf \( E \) and internal resistance \( r \) is connected to a rheostat. When a current of 2A is drawn from the battery, the potential difference across the rheostat is 5V. The potential difference becomes 4V when a current of 4A is drawn from the battery. Calculate the value of \( E \) and \( r \).
Let $ f: \mathbb{R} \to \mathbb{R} $ be a twice differentiable function such that $$ f''(x)\sin\left(\frac{x}{2}\right) + f'(2x - 2y) = (\cos x)\sin(y + 2x) + f(2x - 2y) $$ for all $ x, y \in \mathbb{R} $. If $ f(0) = 1 $, then the value of $ 24f^{(4)}\left(\frac{5\pi}{3}\right) $ is: