Let $ A = \begin{bmatrix} 2 & 2 + p & 2 + p + q \\4 & 6 + 2p & 8 + 3p + 2q \\6 & 12 + 3p & 20 + 6p + 3q \end{bmatrix} $ If $ \text{det}(\text{adj}(\text{adj}(3A))) = 2^m \cdot 3^n, \, m, n \in \mathbb{N}, $ then $ m + n $ is equal to:
Two projectile protons \( P_1 \) and \( P_2 \), both with spin up (along the \( +z \)-direction), are scattered from another fixed target proton \( T \) with spin up at rest in the \( xy \)-plane, as shown in the figure. They scatter one at a time. The nuclear interaction potential between both the projectiles and the target proton is \( \hat{\lambda} \vec{L} \cdot \vec{S} \), where \( \vec{L} \) is the orbital angular momentum of the system with respect to the target, \( \vec{S} \) is the spin angular momentum of the system, and \( \lambda \) is a negative constant in appropriate units. Which one of the following is correct?

The figure shows an opamp circuit with a 5.1 V Zener diode in the feedback loop. The opamp runs from \( \pm 15 \, {V} \) supplies. If a \( +1 \, {V} \) signal is applied at the input, the output voltage (rounded off to one decimal place) is:

A wheel of mass \( 4M \) and radius \( R \) is made of a thin uniform distribution of mass \( 3M \) at the rim and a point mass \( M \) at the center. The spokes of the wheel are massless. The center of mass of the wheel is connected to a horizontal massless rod of length \( 2R \), with one end fixed at \( O \), as shown in the figure. The wheel rolls without slipping on horizontal ground with angular speed \( \Omega \). If \( \vec{L} \) is the total angular momentum of the wheel about \( O \), then the magnitude \( \left| \frac{d\vec{L}}{dt} \right| = N(MR^2 \Omega^2) \). The value of \( N \) (in integer) is:
