Step 1: Recall the Carnot efficiency.
The efficiency of a Carnot engine is \[ \eta_{\text{Carnot}} = 1 - \frac{T_c}{T_h}, \] which depends only on the absolute temperatures \(T_h\) (source) and \(T_c\) (sink). Hence statement (B), which claims independence from these temperatures, is not correct.
Step 2: Reversibility and optimality.
A Carnot engine is by definition a reversible engine and, for given \(T_h\) and \(T_c\), has the maximum possible efficiency among all engines. Therefore, statements (A) and (C) are correct.
Step 3: Entropy change over a reversible cycle.
For a completely reversible cycle involving the engine and the two reservoirs, the total entropy change of the combined system (engine + reservoirs) over a full cycle is \[ \Delta S_{\text{total}} = 0. \] Thus the total entropy does not increase at the end of each cycle. Hence statement (D) is not correct.
Final Answer: \[ \boxed{\;\; \text{Correct statements: (A) and (C)} \;} \]

The figures I, II, and III are parts of a sequence. Which one of the following options comes next in the sequence at IV?

A color model is shown in the figure with color codes: Yellow (Y), Magenta (M), Cyan (Cy), Red (R), Blue (Bl), Green (G), and Black (K). Which one of the following options displays the color codes that are consistent with the color model?

Consider a process with transfer function: \[ G_p = \frac{2e^{-s}}{(5s + 1)^2} \] A first-order plus dead time (FOPDT) model is to be fitted to the unit step process reaction curve (PRC) by applying the maximum slope method. Let \( \tau_m \) and \( \theta_m \) denote the time constant and dead time, respectively, of the fitted FOPDT model. The value of \( \frac{\tau_m}{\theta_m} \) is __________ (rounded off to 2 decimal places).
Given: For \( G = \frac{1}{(\tau s + 1)^2} \), the unit step output response is: \[ y(t) = 1 - \left(1 + \frac{t}{\tau}\right)e^{-t/\tau} \] The first and second derivatives of \( y(t) \) are: \[ \frac{dy(t)}{dt} = \frac{t}{\tau^2} e^{-t/\tau} \] \[ \frac{d^2y(t)}{dt^2} = \frac{1}{\tau^2} \left(1 - \frac{t}{\tau}\right) e^{-t/\tau} \]