The modulation index \( m \) is given by the ratio of the peak voltage of the modulating signal \( V_m \) to the peak voltage of the carrier wave \( V_c \): \[ m = \frac{V_m}{V_c} \] Given that \( m = 0.75 \) and \( V_c = 16 \, {V} \), we can solve for \( V_m \): \[ 0.75 = \frac{V_m}{16} \] \[ V_m = 0.75 \times 16 = 12 \, {V} \] Thus, the peak voltage of the modulating signal is 12 V.
Final Answer: 12 V.
Identify the correct truth table of the given logic circuit. 
Find the correct combination of A, B, C and D inputs which can cause the LED to glow. 
Select correct truth table. 

Which of the following are ambident nucleophiles?
[A.] CN$^{\,-}$
[B.] CH$_{3}$COO$^{\,-}$
[C.] NO$_{2}^{\,-}$
[D.] CH$_{3}$O$^{\,-}$
[E.] NH$_{3}$
Identify the anomers from the following.

The standard Gibbs free energy change \( \Delta G^\circ \) of a cell reaction is \(-301 { kJ/mol}\). What is \( E^\circ \) in volts?
(Given: \( F = 96500 { C/mol}\), \( n = 2 \))