In amplitude modulation (AM), the modulation index (\(\mu\)) describes the extent of amplitude variation relative to the unmodulated carrier.
It is defined in terms of the maximum amplitude (\(A_{max}\)) and minimum amplitude (\(A_{min}\)) of the modulated wave as follows: \[ \mu = \frac{A_{max} - A_{min}}{A_{max} + A_{min}} \] Given values from the problem statement are:
- Maximum amplitude, \(A_{max} = 16 \, V\)
- Minimum amplitude, \(A_{min} = 4 \, V\)
Substituting these values into the modulation index formula: \[ \mu = \frac{16 \, V - 4 \, V}{16 \, V + 4 \, V} = \frac{12 \, V}{20 \, V} = \frac{12}{20} = \frac{3}{5} = 0.6 \] The modulation index is 0.6. Comparing this with the given options, option (3) matches our calculated value.
Correct Answer: (3) 0.6
Young double slit arrangement is placed in a liquid medium of 1.2 refractive index. Distance between the slits and screen is 2.4 m.
Slit separation is 1 mm. The wavelength of incident light is 5893 Å. The fringe width is:
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: \( F = 96500 { C/mol}\), \( n = 2 \))