The voltage across an inductor ' L ' is
V = L · dI/dt .
I(t) = I(max) sin(ωt)
dI/dt = I(max) ω cos(ωt)
V = L · ω · I(max) cos(ωt)
L = 1.34 x 10⁻² H
ω = 2π · 60 = 377 /sec
I(max) = 4.80 A
V = L · ω · I(max) cos(ωt)
V = (1.34 x 10⁻² H) · (377 / sec) · (4.8 A) · cos(377 t)
<em>V = 24.25 cos(377 t)</em>
V is an AC voltage with peak value of 24.25 volts and frequency = 60 Hz.
The expression of the electric flux is

Here,
Q = Total charge enclosed in the closed surface
= Permittivity due to free space
Rearranging to find the charge,

Replacing with our values we have finally



The charge enclosed by the box is 0.1684nC
The sign of the charge can be decided by using the direction of the flux. The charge enclosed by the cube can be calculated by using the electric flux and the permitivity of free space.
In a string of length L, the wavelength of the n-th harmonic of the standing wave produced in the string is given by:

The length of the string in this problem is L=3.5 m, therefore the wavelength of the 1st harmonic of the standing wave is:

The wavelength of the 2nd harmonic is:

The wavelength of the 4th harmonic is:

It is not possible to find any integer n such that
, therefore the correct options are A, B and D.