Net electric flux through a cube is the sum of fluxes through its six faces. Consider a cube as shown in figure,having sides of length L = 10.0 cm. The electric field is uniform, has a magnitude E = 4.00 × 103NC-1 and isparallel to the xy plane at an angle of 37o measured from the +x -axis towards the +y -axis.(i) Electric flux passing through surface S6 isa) -24 Nm2 C-1 b) 32 Nm2 C-1 c) -32 Nm2 C-1 d) 24 Nm2 C-1(ii) Electric flux passing through surface S1 isa) -32 Nm2 C-1 b) -24 Nm2 C-1 c) 32 Nm2 C-1 d) 24 Nm2 C-1(iii) The surfaces that have zero flux area) S2 and S4 b) S3 and S6 c) S1 and S2 d) S1 and S3(iv) The total net electric flux through all faces of the cube isa) 24 Nm2 C-1 b) 8 Nm2 C-1 c) -8 Nm2 C-1 d) zero30. Read the text carefully and answer the question.A transformer is essentially an a c. device. lt cannot work on d.c. It changes alternating voltages or currents.It does not affect the frequency of a.c. It is based on the phenomenon of mutual induction. A transformeressentially consists of two coils of insulated copper wire having a different number of turns and wound onthe same soft iron core.The number of turns in the primary and secondary coils of an ideal transformer are 2000 and 50 respectively.The primary coil is connected to the main supply of 120 V and secondary coil is connected to a bulb ofresistance 0.6 Ω.1. What will be the value of voltage across the secondary coil'?2. Find the value of the current in the bulb.3. What Will be the value of current in the primary coil'?ORCalculate the power in primary coi

The question seems to be cut off at the end, but I'll answer what I can.

(i) The electric flux through a surface is given by the dot product of the electric field and the area vector. For surface S6, the area vector points in the -z direction, while the electric field lies in the xy plane. Therefore, the angle between them is 90 degrees, and the dot product (and hence the flux) is zero. So none of the given options are correct.

(ii) For surface S1, the area vector points in the -x direction. The angle between the electric field and the area vector is 180 - 37 = 143 degrees. The flux is then E * A * cos(143), which gives -32 Nm2 C-1. So the correct answer is (b).

(iii) The surfaces that have zero flux are those where the electric field is perpendicular to the area vector. This is the case for surfaces S2 and S4, so the correct answer is (a).

(iv) The total net electric flux through a cube is zero, because the electric field lines that enter the cube also exit it. So the correct answer is (d).

For the transformer question:

1. The voltage across the secondary coil can be found using the transformer equation Vp/Vs = Np/Ns, where Vp and Vs are the primary and secondary voltages, and Np and Ns are the number of turns in the primary and secondary coils. Plugging in the given values gives Vs = Vp * Ns / Np = 120V * 50 / 2000 = 3V.

2. The current in the bulb can be found using Ohm's law I = V/R, where V is the voltage across the bulb and R is its resistance. Plugging in the given values gives I = 3V / 0.6Ω = 5A.

3. The current in the primary coil can be found using the transformer equation Ip/Is = Ns/Np, where Ip and Is are the primary and secondary currents. Plugging in the given values gives Ip = Is * Np / Ns = 5A * 2000 / 50 = 200A.