A point charge is placed just outside an imaginary hemispherical surface of radius as shown in the figure. Which of the following statements is/are correct?
The electric flux passing through the curved surface of the hemisphere is
Total flux through the curved and the flat surfaces is
The component of the electric field normal to the flat surface is constant over the surface
The circumference of the flat surface is an equipotential
One mole of a monatomic ideal gas is taken along two cyclic processes and as shown in the PV diagram. The processes involved are purely isochoric, isobaric, isothermal or adiabatic.
Match the paths in List I with the magnitudes of the work done in List II and select the correct answer using the codes given below the lists.
In this note we discuss isothermal, adiabatic, isobaric, and isochoric processes with an ideal gas. We begin by recalling a few basic principles from thermodynamics, and their application to a container of ideal gas, as shown in the figure below.
When the gas is at an absolute temperature , the molecules of the gas move around inside the container and bounce off its walls. Each gas molecule suffers a change in momentum upon such a collision, and the totality of all those collisions per unit time exerts a force on the walls of the container. …
A monochromatic light is traveling in a medium of refractive index . It enters a stack of glass layers from the bottom side at an angle . The interfaces of the glass layers are parallel to each other. The refractive indices of different glass layers are monotonically decreasing as , where is the refractive index of the slab and (see the figure). The ray is refracted out parallel to the interface between the and slabs from the right side of the stack. What is the value of ?
For the resistance network shown in the figure, choose the correct option(s).
The current through is zero.
A.
The potential at is less than that at .
A.
Solution
We can speed up the solution by noticing the mirror symmetry of the circuit. Let the current split into two parts and at the node (see figure). Similarly, let the currents going into node be and . If we were to reverse the polarity of the battery connected to the circuit, all currents would simply reverse direction, but their magnitudes would remain the …