Electric flux through a hemisphere

JEE Advanced 2017 Paper 2, Question 10

A point charge +Q is placed just outside an imaginary hemispherical surface of radius {\rm R} as shown in the figure. Which of the following statements is/are correct?

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  1. The electric flux passing through the curved surface of the hemisphere is -\frac{Q}{2 \varepsilon_{0}}\left(1-\frac{1}{\sqrt{2}}\right)
  2. Total flux through the curved and the flat surfaces is \frac{Q}{\varepsilon_{0}}
  3. The component of the electric field normal to the flat surface is constant over the surface
  4. The circumference of the flat surface is an equipotential

Related Problems:
Flux from a charged shell
Conducting wire in a magnetic field

Solution

In problems like these a beginner may …

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Cyclic processes on a PV diagram

JEE Advanced 2013 Paper 2, Question 20

One mole of a monatomic ideal gas is taken along two cyclic processes E \rightarrow F \rightarrow G \rightarrow E and E \rightarrow F \rightarrow H \rightarrow E as shown in the PV diagram. The processes involved are purely isochoric, isobaric, isothermal or adiabatic.

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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.

  1. G \to E
  2. G \to H
  3. F \to H
  4. F \to G
  1. 160 P_0 V_0 \ln 2
  2. 36 P_0 V_0
  3. 24 P_0 V_0
  4. 31 P_0 V_0

Codes:

    \begin{equation*}   \begin{matrix}         & {\rm P} & {\rm Q} & {\rm R} & {\rm S} \\     {\rm (A)} & 4 & 3 & 2 & 1 \\     {\rm (B)} & 4 & 3 & 1 & 2 \\     {\rm (C)} & 3 & 1 & 2 & 4 \\     {\rm (D)} & 1 & 3 & 2 & 4   \end{matrix} \end{equation*}

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Thermodynamic processes on an ideal gas

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.

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A container of ideal gas.

When the gas is at an absolute temperature T > 0, 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. …

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Progressive refraction through glass layers

JEE Advanced 2017 Paper 1, Question 10

A monochromatic light is traveling in a medium of refractive index n=1.6. It enters a stack of glass layers from the bottom side at an angle \theta=30^{\circ}. The interfaces of the glass layers are parallel to each other. The refractive indices of different glass layers are monotonically decreasing as n_{m}=n-m \Delta n, where n_{m} is the refractive index of the m^{\rm th} slab and \Delta n=0.1 (see the figure). The ray is refracted out parallel to the interface between the (m-1)^{\rm th} and m^{\rm th} slabs from the right side of the stack. What is the value of m?

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Solution

Consider the trajectory of …

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Resistor network with mirror symmetry

JEE Advanced 2012 Paper 1, Question 12

For the resistance network shown in the figure, choose the correct option(s).

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  1. The current through PQ is zero.
  2. I_1 = 3 A.
  3. The potential at S is less than that at Q.
  4. I_2 = 2 A.

Solution

We can speed up the solution by noticing the mirror symmetry of the circuit. Let the current I_1 split into two parts I_a and I_b at the node X (see figure). Similarly, let the currents going into node Y be I_a' and I_b'. If we were to reverse the polarity of the 12 \, {\rm V} battery connected to the circuit, all currents would simply reverse direction, but their magnitudes would remain the …

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