JEE First – Page 4 – Physics concepts and problems for the JEE

Electric flux through a hemisphere

March 24th, 2022jeefirst0 Comments

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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The electric flux passing through the curved surface of the hemisphere is $-\frac{Q}{2 \varepsilon_{0}}\left(1-\frac{1}{\sqrt{2}}\right)$
Total flux through the curved and the flat surfaces is $\frac{Q}{\varepsilon_{0}}$
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

Solution

In problems like these a beginner may …

Cyclic processes on a PV diagram

March 22nd, 2022jeefirst0 Comments

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.

$G \to E$
$G \to H$
$F \to H$
$F \to G$

$160 P_0 V_0 \ln 2$
$36 P_0 V_0$
$24 P_0 V_0$
$31 P_0 V_0$

Codes:

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

Thermodynamic processes on an ideal gas

March 19th, 2022jeefirst0 Comments

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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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. …

Progressive refraction through glass layers

March 4th, 2022jeefirst0 Comments

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 …

Resistor network with mirror symmetry

February 26th, 2022jeefirst0 Comments

JEE Advanced 2012 Paper 1, Question 12

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

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