JEE First – Page 7 – Physics concepts and problems for the JEE

Circular orbit in a harmonic potential

January 21st, 2022jeefirst0 Comments

JEE Advanced 2018 Paper 1, Question 1

The potential energy of a particle of mass $m$ at a distance $r$ from a fixed point $O$ is given by $V(r)=k r^{2} / 2$ , where $k$ is a positive constant of appropriate dimensions. This particle is moving in a circular orbit of radius $R$ about the point $O$ . If $v$ is the speed of the particle and $L$ is the magnitude of its angular momentum about $O$ , which of the following statements is (are) true?

$v=\sqrt{\frac{k}{2 m}} R$
$v=\sqrt{\frac{k}{m}} R$
$L=\sqrt{m k} R^{2}$
$L=\sqrt{\frac{m k}{2}} R^{2}$

Solution

The force due to the given potential is

(1) $\begin{equation*} {\bf F} = -\frac{\partial V}{\partial r} \hat{\bf r} = -k r \hat{\bf r} . \end{equation*}$

The mass also experiences a centrifugal force due to its circular motion,

(2) $\begin{equation*} {\bf F}_{\rm cent} = \frac{m v^2}{r} \hat{\bf r} \end{equation*}$

For the …

Total internal reflection from a prism

January 20th, 2022jeefirst0 Comments

JEE Advanced 2019 Paper 2, Question 11

A monochromatic light is incident from air on a refracting surface of a prism of angle $75^{\circ}$ and refractive index $n_{0}=\sqrt{3}$ . The other refracting surface of the prism is coated by a thin film of material of refractive index $n$ as shown in figure. The light suffers total internal reflection at the coated prism surface for an incidence angle of $\theta \leq 60^{\circ}$ . What is the value of $n^{2}$ ?

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Solution

First we trace the given incident ray through the prism to find the angle at which it is incident on the opposite face (see figure below). For …

A spherical capacitor

January 20th, 2022jeefirst0 Comments

A spherical conducting shell with radius $b$ is concentric with a conducting ball with radius $a$ , with $a<b$ .

Compute the capacitance $C = Q / \Delta \phi$ when the shell is grounded and the ball has charge $Q$ .
Compute the capacitance when the ball is grounded and the shell has charge $Q$ .
Compute the full matrix of coefficients of capacitance for the two conductors.
Considering these conductors as a capacitor, determine its capacitance. That is, assign equal and opposite charges $\pm Q$ to the shell and the ball, and compute $C = Q / \Delta \phi$ .

Related Problem: Insulating spherical shell with a hole

Solution

(a) First, we ground the shell and give the …

Mirror on a spring

January 19th, 2022jeefirst0 Comments

JEE Advanced 2019 Paper 2, Question 12

A perfectly reflecting mirror of mass $M$ mounted on a spring constitutes a spring-mass system of angular frequency $\Omega$ such that $\frac{4 \pi M \Omega}{h} = 10^{24} \, {\rm m}^{-2}$ with $h$ as Planck’s constant. $N$ photons of wavelength $\lambda = 8 \pi \times 10^{-6} \, {\rm m}$ strike the mirror simultaneously at normal incidence such that the mirror gets displaced by $1 \mu{\rm m}$ .
If the value of $N$ is $x \times 10^{12}$ , what is the value of $x$ ?

[Consider the spring as massless]

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Solution

The momentum carried by a single photon is given by the de Broglie relation $p_{\rm ph} = h/\lambda$ . The total momentum carried by $N$ photons is therefore $N \times p_{\rm ph} = Nh/\lambda$ . All the photons hit the mirror simultaneously and are reflected …

Dipole in a uniform electric field

January 18th, 2022jeefirst0 Comments

JEE Advanced 2019 Paper 2, Question 4

An electric dipole with dipole moment $\frac{p_{0}}{\sqrt{2}}(\hat{i}+\hat{j})$ is held fixed at the origin $O$ in the presence of an uniform electric field of magnitude $E_{0}$ . If the potential is constant on a circle of radius $R$ centered at the origin as shown in figure, then the correct statement(s) is/are:

( $\varepsilon_{0}$ is permittivity of free space. $R \gg$ dipole size)

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$R=\left(\frac{p_{0}}{4 \pi \epsilon_{0} E_{0}}\right)^{1 / 3}$
Total electric field at point $A$ is ${\bf E}^A=\sqrt{2} E_{0}(\hat{i}+\hat{j})$
Total electric field at point $B$ is ${\bf E}^B=0$
The magnitude of total electric field on any two points of the circle will be same.

Solution

The potential due to the dipole kept at the origin …