JEE Main 2023 Question Paper (06-Apr-2023) (Shift 2)

Question 1

The work functions of Aluminium and Gold are $4.1 \quad \mathrm{eV}$ and $5.1 \quad \mathrm{eV}$ respectively. The ratio of the slope of the stopping potential versus frequency plot for Gold to that of Aluminium is

Accepted Answer

$1$

Answer

$1.24$

Answer

$2$

Answer

$1.5$

Question 2

The weight of a body on the surface of the Earth is $100 \quad \mathrm{N}$. The gravitational force on it when taken at a height from the surface of the Earth equal to one-fourth the radius of the Earth is:

Accepted Answer

$64 \quad \mathrm{N}$

Answer

$25 \quad \mathrm{N}$

Answer

$50 \quad \mathrm{N}$

Answer

$100 \quad \mathrm{N}$

Question 3

A 2 meter long scale with least count of $0.2 \quad \mathrm{cm}$ is used to measure the locations of objects on an optical bench. While measuring the focal length of a convex lens, the object pin and the convex lens are placed at $80 \quad \mathrm{cm}$ mark and $1 \quad \mathrm{m}$ mark, respectively. The image of the object pin on the other side of the lens coincides with the image pin that is kept at $180 \quad \mathrm{cm}$ mark. The $\%$ error in the estimation of focal length is:

Accepted Answer

$1.70$

Answer

$0.85$

Answer

$1.02$

Answer

$0.51$

Question 4

A capacitor of capacitance $150.0\quad \mu \mathrm{F}$ is connected to an alternating source of emf given by $E = 36 \quad \mathrm{sin}\left(120\pi t\right) \quad \mathrm{V}$. The maximum value of current in the circuit is approximately equal to:

Accepted Answer

$2 \quad \mathrm{A}$

Answer

$ \sqrt{2} \quad \mathrm{A} $

Answer

$2\sqrt{2} \quad \mathrm{A}$

Answer

$ \frac{1}{\sqrt{2}} \quad \mathrm{A} $

Question 5

The energy density associated with the electric field $\overrightarrow{E}$ and magnetic field $\overrightarrow{B}$ of an electromagnetic wave in free space is given by ($\epsilon_0$ — permittivity of free space, $\mu_0$ — permeability of free space)

Accepted Answer

${U}_{E}=\frac{\epsilon_{0} E^{2}}{2}, \quad {U}_{B}=\frac{B^{2}}{2 \mu_{0}}$

Answer

${U}_{E}=\frac{E^{2}}{2\epsilon_{0}}, \quad {U}_{B}=\frac{B^{2}}{2\mu_{0}}$

Answer

${U}_{E}=\frac{\epsilon_{0} E^{2}}{2}, \quad {U}_{B}=\frac{\mu_{0} B^{2}}{2}$

Answer

${U}_{E}=\frac{E^{2}}{2\epsilon_{0}}, \quad {U}_{B}=\frac{\mu_{0} B^{2}}{2}$