All Exams Test series for 1 year @ ₹349 only

JEE Main 2023 Question Paper with Solutions (13-Apr-2023) (Shift 1); Download PDF

JEE Main 2023 Question Paper with Solutions for Apr 13, 2023 Shift 1 is available here for PDF download and can also be attempted in a test format for practice. JEE Main Paper 1 (B.E./B.Tech) is conducted in CBT mode and includes Physics, Chemistry, and Mathematics.

Free
JEE Main 2023 Question Paper (13-Apr-2023) (Shift 1)
180 Minutes
90 Questions
300 Marks
English
Showing 1 - 5 of 20 questions
Page 1 of 4

Q1.

A disc is rolling without slipping on a surface. The radius of the disc is \( R \). At \( t = 0 \), the topmost point on the disc is \( A \) as shown in the figure. When the disc completes half of its rotation, the displacement of point \( A \) from its initial position is

Q2.

Different combinations of $3$ resistors of equal resistance $R$ are shown in the figures. The increasing order for power dissipation is:

(A)

(B)

(C)

(D)

Q3.

A planet having mass $9 \quad {\mathrm{M}}_{\mathrm{e}}$ and radius $4 {\mathrm{R}}_{\mathrm{e}},$ where ${\mathrm{M}}_{\mathrm{e}}$ and ${\mathrm{R}}_{\mathrm{e}}$ are the mass and radius of Earth respectively, has escape velocity in $\mathrm{km} \quad \mathrm{s}^{-1}$ given by: (Given escape velocity on Earth $V_{\mathrm{e}} = 11.2 \times 10^{3} \quad \mathrm{m} \quad \mathrm{s}^{-1}$)

Q4.

${}_{92}^{238}\mathrm{A} \to {}_{90}^{234}\mathrm{B} + {}_{2}^{4}\mathrm{D} + \mathrm{Q}$

In the given nuclear reaction, the approximate amount of energy released will be:

[Given, mass of ${}_{92}^{238}\mathrm{A} = 238.05079 \times 931.5 \quad \mathrm{MeV} \quad \mathrm{c}^{-2}$, mass of ${}_{90}^{234}\mathrm{B} = 234.04363 \times 931.5 \quad \mathrm{MeV} \quad \mathrm{c}^{-2}$, mass of ${}_{2}^{4}\mathrm{D} = 4.00260 \times 931.5 \quad \mathrm{MeV} \quad \mathrm{c}^{-2}$]

Q5.

The difference between threshold wavelengths for two metal surfaces $A$ and $B$ having work functions ${\varphi}_{A} = 9\,\mathrm{eV}$ and ${\varphi}_{B} = 4.5\,\mathrm{eV}$ in $\mathrm{nm}$ is:

Given, $hc = 1242 \mathrm{eV} \cdot \mathrm{nm}$

Page 1 of 4
Need Expert Advice?