1
JEE Main 2026 (Online) 22nd January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

The volume of an ideal gas increases 8 times and temperature becomes $(1 / 4)^{\text {th }}$ of initial temperature during a reversible change. If there is no exchange of heat in this process $(\Delta \mathrm{Q}=0)$ then identify the gas from the following options (Assuming the gases given in the options are ideal gases) :

A

$\mathrm{NH}_3$

B

$\mathrm{O}_2$

C

$\mathrm{CO}_2$

D

He

2
JEE Main 2026 (Online) 22nd January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

The minimum frequency of photon required to break a particle of mass 15.348 amu into $4 \alpha$ particles is $\_\_\_\_$ kHz .

[mass of He nucleus = $4.002 \mathrm{amu}, 1 \mathrm{amu}=1.66 \times 10^{-27} \mathrm{~kg}, \mathrm{~h}=6.6 \times 10^{-34} \mathrm{~J} . \mathrm{s}$ and $\mathrm{c}=3 \times 10^8 \mathrm{~m} / \mathrm{s}$ ]

A

$14.94 \times 10^{20}$

B

$9 \times 10^{19}$

C

$9 \times 10^{20}$

D

$14.94 \times 10^{19}$

3
JEE Main 2026 (Online) 22nd January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Six point charges are kept $60^{\circ}$ apart from each other on the circumference of a circle of radius $R$ as shown in figure. The net electric field at the center of the circle is $\_\_\_\_$ .

( $\epsilon_0$ is permittivity of free space)

JEE Main 2026 (Online) 22nd January Morning Shift Physics - Electrostatics Question 23 English

A

$-\left(\frac{5 Q}{8 \pi \epsilon_0 R^2}\right)(\hat{i}-3 \hat{j})$

B

$\frac{Q}{4 \pi \in_{\mathrm{o}} R^2}(\sqrt{3} \hat{i}-\hat{j})$

C

$-\frac{\mathrm{Q}}{4 \pi \in_{\mathrm{o}} R^2}(\sqrt{3} \hat{i}-\hat{j})$

D

$-\frac{5 Q}{8 \pi \epsilon_{\mathrm{o}} R^2}(\hat{i}+\sqrt{3} \hat{j})$

4
JEE Main 2026 (Online) 22nd January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

$7.9 \mathrm{MeV} \alpha$-particle scatters from a target material of atomic number 79 . From the given data the estimated diameter of nuclei of the target material is (approximately) $\_\_\_\_$ m.

$$ \left[\frac{1}{4 \pi \epsilon_{\mathrm{o}}}=9 \times 10^9 \mathrm{Nm}^2 / \mathrm{C}^2 \text { and electron charge }=1.6 \times 10^{-19} \mathrm{C}\right] $$

A

$2.88 \times 10^{-14}$

B

$5.76 \times 10^{-14}$

C

$1.44 \times 10^{-13}$

D

$1.69 \times 10^{-12}$

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