1
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

$1.69 \times 10^{-12}$

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

Net gravitational force at the center of a square is found to be $F_1$ when four particles having mass $M, 2 M, 3 M$ and $4 M$ are placed at the four corners of the square as shown in figure and it is $F_2$ when the positions of $3 M$ and $4 M$ are interchanged. The ratio $\frac{F_1}{F_2}$ is $\frac{\alpha}{\sqrt{5}}$. The value of $\alpha$ is $\_\_\_\_$ .

JEE Main 2026 (Online) 22nd January Morning Shift Physics - Gravitation Question 9 English
A

2

B

$2 \sqrt{5}$

C

1

D

3

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

A meter bridge with two resistances $R_1$ and $R_2$ as shown in figure was balanced (null point) at 40 cm from the point $P$. The null point changed to 50 cm from the point $P$, when $16 \Omega$ resistance is connected in parallel to $R_2$. The values of resistances $R_1$ and $R_2$ are $\_\_\_\_$ .

JEE Main 2026 (Online) 22nd January Morning Shift Physics - Current Electricity Question 19 English
A

$R_2=8 \Omega, R_1=\frac{16}{3} \Omega$

B

$R_2=12 \Omega, R_1=\frac{12}{3} \Omega$

C

$R_2=4 \Omega, R_1=\frac{4}{3} \Omega$

D

$R_2=16 \Omega, R_1=\frac{16}{3} \Omega$

4
JEE Main 2026 (Online) 22nd January Morning Shift
Numerical
+4
-1
Change Language

Inductance of a coil with $10^4$ turns is 10 mH and it is connected to a dc source of 10 V with internal resistance of $10 \Omega$. The energy density in the inductor when the current reaches $\left(\frac{1}{e}\right)$ of its maximum value is $\alpha \pi \times \frac{1}{e^2} \mathrm{~J} / \mathrm{m}^3$. The value of $\alpha$ is

$\_\_\_\_$ .

$$ \left(\mu_0=4 \pi \times 10^{-7} \mathrm{Tm} / \mathrm{A}\right) . $$

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