1
JEE Main 2025 (Online) 23rd January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Consider a circular disc of radius 20 cm with centre located at the origin. A circular hole of radius 5 cm is cut from this disc in such a way that the edge of the hole touches the edge of the disc. The distance of centre of mass of residual or remaining disc from the origin will be

A
1.5 cm
B
2.0 cm
C
0.5 cm
D
1.0 cm
2
JEE Main 2025 (Online) 23rd January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Refer to the circuit diagram given in the figure. which of the following observations are correct?

A. Total resistance of circuit is $6 \Omega$

B. Current in Ammeter is 1 A

C. Potential across $A B$ is 4 Volts.

D. Potential across CD is 4 Volts

E. Total resistance of the circuit is $8 \Omega$.

Choose the correct answer from the options given below:

JEE Main 2025 (Online) 23rd January Morning Shift Physics - Semiconductor Question 14 English

A
B, C and E Only
B
A, C and D Only
C
A, B and C Only
D
A, B and D Only
3
JEE Main 2025 (Online) 23rd January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

The electric field of an electromagnetic wave in free space is $\overrightarrow{\mathrm{E}}=57 \cos \left[7.5 \times 10^6 \mathrm{t}-5 \times 10^{-3}(3 x+4 y)\right](4 \hat{i}-3 \hat{j}) N / C$. The associated magnetic field in Tesla is

A
$\overrightarrow{\mathrm{B}}=\frac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 \mathrm{t}-5 \times 10^{-3}(3 x+4 y)\right](\hat{k})$
B
$\overrightarrow{\mathrm{B}}=\frac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 \mathrm{t}-5 \times 10^{-3}(3 x+4 y)\right](5 \hat{k})$
C
$\overrightarrow{\mathrm{B}}=-\frac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 \mathrm{t}-5 \times 10^{-3}(3 x+4 y)\right](\hat{k})$
D
$\overrightarrow{\mathrm{B}}=-\frac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 \mathrm{t}-5 \times 10^{-3}(3 x+4 y)\right](5 \hat{k})$
4
JEE Main 2025 (Online) 23rd January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

The electric flux is $\phi=\alpha \sigma+\beta \lambda$ where $\lambda$ and $\sigma$ are linear and surface charge density, respectively. $\left(\frac{\alpha}{\beta}\right)$ represents

A
displacement
B
charge
C
electric field
D
area
JEE Main Papers
2023
2021
EXAM MAP