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

A spherical surface of radius of curvature $R$, separates air from glass (refractive index $=1.5$ ). The centre of curvature is in the glass medium. A point object ' $O$ ' placed in air on the optic axis of the surface, so that its real image is formed at 'I' inside glass. The line OI intersects the spherical surface at $P$ and $P O=P I$. The distance $P O$ equals to

A
5R
B
2R
C
1.5R
D
3R
2
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
3
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
4
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
$\overrightarrow{\mathrm{B}}=-\frac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 \mathrm{t}-5 \times 10^{-3}(3 x+4 y)\right](5 \hat{k})$
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