1
JEE Main 2025 (Online) 29th January Evening Shift
MCQ (Single Correct Answer)
+4
-1
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Match List - I with List - II.

List - I List - II
(A) Magnetic induction (I) Ampere meter2
(B) Magnetic intensity (II) Weber
(C) Magnetic flux (III) Gauss
(D) Magnetic moment (IV) Ampere meter

Choose the correct answer from the options given below:

A

(A)-(III), (B)-(IV), (C)-(II), (D)-(I)

B

(A)-(III), (B)-(II), (C)-(I), (D)-(IV)

C

(A)-(III), (B)-(IV), (C)-(I), (D)-(II)

D

(A)-(I), (B)-(II), (C)-(III), (D)-(IV)

2
JEE Main 2025 (Online) 29th January Evening Shift
MCQ (Single Correct Answer)
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JEE Main 2025 (Online) 29th January Evening Shift Physics - Geometrical Optics Question 40 English

Two concave refracting surfaces of equal radii of curvature and refractive index 1.5 face each other in air as shown in figure. A point object O is placed midway, between P and B. The separation between the images of O, formed by each refracting surface is :

A

0.124R

B

0.114R

C

0.411R

D

0.214R

3
JEE Main 2025 (Online) 29th January Evening Shift
MCQ (Single Correct Answer)
+4
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The truth table for the circuit given below is:

JEE Main 2025 (Online) 29th January Evening Shift Physics - Semiconductor Question 17 English
A
ABY
000
111
101
011
B
ABY
000
101
010
110
C
ABY
000
100
110
011
D
ABY
000
011
101
110
4
JEE Main 2025 (Online) 29th January Evening Shift
MCQ (Single Correct Answer)
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JEE Main 2025 (Online) 29th January Evening Shift Physics - Heat and Thermodynamics Question 47 English

A poly-atomic molecule $\left(C_V=3 R, C_P=4 R\right.$, where $R$ is gas constant) goes from phase space point $\mathrm{A}\left(\mathrm{P}_{\mathrm{A}}=10^5 \mathrm{~Pa}, \mathrm{~V}_{\mathrm{A}}=4 \times 10^{-6} \mathrm{~m}^3\right)$ to point $\mathrm{B}\left(\mathrm{P}_{\mathrm{B}}=5 \times 10^4 \mathrm{~Pa}, \mathrm{~V}_{\mathrm{B}}=6 \times 10^{-6} \mathrm{~m}^3\right)$ to point $\mathrm{C}\left(\mathrm{P}_{\mathrm{C}}=10^4\right.$ $\mathrm{Pa}, \mathrm{V}_C=8 \times 10^{-6} \mathrm{~m}^3$ ). A to $B$ is an adiabatic path and $B$ to $C$ is an isothermal path.

The net heat absorbed per unit mole by the system is :

A
$500 \mathrm{R}(\ln 3+\ln 4)$
B
$450 \mathrm{R}(\ln 4-\ln 3)$
C
$500 \mathrm{R} \ln 2$
D
$400 \mathrm{R} \ln 4$
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