1
JEE Main 2020 (Online) 8th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
In finding the electric field using Gauss Law the formula $$\left| {\overrightarrow E } \right| = {{{q_{enc}}} \over {{\varepsilon _0}\left| A \right|}}$$ is applicable. In the formula $${{\varepsilon _0}}$$ is permittivity of free space, A is the area of Gaussian surface and qenc is charge enclosed by the Gaussian surface. The equation can be used in which of the following situation?
A
Only when $$\left| {\overrightarrow E } \right|$$ = constant on the surface.
B
For any choice of Gaussian surface.
C
Only when the Gaussian surface is an equipotential surface.
D
Only when the Gaussian surface is an equipotential surface and $$\left| {\overrightarrow E } \right|$$ is constant on the surface.
2
JEE Main 2020 (Online) 8th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Consider two solid spheres of radii R1 = 1m, R2 = 2m and masses M1 and M2, respectively. The gravitational field due to sphere (1) and (2) are shown. The value of $${{{M_1}} \over {{M_2}}}$$ is : JEE Main 2020 (Online) 8th January Morning Slot Physics - Gravitation Question 138 English
A
$${2 \over 3}$$
B
$${1 \over 6}$$
C
$${1 \over 2}$$
D
$${1 \over 3}$$
3
JEE Main 2020 (Online) 8th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The plot that depicts the behavior of the mean free time t (time between two successive collisions) for the molecules of an ideal gas, as a function of temperature (T), qualitatively, is:
(Graphs are schematic and not drawn to scale)
A
JEE Main 2020 (Online) 8th January Morning Slot Physics - Heat and Thermodynamics Question 261 English Option 1
B
JEE Main 2020 (Online) 8th January Morning Slot Physics - Heat and Thermodynamics Question 261 English Option 2
C
JEE Main 2020 (Online) 8th January Morning Slot Physics - Heat and Thermodynamics Question 261 English Option 3
D
JEE Main 2020 (Online) 8th January Morning Slot Physics - Heat and Thermodynamics Question 261 English Option 4
4
JEE Main 2020 (Online) 8th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Photon with kinetic energy of 1MeV moves from south to north. It gets an acceleration of 1012 m/s2 by an applied magnetic field (west to east). The value of magnetic field : (Rest mass of proton is 1.6 × 10–27 kg) :
A
0.71mT
B
7.1mT
C
0.071mT
D
71mT
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