1
JEE Main 2020 (Online) 3rd September Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A particle is moving unidirectionally on a horizontal plane under the action of a constant power supplying energy source. The displacement (s) - time (t) graph that describes the motion of the particle is (graphs are drawn schematically and are not to scale) :
A
JEE Main 2020 (Online) 3rd September Evening Slot Physics - Work Power & Energy Question 78 English Option 1
B
JEE Main 2020 (Online) 3rd September Evening Slot Physics - Work Power & Energy Question 78 English Option 2
C
JEE Main 2020 (Online) 3rd September Evening Slot Physics - Work Power & Energy Question 78 English Option 3
D
JEE Main 2020 (Online) 3rd September Evening Slot Physics - Work Power & Energy Question 78 English Option 4
2
JEE Main 2020 (Online) 3rd September Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Amount of solar energy received on the earth’s surface per unit area per unit time is defined a solar constant. Dimension of solar constant is :
A
MLT–2
B
ML0T–3
C
M2L0T–1
D
ML2T–2
3
JEE Main 2020 (Online) 3rd September Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Which of the following will NOT be observed when a multimeter (operating in resistance measuring mode) probes connected across a component, are just reversed?
A
Multimeter shows NO deflection in both cases i.e. before and after reversing the probes if the chosen component is metal wire.
B
Multimeter shows a deflection, accompanied by a splash of light out of connected component in one direction and NO deflection on reversing the probes if the chosen component is LED.
C
Multimeter shows an equal deflection in both cases i.e. before and after reversing the probes if the chosen component is resistor.
D
Multimeter shows NO deflection in both cases i.e. before and after reversing the probes if the chosen component is capacitor.
4
JEE Main 2020 (Online) 3rd September Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A uniform rod of length ‘$$l$$’ is pivoted at one of its ends on a vertical shaft of negligible radius. When the shaft rotates at angular speed $$\omega $$ the rod makes an angle $$\theta $$ with it (see figure). To find $$\theta $$ equate the rate of change of angular momentum (direction going into the paper) $${{m{l^2}} \over {12}}{\omega ^2}\sin \theta \cos \theta $$ about the centre of mass (CM) to the torque provided by the horizontal and vertical forces FH and FV about the CM. The value of $$\theta $$ is then such that : JEE Main 2020 (Online) 3rd September Evening Slot Physics - Rotational Motion Question 116 English
A
$$\cos \theta = {{2g} \over {3l{\omega ^2}}}$$
B
$$\cos \theta = {{3g} \over {2l{\omega ^2}}}$$
C
$$\cos \theta = {g \over {2l{\omega ^2}}}$$
D
$$\cos \theta = {g \over {l{\omega ^2}}}$$
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