1
JEE Main 2019 (Online) 9th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
If the angular momentum of a planet of mass m, moving around the Sun in a circular orbit is L, about the center of the Sun, its areal velocity is :
A
$${L \over m}$$
B
$${4L \over m}$$
C
$${L \over 2m}$$
D
$${2L \over m}$$
2
JEE Main 2019 (Online) 9th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If, for an n-type semiconductor, the density of electrons is 1019 m$$-$$3 and their mobility is 1.6 m2/(V.s) then the resistivity of the semiconductor (since it is an n-type semiconductor contribution of holes is ignored) is close to :
A
$$2\,\Omega $$m
B
4$$\,\Omega $$m
C
0.4 $$\,\Omega $$m
D
0.2 $$\,\Omega $$m
3
JEE Main 2019 (Online) 9th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
For a uniformly charged ring of radius R, the electric field on its axis has the largest magnitude at a distance h from its center. Then value of h is :
A
$${R \over {\sqrt 5 }}$$
B
$${R \over {\sqrt 2 }}$$
C
R
D
R$$\sqrt 2 $$
4
JEE Main 2019 (Online) 9th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Two masses m and $${m \over 2}$$ are connected at the two ends of a massless rigid rod of length l. The rod is suspended by a thin wire of torsional constant k, at the centre of mass of the rod-mass system(see figure). Because of torsional constant k, the restoring torque is $$\tau $$ = k$$\theta $$ for angular displacement $$\theta $$. If the rod is rotated by $$\theta $$0 and released, the tension in it when it passes through its mean position will be :

JEE Main 2019 (Online) 9th January Morning Slot Physics - Center of Mass and Collision Question 110 English
A
$${{3k{\theta _0}^2} \over l}$$
B
$${{2k{\theta _0}^2} \over l}$$
C
$${{k{\theta _0}^2} \over l}$$
D
$${{k{\theta _0}^2} \over {2l}}$$

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