1
JEE Main 2021 (Online) 24th February Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
A particle is projected with velocity v0 along x-axis. A damping force is acting on the particle which is proportional to the square of the distance from the origin i.e. ma = $$-$$ $$\alpha$$x2. The distance at which the particle stops :
A
$${\left[ {{{3mv_0^2} \over {2\alpha }}} \right]^{{1 \over 3}}}$$
B
$${\left( {{{2{v_0}} \over {3\alpha }}} \right)^{{1 \over 3}}}$$
C
$${\left( {{{3v_0^2} \over {2\alpha }}} \right)^{{1 \over 2}}}$$
D
$${\left( {{{2v_0^2} \over {3\alpha }}} \right)^{{1 \over 2}}}$$
2
JEE Main 2020 (Online) 6th September Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A particle moving in the xy plane experiences a velocity dependent force
$$\overrightarrow F = k\left( {{v_y}\widehat i + {v_x}\widehat j} \right)$$ , where vx and vy are the
x and y components of its velocity $$\overrightarrow v $$ . If $$\overrightarrow a $$ is the
acceleration of the particle, then
which of the following statements is true for the particle?
A
kinetic energy of particle is constant in time
B
quantity $$\overrightarrow v \times \overrightarrow a $$ is constant in time
C
quantity $$\overrightarrow v .\overrightarrow a $$ is constant in time
D
$$\overrightarrow F $$ arises due to a magnetic field
3
JEE Main 2020 (Online) 6th September Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
An insect is at the bottom of a hemispherical ditch of radius 1 m. It crawls up the ditch but starts slipping after it is at height h from the bottom. If the coefficient of friction between the ground and the insect is 0.75, then h is :
(g = 10 ms–2)
A
0.45 m
B
0.60 m
C
0.20 m
D
0.80 m
4
JEE Main 2020 (Online) 5th September Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A spaceship in space sweeps stationary interplanetary dust. As a result, its mass
increases at a rate $${{dM\left( t \right)} \over {dt}}$$ = bv2(t), where v(t) is its instantaneous velocity. The instantaneous acceleration of the satellite is :
A
-bv3(t)
B
$$ - {{2b{v^3}} \over {M\left( t \right)}}$$
C
$$ - {{b{v^3}} \over {M\left( t \right)}}$$
D
$$ - {{b{v^3}} \over {2M\left( t \right)}}$$
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