1
JEE Main 2019 (Online) 10th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
A plane is inclined at an angle $$\alpha $$ = 30° with respect to the horizontal. A particle is projected with a speed u = 2 ms–1 , from the base of the plane, making an angle $$\theta $$ = 15° with respect to the plane as shown in the figure. the distance from the base, at which the particle hits the plane is close to :
(Take g = 10 ms –2) JEE Main 2019 (Online) 10th April Evening Slot Physics - Motion Question 100 English
A
14 cm
B
18 cm
C
20 cm
D
26 cm
2
JEE Main 2019 (Online) 10th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
A ball is thrown upward with an initial velocity V0 from the surface of the earth. The motion of the ball is affected by a drag force equal to m$$\gamma $$u2 (where m is mass of the ball, u is its instantaneous velocity and $$\gamma $$ is a constant). Time taken by the ball to rise to its zenith is :
A
$${1 \over {\sqrt {\gamma g} }}{\tan ^{ - 1}}\left( {\sqrt {{\gamma \over g}} {V_0}} \right)$$
B
$${1 \over {\sqrt {\gamma g} }}{ln}\left( 1+ {\sqrt {{\gamma \over g}} {V_0}} \right)$$
C
$${1 \over {\sqrt {\gamma g} }}{\sin ^{ - 1}}\left( {\sqrt {{\gamma \over g}} {V_0}} \right)$$
D
$${1 \over {\sqrt {2\gamma g} }}{\tan ^{ - 1}}\left( {\sqrt {{2\gamma \over g}} {V_0}} \right)$$
3
JEE Main 2019 (Online) 9th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
The position of a particle as a function of time t, is given by
x(t) = at + bt2 – ct3
where a, b and c are constants. When the particle attains zero acceleration, then its velocity will be :
A
$$a + {{{b^2}} \over {c}}$$
B
$$a + {{{b^2}} \over {4c}}$$
C
$$a + {{{b^2}} \over {3c}}$$
D
$$a + {{{b^2}} \over {2c}}$$
4
JEE Main 2019 (Online) 9th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
The position vector of a particle changes with time according to the relation $$\overrightarrow r (t) = 15{t^2}\widehat i + (4 - 20{t^2})\widehat j$$
What is the magnitude of the acceleration at t = 1 ?
A
50
B
25
C
40
D
100
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