1
JEE Main 2021 (Online) 16th March Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
A mosquito is moving with a velocity $$\overrightarrow v = 0.5{t^2}\widehat i + 3t\widehat j + 9\widehat k$$ m/s and accelerating in uniform conditions. What will be the direction of mosquito after 2 s?
A
$${\tan ^{ - 1}}\left( {{\sqrt {85} } \over 6}\right)$$ from y-axis
B
$${\tan ^{ - 1}}\left( {{5 \over 2}} \right)$$ from y-axis
C
$${\tan ^{ - 1}}\left( {{2 \over 3}} \right)$$ from x-axis
D
$${\tan ^{ - 1}}\left( {{5 \over 2}} \right)$$ from x-axis
2
JEE Main 2021 (Online) 16th March Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
The velocity-displacement graph describing the motion of bicycle is shown in the figure.

JEE Main 2021 (Online) 16th March Morning Shift Physics - Motion Question 110 English
The acceleration-displacement graph of the bicycle's motion is best described by :
A
JEE Main 2021 (Online) 16th March Morning Shift Physics - Motion Question 110 English Option 1
B
JEE Main 2021 (Online) 16th March Morning Shift Physics - Motion Question 110 English Option 2
C
JEE Main 2021 (Online) 16th March Morning Shift Physics - Motion Question 110 English Option 3
D
JEE Main 2021 (Online) 16th March Morning Shift Physics - Motion Question 110 English Option 4
3
JEE Main 2021 (Online) 26th February Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
A scooter accelerates from rest for time t1 at constant rate a1 and then retards at constant rate a2 for time t2 and comes to rest. The correct value of $${{{t_1}} \over {{t_2}}}$$ wil be :
A
$${{{a_1} + {a_2}} \over {{a_2}}}$$
B
$${{{a_1} + {a_2}} \over {{a_1}}}$$
C
$${{{a_2}} \over {{a_1}}}$$
D
$${{{a_1}} \over {{a_2}}}$$
4
JEE Main 2021 (Online) 26th February Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
The trajectory of a projectile in a vertical plane is y = $$\alpha$$x $$-$$ $$\beta$$x2, where $$\alpha$$ and $$\beta$$ are constants and x & y are respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projection $$\theta$$ and the maximum height attained H are respectively given by :
A
$${\tan ^{ - 1}}\alpha ,{{{\alpha ^2}} \over {4\beta }}$$
B
$${\tan ^{ - 1}}\alpha ,{{4{\alpha ^2}} \over \beta }$$
C
$${\tan ^{ - 1}}\left( {{\beta \over \alpha }} \right),{{{\alpha ^2}} \over \beta }$$
D
$${\tan ^{ - 1}}\beta ,{{{\alpha ^2}} \over {2\beta }}$$
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