1
JEE Main 2020 (Online) 8th January Evening Slot
+4
-1
A particle moves such that its position vector $$\overrightarrow r \left( t \right) = \cos \omega t\widehat i + \sin \omega t\widehat j$$ where $$\omega$$ is a constant and t is time. Then which of the following statements is true for the velocity $$\overrightarrow v \left( t \right)$$ and acceleration $$\overrightarrow a \left( t \right)$$ of the particle :
A
$$\overrightarrow v$$ and $$\overrightarrow a$$ both are perpendicular to $$\overrightarrow r$$
B
$$\overrightarrow v$$ and $$\overrightarrow a$$ both are parallel to $$\overrightarrow r$$
C
$$\overrightarrow v$$ is perpendicular to $$\overrightarrow r$$ and $$\overrightarrow a$$ is directed towards the origin
D
$$\overrightarrow v$$ is perpendicular to $$\overrightarrow r$$ and $$\overrightarrow a$$ is directed away from the origin
2
JEE Main 2019 (Online) 12th April Evening Slot
+4
-1
Two particles are projected from the same point with the same speed u such that they have the same range R, but different maximum heights, h1 and h2. Which of the following is correct ?
A
R2 = h1h2
B
R2 = 16 h1h2
C
R2 = 4 h1h2
D
R2 = 2h1h2
3
JEE Main 2019 (Online) 12th April Morning Slot
+4
-1
A shell is fired from a fixed artillery gun with an initial speed u such that it hits the target on the ground at a distance R from it. If t1 and t2 are the values of the time taken by it to hit the target in two possible ways, the product t1t2 is -
A
$${{2R} \over g}$$
B
$${R \over g}$$
C
$${R \over {2g}}$$
D
$${R \over {4g}}$$
4
JEE Main 2019 (Online) 12th April Morning Slot
+4
-1
The trajectory of a projectile near the surface of the earth is given as y = 2x – 9x2 . If it were launched at an angle $$\theta$$0 with speed v0 then (g = 10 ms–2) :
A
$${\theta _0} = {\cos ^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)$$ and $${v_0} = {5 \over 3}$$ ms-1
B
$${\theta _0} = {\cos ^{ - 1}}\left( {{2 \over {\sqrt 5 }}} \right)$$ and $${v_0} = {3 \over 5}$$ ms-1
C
$${\theta _0} = {\sin ^{ - 1}}\left( {{2 \over {\sqrt 5 }}} \right)$$ and $${v_0} = {3 \over 5}$$ ms-1
D
$${\theta _0} = {\sin ^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)$$ and $${v_0} = {5 \over 3}$$ ms-1
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