JEE Main 2019 (Online) 12th April Morning Slot | Motion Question 98 | Physics

JEE Main 2019 (Online) 12th April Morning Slot

MCQ (Single Correct Answer)

-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 -

$${{2R} \over g}$$

$${R \over g}$$

$${R \over {2g}}$$

$${R \over {4g}}$$

JEE Main 2019 (Online) 12th April Morning Slot

MCQ (Single Correct Answer)

-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) :

$${\theta _0} = {\cos ^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)$$ and $${v_0} = {5 \over 3}$$ ms^-1

$${\theta _0} = {\cos ^{ - 1}}\left( {{2 \over {\sqrt 5 }}} \right)$$ and $${v_0} = {3 \over 5}$$ ms^-1

$${\theta _0} = {\sin ^{ - 1}}\left( {{2 \over {\sqrt 5 }}} \right)$$ and $${v_0} = {3 \over 5}$$ ms^-1

$${\theta _0} = {\sin ^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)$$ and $${v_0} = {5 \over 3}$$ ms^-1

JEE Main 2019 (Online) 10th April Evening Slot

MCQ (Single Correct Answer)

-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

JEE Main 2019 (Online) 10th April Evening Slot Physics - Motion Question 100 English

14 cm

18 cm

20 cm

26 cm

JEE Main 2019 (Online) 10th April Morning Slot

MCQ (Single Correct Answer)

-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 :

$${1 \over {\sqrt {\gamma g} }}{\tan ^{ - 1}}\left( {\sqrt {{\gamma \over g}} {V_0}} \right)$$

$${1 \over {\sqrt {\gamma g} }}{ln}\left( 1+ {\sqrt {{\gamma \over g}} {V_0}} \right)$$

$${1 \over {\sqrt {\gamma g} }}{\sin ^{ - 1}}\left( {\sqrt {{\gamma \over g}} {V_0}} \right)$$

$${1 \over {\sqrt {2\gamma g} }}{\tan ^{ - 1}}\left( {\sqrt {{2\gamma \over g}} {V_0}} \right)$$

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