JEE MAIN - Physics (2016 - 9th April Morning Slot - No. 21)
A rocket is fired vertically from the earth with an acceleration of 2g, where g is the
gravitational acceleration. On an inclined plane inside the rocket, making an angle $$\theta $$ with the horizontal, a point object of mass m is kept. The minimum coefficient of friction $$\mu $$min between the mass and the inclined surface such that the mass does not move is :
tan$$\theta $$
2tan$$\theta $$
3tan$$\theta $$
tan2$$\theta $$
Explanation
Rocket is moving upward with acceleration 2g and gravitation acceleration is g downward direction.
So, acceleration experienced by the point object,
= 2g $$-$$ ($$-$$ g) = 3g
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At equilibrium,
N = 3mgcos$$\theta $$
$$\mu $$N = 3mgsin$$\theta $$
$$ \Rightarrow $$ $$\mu $$ (3mgcos$$\theta $$) = 3mg sin$$\theta $$
$$ \Rightarrow $$ $$\mu $$ = tan$$\theta $$
So, acceleration experienced by the point object,
= 2g $$-$$ ($$-$$ g) = 3g
At equilibrium,
N = 3mgcos$$\theta $$
$$\mu $$N = 3mgsin$$\theta $$
$$ \Rightarrow $$ $$\mu $$ (3mgcos$$\theta $$) = 3mg sin$$\theta $$
$$ \Rightarrow $$ $$\mu $$ = tan$$\theta $$
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