$A$ block of mass $5\,kg$ and surface area $2\,m^2$ just begins to slide down an inclined plane when the angle of inclination is $30^{\circ}$. Keeping the mass the same,the surface area of the block is doubled. The angle at which this starts sliding down is:

  • A
    $30^{\circ}$
  • B
    $60^{\circ}$
  • C
    $15^{\circ}$
  • D
    None of these

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The force required to move a body up a rough inclined plane is double the force required to prevent the body from sliding down the plane. The coefficient of friction, when the angle of inclination of the plane is $60^{\circ}$ is

The time taken by a block of mass $m$ to slide down from the highest point to the lowest point on a rough inclined plane is $50\%$ more compared to the time taken by the same block on an identical inclined smooth plane. Both inclined planes are at $45^\circ$ with the horizontal. The coefficient of kinetic friction between the rough inclined surface and the block is . . . . . . .

$A$ rectangular box lies on a rough inclined surface. The coefficient of friction between the surface and the box is $\mu$. Let the mass of the box be $m$.
$(a)$ At what angle of inclination $\theta$ of the plane to the horizontal will the box just start to slide down the plane?
$(b)$ What is the force acting on the box down the plane,if the angle of inclination of the plane is increased to $\alpha > \theta$?
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