A cyclist riding the bicycle at a speed of $14\sqrt 3 ms^{-1}$takes a turn around a circular road of radius $20\sqrt 3 $ m without skidding. Given $g = 9.8 ms^{-2},$ what is his inclination to the vertical ....... $^o$
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Starting from rest a body slides down a $45^o$ inclined plane in twice the time it takes to slide down the same distance in the absence of friction. The co-efficient of friction between the body and the inclined plane is
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A cyclist riding the bicycle at a speed of $14 \sqrt{3} \,m / s$ takes a turn around a circular road of radius $20 \sqrt{3} \,m$ without skidding. What is his inclination to the vertical?
Abody is placed on a rough inclined plane of inclination $\theta$ .As the angle $\theta$ is increased from $0^o$ to $90^o$ the contact force between the block and the plane
In the figure shown, horizontal force $F_1$ is applied on a block but the block does not slide. Then as the magnitude of vertical force $F_2$ is increased from zero the block begins to slide; the correct statement is
A body of mass $1\, kg$ rests on a horizontal floor with which it has a coefficient of static friction $\frac{1}{\sqrt{3}}$. It is desired to make the body move by applying the minimum possible force $F\, N$. The value of $F$ will be the Nearest Integer) [Take $g =10 \,ms ^{-2}$ ]