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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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A block takes twice as much time to slide down a rough $45^{\large\circ}$ inclined plane as it takes to slide down an identical smooth $45^{\large\circ}$ inclined plane.The coefficient of kinetic friction between the block and the rough inclined plane is

$\begin{array}{1 1}0.25\\0.50\\0.75\\1.0\end{array} $

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Answer : 0.75
The accelerations of the block sliding down a smooth and rough $45^{\large\circ}$ inclined planes respectively are
$a_1=g\sin 45^{\large\circ}=\large\frac{g}{\sqrt 2}$
$a_2=g(\sin 45^{\large\circ}-\mu_k\cos 45^{\large\circ})=\large\frac{g}{\sqrt 2}$$(1-\mu_k)$
Where $\mu_k$ is the coefficient of kinetic friction.
Now,we know that the square of the time of slide is inversely proportional to the acceleration.
Therefore $\large\frac{t_2^2}{t_1^2}=\frac{a_1}{a_2}=\frac{1}{1-\mu_k}$
Since $t_2=2t_1$,we have
$4=\large\frac{1}{1-\mu_k}$
$\mu_k=0.75$
Hence the correct answer is 0.75
answered Aug 27, 2014 by sreemathi.v
 

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