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:
Difficult
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Let acceleration in $\mathrm{I}^{\mathrm{st}}$ case is $\mathrm{a}_{1}$ and that in second case is $a_2$
Now, $\frac{1}{2} a_{1} t^{2}=\frac{1}{2} a_{2}(2 t)^{2} \Rightarrow a_{2}=\frac{a_{1}}{4}$ $...(i)$
Clearly $\quad \mathrm{a}_{1}=\frac{\mathrm{mg} \sin \theta}{\mathrm{m}}=\mathrm{g} \sin \theta$ $...(ii)$
and $a_{2}=\frac{m g \sin \theta-\mu m g \cos \theta}{m}=g \sin \theta-\mu g \cos \theta$ $...(iii)$
From $( i ),$ $(ii)$ and $(iii),$
we get $\mu=0.75$
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