Question

$$\begin{gathered} 30.Anobjecttakestime{t}_{1}toreachthebottom,whenitisplacedonthesurfaceofasmoothinclinedplaneofinclination\theta .Ifsameobjecttakestime{t}_{2}toreachthebottom,whenitisallowedtoslidedownaroughinclinedplaneofinclination\theta (\mu =coefficientoffrictionbetweenobjectandinclinedplane).theratioof\frac{t_1}{t_2} \\ 1.\sqrt{\frac{\sin \theta +\mu \cos \theta }{\cos \theta }} \\ 2.\sqrt{\frac{\sin \theta -\mu \cos \theta }{\cos \theta }} \\ 3.\sqrt{\frac{\sin \theta -\mu \cos \theta }{\sin \theta }} \\ 4.\sqrt{\frac{\sin \theta +\mu \cos \theta }{\sin \theta }} \end{gathered}$$

Answer 1

Dear Student,

$$\begin{gathered} accnwhenonsmoothplane \\ a=g\sin \theta \\ letthelengthbeLoftheplank \\ L=\frac{g\sin \theta {\left(t_1\right)}^2}{2} \\ \left(t_1\right)=\sqrt{\frac{2L}{g\sin \theta }} \\ fortheroughplane. \\ a=g\sin \theta -\mu g\cos \theta \\ L=\frac{\left(g\sin \theta -\mu g\cos \theta \right){\left(t_2\right)}^2}{2} \\ {t}_2=\sqrt{\frac{2L}{g\sin \theta -\mu g\cos \theta }} \\ \frac{t_1}{t_2}=\frac{\sqrt{\frac{2L}{g\sin \theta }}}{\sqrt{\frac{2L}{g\sin \theta -\mu g\cos \theta }}}=\sqrt{\frac{\sin \theta -\mu \cos \theta }{\sin \theta }} \\ regards \end{gathered}$$