Question

A block of mass $m$ is put on a rough inclined plane of inclination $\theta$, and is tied with a light thread shown. Inclination $\theta$ is increased gradually from $\theta =0^{\circ }to{90}^{\circ }.$ Match the columns according to corresponding curve.

• A. $A-q,B-s,C-r,D-p$
• B. $A-s,B-p,C-r,D-q$
• C. $A-r,B-p,C-p,D-q$
• D. $A-p,B-q,C-s,D-r$

Answer 1

The correct option is A $A-q,B-s,C-r,D-p$


A) Tension comes into picture only after limiting friction or angle of repose $({\theta }_R=ta{n}^{-1}(\mu \left)\right)$ is reached.
Thus if $\theta <{\theta }_R,T=0$.
$If\theta >{\theta }_R,T=mgsin\theta -\mu mgcos\theta$
Thus $A\to q.$

B) Normal reaction $N=mgcos\theta .$
Thus $B\to s.$


C) Friction force for $\theta \le {\theta }_R,$ $f_s=mgsin\theta$
As $\theta$ increases, $f_s$ increases.
For $\theta >{\theta }_R;f_l=\mu N=\mu mgcos\theta$ and $cos\theta$ is decreasing
Thus $C\to r$




D)Net reaction force for $\theta \le {\theta }_R$
$R=\sqrt{N^2+f_s^2}$
$R=\sqrt{m^2{g}^{2}co{s}^2\theta +m^2{g}^{2}si{n}^2\theta }$
$R=mg$ (constant)
For $\theta >{\theta }_R$
$R=\sqrt{m^2{g}^{2}co{s}^2\theta +{\mu }^2{m}^2{g}^{2}co{s}^2\theta }$
$R=mg\sqrt{1+{\mu }^2}cos\theta$
As $\theta$ increases, $cos\theta$ decreases and hence $R$ decreases.
Thus D $\to p.$