Question

Assertion :Two identical blocks are placed over a rough inclined plane. One block is given an upward velocity and the other in downward direction. If $\mu =\frac{1}{3}$ and $\theta ={30}^{\circ }$ the ratio of magnitudes of accelerations of two blocks is 2 : 1. Reason: The desired ratio is $\frac{1+\mu }{1-\mu }.$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is D Both Assertion and Reason are incorrect

Block moving upwards: Component of gravity Force & Friction both will be downwards.
Therefore acceleration, $a=(mgsin\theta +f)$
$a=(mgsin\theta +\mu N)$
$a=(mgsin\theta +\mu mgcos\theta )$

$a=g(sin30+\mu cos30)$
$a=g(\frac{1}{2}+\frac{1}{3}\frac{\sqrt{3}}{2})=g(\frac{1}{2}+\frac{1}{2\sqrt{3}})a=g\left(\frac{\sqrt{3}+1}{2\sqrt{3}}\right)=gsin30(1+\mu cot30)$
Block moving downwards: Component of gravity force & Ffriction will be downwards & upwards respectively.
Therefore acceleration, $a=(mgsin\theta -\mu N)/ma=(mgsin\theta -\mu mgcos\theta )/ma=g(sin30-\mu cos30)a=g(\frac{1}{2}-\frac{1}{3}\frac{\sqrt{3}}{2})=g(\frac{1}{2}-\frac{1}{2\sqrt{3}})a=g\left(\frac{\sqrt{3}-1}{2\sqrt{3}}\right)=gsin30(1-\mu cot30)$
Ratio is: $\frac{gsin30(1+\mu cot30)}{gsin30(1-\mu cot30)}=\frac{(1+\mu cot30)}{(1-\mu cot30)}=\frac{(1+1/\sqrt{3})}{(1-1/\sqrt{3})}=\frac{\sqrt{3}+1}{\sqrt{3}-1}=\frac{3+1+2\sqrt{3}}{2}=2+\sqrt{3}$
Therefore both Assertion and Reason are incorrect.