QJEEAdv17Ph70a4: JEE Advanced 2017 Physics - Class 11 JEEAdv17Ph70a4 Solution: A rigid uniform bar AB of length L is slipping from its…

We discuss the options as follows:


$\bullet$ For option (A) : When the bar makes an angle $\theta$ with the floor, the height of the midpoint is ${L \over 2}\cos \theta$ . this is height of centre of mass.

A force mg acts vertically downward. Thus, the midpoint of bar falls vertically downward. Hence, option (A) is correct.

$\bullet$ For option (B) : Here, we have $x = {L \over 2}\sin \theta$ and $y = L\cos \theta$ ; therefore,

$\sin \theta = {x \over {L/2}}$ and $\cos \theta = {y \over L}$

Using the condition ${\sin ^2}\theta + {\cos ^2}\theta = 1$ , we get

${{{x^2}} \over {{{(L/2)}^2}}} + {{{y^2}} \over {{L^2}}} = 1$ (equation of ellipse)

Thus, trajectory of point A is not parabola. Hence, option (B) is incorrect.

$\bullet$ For option (C) : The torque acting about the point of contact with floor, that is, at point B is

$\tau = \overrightarrow F \times \overrightarrow r = mg{L \over 2}\sin \theta$

Thus, the torque is proportional to sin $\theta$ . Hence, option (C) is correct.

$\bullet$ For option (D) : The displacement of midpoint is

${L \over 2} - {L \over 2}\cos \theta = {L \over 2}(1 - \cos \theta )$

That is,

Displacement $\propto$ (1 $-$ cos $\theta$ )

Hence, option (D) is correct.