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  • In the given figure the distance PQ is constant. SQ is a vertical line passing through point R. A particle is kept at R and the plane PR is such that an

In the given figure the distance PQ is constant. SQ is a vertical line passing through point R. A particle is kept at R and the plane PR is such that angle θ can be varied such that R lies on line SQ. The time taken by particle to come down varies, as the θ increases

Option: 1

Decreases continuously


Option: 2

Increases continuously


Option: 3

Increases then decreases


Option: 4

decreases then increases


Answers (1)

best_answer

Let distance PR is covered by the particle in time t.

\begin{aligned} \Rightarrow \mathrm{PR} &=0+\frac{1}{2} \mathrm{~g} \sin \theta \cdot \mathrm{t}^{2} \\ &=\frac{1}{2} \mathrm{gt}^{2} \sin \theta \end{aligned}

Further PR=\frac{P Q}{\cos \theta} \quad (Given PQ= constant )
\Rightarrow P Q=\frac{1}{2} g t^{2} \sin \theta \cos \theta =\frac{1}{4} \mathrm{gt}^{2} \sin 2 \theta \Rightarrow \mathrm{t}=2 \sqrt{\frac{\mathrm{PQ}}{\mathrm{g} \sin 2 \theta}}

\therefore \quad \mathrm{t} \propto \frac{1}{\sqrt{\sin 2 \theta}}
So as \theta increases, \sin 2 \theta first increases and then decreases. Hence t first decreases and then increases.

Posted by

avinash.dongre

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