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  • A particle of mass m is suspended from a ceilling through a string of length L. The particle moves in a horizontal circle of radius r such that <img alt="r= \frac{L}{\sqrt{2}}" src="https://entrancecorner.codecogs.com/gif.latex?r%3D%20%5Cfrac%7BL%7D%7B%5C

A particle of mass m is suspended from a ceilling through a string of length L. The particle moves in a horizontal circle of radius r such that r= \frac{L}{\sqrt{2}} .The speed of particle will be :
Option: 1 \sqrt{rg}
Option: 2 \sqrt{2rg}
Option: 3 \sqrt{\frac{rg}{2}}
Option: 4 2\sqrt{rg}

Answers (1)

best_answer


\sin \theta= \frac{r}{L}= \frac{1}{\sqrt{2}}
\theta= 45\degree
\tan \theta= \frac{\frac{mv^{2}}{r}}{mg}
1= \frac{v^{2}}{rg}
v^{2}= rg
v= \sqrt{rg}
The correct option is (1)

Posted by

vishal kumar

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