question_answer

A heavy particle hanging from a string of length \[l\]is projected horizontally with speed \[\sqrt{2gl}\]. The speed of the particle at the point where the tension in the string equals weight of the particle

A) \[\sqrt{2gl}\]     

B) \[\sqrt{3gl}\]

C) \[\sqrt{gl/2}\]    

D) \[\sqrt{gl/3}\]

Correct Answer: D

Solution :

[d] \[T-mg\cos \theta =\frac{m{{v}^{2}}}{R}\] Given \[T=mg\] \[mg-mg\cos \theta =\frac{m{{v}^{2}}}{R}\] \[g(1-cos\theta )=\frac{{{v}^{2}}}{R}\] C.O.M.E. at A and B; \[\Delta K+\Delta U=0\] \[\left( \frac{1}{2}m{{v}^{2}}-\frac{1}{2}m{{u}^{2}} \right)+mg(R-Rcos\theta )=0\] \[\Rightarrow {{v}^{2}}-{{u}^{2}}=-2gR(1-cos\theta )\] \[\Rightarrow {{v}^{2}}-{{(\sqrt{gl})}^{2}}=-2{{v}^{2}}\] \[3{{v}^{2}}=gl\Rightarrow v=\sqrt{\frac{gl}{3}}\]