question_answer

A stone is tied to a string of length \[l\] and is whirled in a vertical circle with the other end of the string as the centre. At a certain instant of time, the stone is at its lowest position and has a speed \[{{u}_{1}}.\]The magnitude of the change in velocity as it reaches a position where the string is horizontal (g being acceleration due to gravity) is

A) \[\sqrt{2({{u}^{2}}-gl)}\]

B) \[\sqrt{{{u}^{2}}-gl}\]

C) \[u-\sqrt{{{u}^{2}}-2gl}\]

D) \[\sqrt{2gl}\] 

Correct Answer: A

Solution :

Using conservation of mechanical energy at initial and final position. \[\Delta K+\Delta U=0\] \[\left( \frac{1}{2}mu{{'}^{2}}-\frac{1}{2}m{{u}^{2}} \right)+(mgl)=0\] or \[u{{'}^{2}}={{u}^{2}}-2gl\] or \[u'=\sqrt{{{u}^{2}}-2gl}\]                (i) So, the magnitude of change in velocity \[\Delta \vec{u}={{\vec{u}}_{f}}-{{\vec{u}}_{i}}={{\vec{u}}_{f}}+({{\vec{u}}_{i}})\] \[|\Delta \vec{u}|=\sqrt{u{{'}^{2}}+{{u}^{2}}}=\sqrt{({{u}^{2}}-2gh)+{{u}^{2}}}\] \[=\sqrt{2({{u}^{2}}-gl)}\]