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. 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 :

 Key Idea : When stone reaches a position where string is horizontal, it attains the energy partially as kinetic and partially as potential. When stone is at its lowest position, it has only kinetic energy, given by \[K=\frac{1}{2}m{{u}^{2}}\] At the horizontal position, it has energy \[E=U+K=\frac{1}{2}m{{}^{2}}+mgl\] According to conservation of energy, \[K=E\] \[\therefore \] \[\frac{1}{2}m{{u}^{2}}=\frac{1}{2}mu{{}^{2}}+mgl\] Or \[\frac{1}{2}mu{{}^{2}}=\frac{1}{2}m{{u}^{2}}-mgl\] Or \[u{{}^{2}}={{u}^{2}}-2gl\] Or \[u=\sqrt{{{u}^{2}}-2gl}\]             ...(i) So, the magnitude of change in velocity \[|\Delta \overrightarrow{u}|=|\overrightarrow{u}|=\sqrt{u{{}^{2}}+{{u}^{2}}+2uu\cos 90{}^\circ }\] \[|\Delta \overrightarrow{u}|=\sqrt{u{{}^{2}}+{{u}^{2}}}\] \[=\sqrt{2({{u}^{2}}+gL)}\]             [from Eq. (i)]