A) \[mgl\,\cos \,\,({{\theta }_{1}}-{{\theta }_{2}})\]
B) \[mgl\,\,(\cos \,{{\theta }_{2}}-\cos {{\theta }_{1}})\]
C) \[mgl\,\,(\cos \,{{\theta }_{1}}-\cos \,{{\theta }_{2}})\]
D) \[mgl\,\,\sin \,({{\theta }_{1}}-{{\theta }_{2}})\]
Correct Answer: B
Solution :
PE at the point B = TE \[=mg\times AP=mg\,(l-l\,\cos \,{{\theta }_{1}})\] \[(\because \,KE=0\,\,at\,B)\] PE at the point \[C=mg\times AQ=mg\,(l-l\,\cos \,{{\theta }_{2}})\] If KE of the bob at the point C is E, then according to law of conservation of energy, \[E+mg\,(l-l\,\cos \,{{\theta }_{2}})=mg\,(l-l\,\cos \,{{\theta }_{1}})\] \[E=mgl\,(\cos \,{{\theta }_{2}}-\cos \,{{\theta }_{1}})\] (Here E = KE)You need to login to perform this action.
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