A frictionless wire AB is fixed on a sphere of radius R. A very small spherical ball slips on this wire. The time taken by this ball to slip from A to B is
A) \[\frac{2\sqrt{gR}}{g\cos \theta }\]
B) \[2\sqrt{gR}.\frac{\cos \theta }{g}\]
C) \[2\sqrt{\frac{R}{g}}\]
D) \[\frac{gR}{\sqrt{g\cos \theta }}\]
Correct Answer: C
Solution :
Acceleration of body along AB is \[g\cos \theta \] Distance travelled in time t sec =\[AB=\frac{1}{2}(g\cos \theta ){{t}^{2}}\] From\[\Delta ABC,\ AB=2R\cos \theta ;\ 2R\cos \theta =\frac{1}{2}g\cos \theta {{t}^{2}}\] Þ \[{{t}^{2}}=\frac{4R}{g}\]or \[t=2\sqrt{\frac{R}{g}}\]
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