Kinematics Question 188
Question: 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
Options:
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 }} $
Show Answer
Answer:
Correct Answer: C
Solution:
Acceleration of body along AB it’s $ 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} $
therefore $ t^{2}=\frac{4R}{g} $ or $ t=2\sqrt{\frac{R}{g}} $
BETA
