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A solid cylinder is released from rest from the top of an inclined plane of inclination  30\degree, and length
60 cm. If the cylinder rolls without slipping, its speed upon reaching the bottom of the inclined plane
is ________  ms^{-1}  (Given g = 10 ms^{-2}  )

Option: 1


Option: 2

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Option: 3

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Option: 4

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Answers (1)

best_answer

 

h = 60 sin 30
\therefore  h = 30 cm

 
v=\sqrt{\frac{2 g h}{1+\frac{k^2}{R^2}}} 
 
 
k^2=\frac{R^2}{2}
 
v=2 \mathrm{~ms}^{-1} 

Posted by

vishal kumar

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