A solid cylinder of mass M and radius R rolls down an inclined plane of height h. The angular velocity of the cylinder when it reaches the bottom of the plane will be :

Options

A. `1/(2R) sqrt(gh)`
B. ` 2/R sqrt(gh)`
C. `2/R sqrt((gh)/3)`
D. `2/R sqrt((gh)/2)`

Answer

C

Explanation

` V = sqrt ((2gh)/(1+(K^(2))/(r^(2)))) = sqrt((2gh)/(1+1/2) ) = sqrt((4gh)/3)`