A solid sphere, resting at the top of a smooth inclined plane of inclination `30^(@)` with the horizontal, rolls down the plane and reaches the bottom, which is at 15.75 m from the top. How much time it will take to reach the bottom ?

Options

A. 2 s
B. 2.5 s
C. 3 s
D. 4 s

Answer

C

Explanation

The downward acceleration of the sphere rolling down the inclined plane is given by
`a=(g sin theta)/(1+(K^(2))/(R^(2)))`
But for a solid sphere, `I=(2)/(5)MR^(2)=MK^(2)`
`therefore" "(K^(2))/(R^(2))=(2)/(5)" "thereore a=(g xx sin 30^(@))/(1+(2)/(5))`
`therefore a=(5)/(7)gxx(1)/(2)=(5)/(7)xx9.8xx(1)/(2)=3.5m//s^(2)`
Using `s=ut+(1)/(2)at^(2),` we get,
`15.75=0+(1)/(2)xx3.5xxt^(2)`
`therefore t^(2)=(15.75xx2)/(3.5)=9" "therefore t=3s`