Correct Answer - A
The acceleration of an object rolling down an inclined plane is given by
`a = (g sin theta)/(1 + I//mr^(2))`
For disc, `I = (1)/(2)mr^(2)`
`:. (I)/(mr^(2)) = (1)/(2), a = (2)/(3)g sin theta = 0.67 g sin theta`
For solid sphere,`I = (2)/(5)mr^(2)`
`:. (I)/(mr^(2)) = (2)/(5), a = (5)/(7) g sin theta = 0.71g sin theta`
Clearly, `a_("solid sphere") gt a_("disc") :. t_("solid sphere") lt t_("disc")`
Hence solid sphere gets to the bottom of the plane first.