
If disk slips on inclined plane, then its acceleration, a1=gsinθ
L=21a1t12
⇒t1=a12L…(i)
If disk rolls on inclined plane, its acceleration,
a2=1+mR2Igsinθ
a2=1+2mR2mR2gsinθ
a2=32gsinθ
Now L=21a2⋅t22
⇒t2=a22L…(ii)
Now, t1t2=a2a1=23
⇒x=2