A ring and a disc are initially at rest, side by side, at the top of an inclined plane which makes an angle 60° with the horizontal. They start to roll without slipping at the same instant of time along the shortest path. If the time difference between their reaching the ground is (2−√3)√10s, then the height of the top of the inclined plane, in meters, is ________. Take g=10ms⁻².

acceleration of a rolling body down an inclined plane
ac = gsinθ/(1 + Ic/MR²)
where Ic is the moment of inertia of the body
aring = gsinθ/2
disc = 2gsinθ/3
using the formula S = Ut + at²/2
hsinθ = 1/2(gsinθ/2)t₁² ⇒ t₁ = √(4h/g)
√(16h/3g)
hsinθ = 1/2(2gsinθ/3)t₂² ⇒ t₂ = √(3h/g)
√(4h/g)
⇒ √(16h/3g) - √(4h/g) = (2 - √3)√10
√h[4/√3 - 2] = (2 - √3)√10
√h[4 - 2√3]/√3 = (2 - √3)√10
√h = √3/2√10
√h = √30/2
⇒ h = 30/4 = 7.5 m
Let the height of the inclined plane be h.
For the ring:
a_r = g sinθ / 2
For the disc:
a_d = 2g sinθ / 3
Time taken by ring to reach the bottom is t_r = √(2h/(a_r sinθ)) = √(4h/(g sinθ))
Time taken by disc to reach the bottom is t_d = √(2h/(a_d sinθ)) = √(3h/(g sinθ))
given that t_r - t_d = (2 - √3)√10
√(4h/g) - √(3h/g) = (2 - √3)√10
√h(2/√g - √3/√g) = (2 - √3)√10
√h (2 - √3)/√g = (2 - √3)√10
√h = √(10g)
Given g = 10 m/s²
√h = √100 = 10
h = 100 m
For θ = 60°
t_r = √(4h/(g sin60°)) = √(8h/(g√3))
t_d = √(3h/(g sin60°)) = √(6h/(g√3))
√(8h/(g√3)) - √(6h/(g√3)) = (2 - √3)√10
√h/√g√3 [2√2 - √6] = (2-√3)√10
√h (2√2 - √6)/√(10√3) = (2-√3)√10
√h = (2 - √3)√10 √(10√3) / (2√2 - √6)
This gives h = 7.5 m