Question

if a disc slides from the top to bottom of an inclined plane it takes time t1 If it rolls it takes time t2 now t2^2/t1^2

Answer 1

t₁² = 2 L/(g SinФ)  ;   t₂² = 3 L/(g SinФ)     =>  t₁² : t₂² = 2 : 3

Let the length of slope = L, mass of the disc = m.  
Moment of Inertia about the center = I = MR² /2.

When the disc slides without rolling (when friction = 0):
      a = g SinФ.    u = 0,     v = a t = g SinФ t
      L = 1/2 g SinФ t₁²    =>   t₁² = 2L/(g SinФ)    ---(1)

When the disc rolls:   friction ≠ 0.
    Friction acts upwards along the point of contact. The disc rolls forward.
    a = g sinФ - f/m      =>   v = a t
    Torque = I α = f R     => R α = 2 f/m      => ω = α t

    Condition for Rolling:  v = R ω   or,   a = R α
    => 3f/m = g sinФ       =>  f = m g SinФ /3
          We know f <= μ N or,  μ m g CosΦ.

Hence, μ >= TanΦ /3   for the disc to roll. Otherwise, the disc only slides.
We assume that the time required from t = 0s, to when the disc starts rolling is very small.

    => a = g SinФ * 2/3
    => L = 1/2 *(g SinФ * 2/3 ) t₂²
    =>  t₂² = 3 L / (gSinФ)

Hence the ratio of squares of times to reach the floor: 2 : 3