Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time?

Let m and r be the respective masses of the hollow cylinder and the solid sphere. 
The moment of inertia of the hollow cylinder about its standard axis, I₁ = mr² 

The moment of inertia of the solid sphere about an axis passing through its centre, I₂ = mr² 

We have the relation, 
τ = Iα 

where, 
α = Angular acceleration, 
τ = Torque, 
I = Moment of inertia, 
For the hollow cylinder, τ₁ = I₁ α₁
For the solid sphere, τ_n = I_n α_n
As an equal torque is applied to both the bodies, τ₁ = τ_2, 

∴        =   =  
            α₂ > α₁                                      ...(i) 

Now, using the relation, 
ω = ω₀ + αt 

where,
ω₀ = Initial angular velocity 
  t = Time of rotation 
 ω = Final angular velocity 
For equal ω₀ and t, we have
    ω ∝ α                                                 … (ii) 

From equations (i) and (ii), 
ω₂ > ω₁ 

Hence, the angular velocity of the solid sphere will be greater than that of the hollow cylinder.