State theorem of perpendicular and parallel …

Let m and r be the respective masses of the hollow cylinder and the solid sphere.
 moment of inertia of the hollow cylinder I₁ = mr²
The moment of inertia of the solid sphere I₂ {tex}= \frac { 2 } { 5 } m r ^ { 2 }{/tex}
We have the relation:
{tex}\tau = I a{/tex}
For the hollow cylinder, {tex}\tau _ { 1 } = I _ { 1 } \alpha _ { 1 }{/tex}
For the solid sphere, {tex}\tau _ { 2 } = I _ { 2 } \alpha _ { 2 }{/tex}
As an equal torque is applied to both the bodies, {tex}\tau _ { 1 } = \tau _ { 2 }{/tex}
{tex}\therefore \frac {\alpha_ { 2 } } { \alpha _ { 1 } } = \frac { I _ { 1} } { I _ { 2 } } = \frac { M r ^ { 2 } } { \frac { 2 } { 5 } M r ^ { 2 } } = \frac { 2 } { 5 }{/tex}
{tex}a _ { 2 } > a _ { 1 }{/tex} ….(i)
Now, using the relation:
{tex}\omega = \omega _ { 0 } + a t{/tex}
{tex}\omega \propto a{/tex} …(ii)
From equations (i) and (ii), we can write:
{tex}\omega _ { 2 } > \omega _ { 1 }{/tex}
Hence, the angular velocity of the solid sphere will be greater than that of the hollow cylinder.