__Question 7.11__:

__Solution__:*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, I

_{1}= mr

^{2}

The moment of inertia of the solid sphere about an axis passing through its centre, I

_{2}= (2/5)mr

^{2}

We have the relation:

τ = Iα

Where,

α = Angular acceleration

τ = Torque

*I*= Moment of inertia

For the hollow cylinder, τ

_{1}= I

_{1}α

_{1}

_{}

For the solid sphere, τ

_{n}= I

_{n}α

_{n}

As an equal torque is applied to both the bodies, τ

_{1 }= τ

_{2}

∴ α

_{2 }

_{}/ α

_{1}= I

_{1}/ I

_{2}= mr

^{2}/ (2/5)mr

^{2}

α

_{2}> α

_{1}...(i)

Now, using the relation:

ω = ω

_{0}+ αt

Where,

*ω*

_{0}= Initial angular velocity

*t*= Time of rotation

*ω*= Final angular velocity

For equal

*ω*

_{0}and

*t*, we have:

*ω*∝ α … (

*ii*)

From equations (

*i*) and (

*ii*), we can write:

*ω*

_{2}

*> ω*

_{1}

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

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