You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Moment of Inertia of Hollow Sphere

The moment of...

Question

The moment of inertia of a fly-wheel is $4 k g m^{2}$ . A torque of $10$ Newton-meter is applied on it. The angular acceleration produced will be :

25 r a d i a n s / s e c^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.25 r a d i a n s / s e c^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2.5 r a d i a n s / s e c^{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Z e r o

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $2.5 r a d i a n s / s e c^{2}$
Using the relation $τ = I α$ we get $α = \frac{10}{4} = 2.5 r a d / s^{2}$

Suggest Corrections

Similar questions

Q. The moment of inertia of a body about a given axis is

1.2 k g m^{2}

. Initially, the body is at rest. In order to produce a rotational kinetic energy of

1500 j o u l e

, an angular acceleration of

25 r a d i a n / s e c^{2}

must be applied about that axis for a duration of

Q. The linear and angular acceleration of a particle are

10 m / s e c^{2}

and

5 r a d / s e c^{2}

respectively it will be at a distance from the axis of rotation-

Q. The moment of inertia of a body about a given axis is

1.2 k g m^{2}

. Initially the body is at rest. In order to produce a rotational KE of

1500 J

, an angular acceleration of

25 r a d / s e c^{2}

must be applied about that axis for a duration of:

The moment of inertia of a body about a given axis is $1.2 k g m^{2}$ . Initially the body is at rest. In order to produce a rotational kinetic energy of 1500J, an angular acceleration of $25 r a d / s e c^{2}$ must be applied about that axis for a duration of:

Q. A particle, situated in an object, moves with angular acceleration of

6 r a d / s e c^{2}

and with

2 \sqrt{2} r a d / s e c

angular velocity. If the radius of the circular path is

1 m

, its total acceleration in

m / s e c^{2}

will be:

Join BYJU'S Learning Program

The moment of inertia of a fly-wheel is 4kg m2. A torque of 10 Newton-meter is applied on it. The angular acceleration produced will be :

The correct option is C 2.5radians/sec2Using the relation τ=Iα we get α=104=2.5rad/s2

The moment of inertia of a fly-wheel is $4 k g m^{2}$ . A torque of $10$ Newton-meter is applied on it. The angular acceleration produced will be :

The correct option is C $2.5 r a d i a n s / s e c^{2}$
Using the relation $τ = I α$ we get $α = \frac{10}{4} = 2.5 r a d / s^{2}$