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A wheel of moment of inertia 2.5 kg .m^2...

A wheel of moment of inertia `2.5 kg .m^2` has an initial angular velocity of `40 rad s^-1`. A constant torque of 10 Nm acts on the wheel. The time during which the wheel is accelerated to `60 rad s^-1` is

A

4s

B

6s

C

5s

D

2.5s

Text Solution

Verified by Experts

The correct Answer is:
C
Given: `MI=2.5 kg m^-2`
`omega_0=40 rad s^-1`
t=10Nm
As t=Ia
`therefore 10=2.5a`
`a=4 rad s^-2`
Now, `omega=omega_0+a t`
`therefore 60=40 +4 times t`
or `20=4t`
`t=5s`
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