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A wheel of moment of inertia 2.5 kgm^(2)...

A wheel of moment of inertia 2.5 `kgm^(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 kgm^(-2), omega = 40 rad s^(-1), tau = 10 Nm`
As `tau = Ialpha, 10 = 2.5 alpha, alpha = 4 rad s^(-2)` `20 = 4t, t=5 s`
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