Moment of inertia of a body about an axis is `4kg-m^(2)`. The body is initially at rest and a torque of 8 N-m starts along the same axis. Work done by the R, about an axis which is a tangent and parallel to its torque in 20 s, in joules, is

Options

A. 40
B. 640
C. 2560
D. 3200

Answer

(d)

Explanation

Moment of inertia,`I=4kg-m`
torque, omega=8 N-m
Time,t=20s
The expression for eork done by the torque is given as,
omega=tautheta
Where,
`theta=(1)/(2)omega^(2)`
and `a=(tau)/(I)=(8)/(4)=2` `(becausetau=Ia)`
Now,
`theta=(1)/(2)(2)(20)^(2)=400`
`therefore omega=(8)xx(400)=3200J`.