A constant torque of 3.14 N mis exerted on a pivoted wheel. If the angular acceleration of the wheel is 4 pi "rad/" s^(2) then the moment of inertia of the wheel is

A constant torque of 3.14 N mis exerted on a pivoted wheel. If the ang

1.	A constant torque of 3.14 N mis exerted on a pivoted wheel. If the angular acceleration of the wheel is 4 pi "rad/" s^(2) then the moment of inertia of the wheel is
Answer» `0.25 KG m^(2)` `2.5 kg m^(2)` `4.5 kg m^(2)` `25 kg m^(2)` SOLUTION :Here , `tau = 3.14 N m , alpha = 4 PI "rad/"s^(2)` As `tau = I alpha` `THEREFORE I = (tau)/(alpha) = (3.14)/(4pi) = (3.14)/(4 xx 3.14) = 0.25 kg m^(2)`