The angular momentum of a flywheel having a moment of inertia of $0.4 {kg-m}^{2}$ decreases from $30$ to $20 {kg-m}^{2} / s$ in a period of $2 seconds$ . The average torque acting on the flywheel during this period is :

10 N.m

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2.5 N.m

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2 N.m

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1.5 N.m

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Solution

The correct option is C $2 N.m$
Given :
Moment of inertia $I = 0.4 {kg-m}^{2}$
$∵$ Angular momentum $L = I ω$
or $Δ L = I (ω_{1} - ω_{2}) = 4 {kg-m}^{2} / s$ and $Δ t = 2 sec$
$∴$ average torque $τ_{a v g} = \frac{Δ L}{Δ t}$
$= \frac{4}{2} = 2 N.m.$

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The angular momentum of a flywheel having a moment of inertia of 0.4 kg-m2 decreases from 30 to 20 kg-m2/s in a period of 2 seconds. The average torque acting on the flywheel during this period is :

The correct option is C 2 N.mGiven : Moment of inertia I=0.4 kg-m2 ∵ Angular momentum L=Iω or Δ L=I(ω1−ω2)=4 kg-m2/s and Δ t=2 sec ∴ average torque τavg=Δ LΔ t =42=2 N.m.

The angular momentum of a flywheel having a moment of inertia of $0.4 {kg-m}^{2}$ decreases from $30$ to $20 {kg-m}^{2} / s$ in a period of $2 seconds$ . The average torque acting on the flywheel during this period is :

The correct option is C $2 N.m$
Given :
Moment of inertia $I = 0.4 {kg-m}^{2}$
$∵$ Angular momentum $L = I ω$
or $Δ L = I (ω_{1} - ω_{2}) = 4 {kg-m}^{2} / s$ and $Δ t = 2 sec$
$∴$ average torque $τ_{a v g} = \frac{Δ L}{Δ t}$
$= \frac{4}{2} = 2 N.m.$