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Byju's Answer

Standard XII

Physics

Superposition of Electric Fields

Two bodies wi...

Question

Two bodies with with moment of inertia $I_{1}$ and $I_{2} (I_{2} > I_{1})$ are rotating with same angular momentum. If $K_{1}$ and $K_{2}$ are their kinetic energies, then

K_{2} > K_{1}

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K_{2} < K_{1}

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K_{1} = K_{2}

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K_{2} \geq K_{1}

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Solution

The correct option is B $K_{2} < K_{1}$
KE= $\frac{L^{2}}{2 I}$
L is constant in this problem
KE=Kinetic energy
L=angular momentum
I=moment of inertia
KE is inversely proportional to I
As $I_{2} > I_{1}$
$K E_{1} > K E_{2}$

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Two bodies with with moment of inertia I1 and I2(I2>I1) are rotating with same angular momentum. If K1 and K2 are their kinetic energies, then

The correct option is B K2<K1KE=L22IL is constant in this problemKE=Kinetic energyL=angular momentumI=moment of inertiaKE is inversely proportional to IAs I2>I1KE1>KE2

Two bodies with with moment of inertia $I_{1}$ and $I_{2} (I_{2} > I_{1})$ are rotating with same angular momentum. If $K_{1}$ and $K_{2}$ are their kinetic energies, then

The correct option is B $K_{2} < K_{1}$
KE= $\frac{L^{2}}{2 I}$
L is constant in this problem
KE=Kinetic energy
L=angular momentum
I=moment of inertia
KE is inversely proportional to I
As $I_{2} > I_{1}$
$K E_{1} > K E_{2}$