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Standard XII

Physics

Moment of Inertia of a Ring

Momentum of i...

Question

Momentum of inertia of a rod of mass $2 k g$ and length $1 m$ about an axis passing through a point $25 c m$ from the center and normal to the length is:

2.9 k g m^{2}

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1.9 k g m^{2}

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0.29 k g m^{2}

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0.19 k g m^{2}

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Solution

The correct option is C $0.29 k g m^{2}$
Moment of inertia= $\frac{m l^{2}}{12} + m r^{2}$
$\frac{2 \times 1^{2}}{12} + 2 \times (\frac{1}{4})^{2}$ =0.29kgm²

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Momentum of inertia of a rod of mass 2 kg and length 1 m about an axis passing through a point 25 cm from the center and normal to the length is:

The correct option is C 0.29 kg m2Moment of inertia=ml212+mr22×1212+2×(14)2=0.29kgm²

Momentum of inertia of a rod of mass $2 k g$ and length $1 m$ about an axis passing through a point $25 c m$ from the center and normal to the length is:

The correct option is C $0.29 k g m^{2}$
Moment of inertia= $\frac{m l^{2}}{12} + m r^{2}$
$\frac{2 \times 1^{2}}{12} + 2 \times (\frac{1}{4})^{2}$ =0.29kgm²