KCET · Physics · Rotational Motion
Moment of inertia of a body about two perpendicular axes \( X \) and \( Y \) in the plane of lamina are \( 20 \mathrm{Kg} m^{2} \) and \( 25 \mathrm{Kg} m^{2} \) respectively. Its moment of inertia about an axis perpendicular to the plane of the lamina and passing through the point of intersection of \( X \) and \( Y \) axes is
- A \( 5 \mathrm{Kg} m^{2} \)
- B \( 45 \mathrm{Kg} m^{2} \)
- C \( 12.5 \mathrm{Kg} m^{2} \)
- D \( 500 \mathrm{Kg} m^{2} \)
Answer & Solution
Correct Answer
(B) \( 45 \mathrm{Kg} m^{2} \)
Step-by-step Solution
Detailed explanation
Given, moment of inertia of body about \( X \) axis \( I_{X}=20 \mathrm{kgm}^{2} \)
Moment of inertia of body about \( Y \) axis \( =I_{Y}=25 \mathrm{kgm}^{2} \)
Then by perpendicular axis theorem, moment of inertia about an axis perpendicular to the plane is the sum of the
moments of inertia of two perpendicular axes through the same point. Therefore
\(I_{Z}=I_{X}+I_{Y}=(20+25) \mathrm{kgm}^{2}=45 \mathrm{kgm}^{2}\)
Hence, moment of inertia about an axis perpendicular to the plane of the lamina and passing through the point of
intersection of \( X \) and \( Y \) axes is \( 45 \mathrm{kgm}^{2} \)
Moment of inertia of body about \( Y \) axis \( =I_{Y}=25 \mathrm{kgm}^{2} \)
Then by perpendicular axis theorem, moment of inertia about an axis perpendicular to the plane is the sum of the
moments of inertia of two perpendicular axes through the same point. Therefore
\(I_{Z}=I_{X}+I_{Y}=(20+25) \mathrm{kgm}^{2}=45 \mathrm{kgm}^{2}\)
Hence, moment of inertia about an axis perpendicular to the plane of the lamina and passing through the point of
intersection of \( X \) and \( Y \) axes is \( 45 \mathrm{kgm}^{2} \)
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