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Let I be the moment of inertia of a uniform square plate about an axis AB that passes through its centre and is parallel to two of its sides. CD is a line in the plane of the plate that passes through the centre of the plate and makes an angle q with AB. The moment of inertia of the plate about the axis CD is then equal to
- a)I
- b)I sin2θ
- c)I cos2θ
- d)I cos2 (θ/2)
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Let I be the moment of inertia of a uniform square plate about an axis...
A'B' ⊥ AB and C' D'⊥ CD
From symmetry IAB = IA'B' and ICD = IC'D'
From theorem of
From theorem of
perpendicular axes,
Izz = IAB + IA'B' = ICD + IC'D'
⇒ 2IAB = 2ICD
∴ IAB = ICD
Most Upvoted Answer
Let I be the moment of inertia of a uniform square plate about an axis...
The moment of inertia of a uniform square plate about an axis passing through its center and parallel to two of its sides is given by the formula:
I = (1/6) * m * a^2
where m is the mass of the plate and a is the length of one side of the square.
To find the moment of inertia about the axis CD, we can use the parallel axis theorem. The parallel axis theorem states that the moment of inertia about an axis parallel to and at a distance d from an axis passing through the center of mass is given by:
I' = I + m * d^2
where I' is the moment of inertia about the new axis.
In this case, the distance between the axis AB and CD is a/2, since CD passes through the center of the plate. Therefore, we can substitute I and d into the parallel axis theorem equation:
I' = (1/6) * m * a^2 + m * (a/2)^2
= (1/6) * m * a^2 + (1/4) * m * a^2
= (1/6 + 1/4) * m * a^2
= (5/12) * m * a^2
Therefore, the moment of inertia of the plate about the axis CD is (5/12) * I, which is equal to option b) I * sin^2(q).
I = (1/6) * m * a^2
where m is the mass of the plate and a is the length of one side of the square.
To find the moment of inertia about the axis CD, we can use the parallel axis theorem. The parallel axis theorem states that the moment of inertia about an axis parallel to and at a distance d from an axis passing through the center of mass is given by:
I' = I + m * d^2
where I' is the moment of inertia about the new axis.
In this case, the distance between the axis AB and CD is a/2, since CD passes through the center of the plate. Therefore, we can substitute I and d into the parallel axis theorem equation:
I' = (1/6) * m * a^2 + m * (a/2)^2
= (1/6) * m * a^2 + (1/4) * m * a^2
= (1/6 + 1/4) * m * a^2
= (5/12) * m * a^2
Therefore, the moment of inertia of the plate about the axis CD is (5/12) * I, which is equal to option b) I * sin^2(q).
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Let I be the moment of inertia of a uniform square plate about an axis AB that passes through its centre and is parallel to two of its sides. CD is a line in the plane of the plate that passes through the centre of the plate and makes an angle q with AB. The moment of inertia of the plate about the axis CD is then equal toa)Ib)I sin2θc)I cos2θd)I cos2 (θ/2)Correct answer is option 'A'. Can you explain this answer?
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Let I be the moment of inertia of a uniform square plate about an axis AB that passes through its centre and is parallel to two of its sides. CD is a line in the plane of the plate that passes through the centre of the plate and makes an angle q with AB. The moment of inertia of the plate about the axis CD is then equal toa)Ib)I sin2θc)I cos2θd)I cos2 (θ/2)Correct answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Let I be the moment of inertia of a uniform square plate about an axis AB that passes through its centre and is parallel to two of its sides. CD is a line in the plane of the plate that passes through the centre of the plate and makes an angle q with AB. The moment of inertia of the plate about the axis CD is then equal toa)Ib)I sin2θc)I cos2θd)I cos2 (θ/2)Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let I be the moment of inertia of a uniform square plate about an axis AB that passes through its centre and is parallel to two of its sides. CD is a line in the plane of the plate that passes through the centre of the plate and makes an angle q with AB. The moment of inertia of the plate about the axis CD is then equal toa)Ib)I sin2θc)I cos2θd)I cos2 (θ/2)Correct answer is option 'A'. Can you explain this answer?.
Let I be the moment of inertia of a uniform square plate about an axis AB that passes through its centre and is parallel to two of its sides. CD is a line in the plane of the plate that passes through the centre of the plate and makes an angle q with AB. The moment of inertia of the plate about the axis CD is then equal toa)Ib)I sin2θc)I cos2θd)I cos2 (θ/2)Correct answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Let I be the moment of inertia of a uniform square plate about an axis AB that passes through its centre and is parallel to two of its sides. CD is a line in the plane of the plate that passes through the centre of the plate and makes an angle q with AB. The moment of inertia of the plate about the axis CD is then equal toa)Ib)I sin2θc)I cos2θd)I cos2 (θ/2)Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let I be the moment of inertia of a uniform square plate about an axis AB that passes through its centre and is parallel to two of its sides. CD is a line in the plane of the plate that passes through the centre of the plate and makes an angle q with AB. The moment of inertia of the plate about the axis CD is then equal toa)Ib)I sin2θc)I cos2θd)I cos2 (θ/2)Correct answer is option 'A'. Can you explain this answer?.
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