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Standard XII

Physics

Physical Pendulum

The moment of...

Question

The moment of inertia of a thin square plate ABCD of uniform thickness about an axis passing through its centre and perpendicular to its plane will be:

I_{3} + I_{2}

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I_{1} - I_{3}

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I_{3} + I_{2} + I_{4}

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I_{1} + I_{2} + I_{3} + I_{4}

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Solution

The correct option is D $I_{1} + I_{2} + I_{3} + I_{4}$
The axis of $I_{1}$ and $I_{2}$ and that of $I_{3}$ and $I_{4}$ are perpendicular to each other .
Using perpendicular axis theorem $I = I_{1} + I_{2} + I_{3} + I_{4}$
$\begin{matrix} \Rightarrow I = 2 I_{1} + 2 I_{2} + 2 I_{3} + 2 I_{4} \Rightarrow I_{1} = I_{2} = I_{3} = I_{4} \end{matrix}$
Hence, Option $D$ is correct answer.

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Similar questions

Q. The moment of inertia of a thin square plate ABCD of uniform thickness about an axis passing through the centre O and perpendicular to the plane of the plate is:
a)

I_{1} + I_{3}

I_{2} + I_{4}

2 I_{1} + I_{3}

I_{1} + 2 I_{3}

Q. A thin square plate

A B C D

of uniform thickness is shown in the figure.

I_{1}, I_{2}, I_{3}

and

I_{4}

are respectively the moments of inertia about axis

1, 2, 3

and

4

which are in the plane of the square plate. Then, the moment of inertia

I_{0}

of the plate about an axis passing through its centre

O

and perpendicular to the plane of the plate will be equal to : (Choose the correct option(s))

A B C D

is a square plate with centre

O

. The moment of inertia of the plate about the axes

1, 2, 3

and

4

are

I_{1}, I_{2}, I_{3} & I_{4}

respectively. It follows that

The moment of inertia of thin square plate ABCD of uniform thickness about an axis passing through the centre O and perpendicular to the plane of the plate is not equal to (Given $I_{1}, I_{2}, I_{3} a n d I_{4}$ respectively, are the moments of inertia about axes 1,2,3and 4 which are in the plane of the plate)

The moment of inertia of a thin square plate ABCD of uniform thickness about an axis passing through the centre O and perpendicular to the plane of the plate can NOT be given by :

where, $I_{1}, I_{2}, I_{3}, I_{4}$ are the moments of inertia of the plates about axes 1,2,3,4 respectively.

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The moment of inertia of a thin square plate ABCD of uniform thickness about an axis passing through its centre and perpendicular to its plane will be:

The correct option is D I1+I2+I3+I4The axis of I1 and I2 and that of I3 and I4 are perpendicular to each other .Using perpendicular axis theorem I=I1+I2+I3+I4⇒I=2I1+2I2+2I3+2I4⇒I1=I2=I3=I4Hence, Option D is correct answer.