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BITSAT 2015 Paper

Section: Physics

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Question : 28 of 150

Marks: +1, -0

A thin rod of length 4l and mass 4m is bent at the points as shown in figure. What is the moment of inertia of the rod about the axis passes through point O and perpendicular to the plane of paper?

$\frac{Ml^3}{3}$
$\frac{10Ml^3}{3}$
$\frac{Ml^2}{12}$
$\frac{Ml^2}{24}$

Solution:

Total moment of inertia

=

I_1+I_2+I_3+I_4

=

2I_1+2I_2

=

2(I_1+I_2)

[

I_3

=

I_1

,

I_1

=

I_4

]

Now,

I_2

=

I_3

=

\frac{MI^2}{3}

Using parallel axes theorem, we have

I =

I_{CM} + Mx^2

and x =

\sqrt{l^2 + \frac{l^2}{4}}

I_1

=

I_4

=

\frac{Ml^2}{12}

+ M

\left[ \sqrt{l^2 \left(\frac{l}{2}\right)^2} \right]^2

Putting all values we get

Moment of inertia, I =

\frac{10Ml^3}{3}

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