A uniform thin bar of mass 6 kg and length 2.4 meter is bent to make an equilateral hexagon. The moment of inertia about an axis passing through the centre of mass and perpendicular to the plane of hexagon is \underline\hspace2cm\times 10 - 1 kg m2.

Rotational Motion PYQ — JEE Main 2021 · 24 Feb · Shift 2 · Q68 with Solution

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Correct answer: 8

Understand the geometry

A thin uniform bar of total length $2.4\,\text{m}$ is bent into an equilateral hexagon.

So each side of the hexagon is

a=\frac{2.4}{6}=0.4\,\text{m}

Total mass is $6\,\text{kg}$ , so mass per side is

m_s=\frac{6}{6}=1\,\text{kg}

Required moment of inertia

We need the moment of inertia of the hexagonal wire about an axis through its centre and perpendicular to its plane.

The hexagon consists of 6 identical thin rods.

So,

I=6\times I_{\text{one side about centre of hexagon}}

Moment of inertia of one side

For one side, use the parallel axis theorem.

(a) MOI of one rod about an axis through its own centre and perpendicular to plane

For a thin rod of length $a$ :

I_{\text{cm, rod}}=\frac{1}{12}m_s a^2

(b) Distance of side from centre of regular hexagon

For a regular hexagon of side $a$ , the apothem is

d=\frac{\sqrt{3}}{2}a

Thus,

d=\frac{\sqrt{3}}{2}(0.4)

So,

d^2=\frac{3}{4}a^2

(c) By parallel axis theorem

I_{\text{one side}}=\frac{1}{12}m_s a^2+m_s d^2

=\frac{1}{12}m_s a^2+m_s\cdot \frac{3}{4}a^2

=m_s a^2\left(\frac{1}{12}+\frac{3}{4}\right)

=m_s a^2\left(\frac{1}{12}+\frac{9}{12}\right)

=m_s a^2\cdot \frac{10}{12} =\frac{5}{6}m_s a^2

Total MOI of hexagon

I=6\times \frac{5}{6}m_s a^2=5m_s a^2

Since $m_s=1\,\text{kg}$ and $a=0.4\,\text{m}$ ,

I=5(1)(0.4)^2

I=5\times 0.16=0.8\,\text{kg m}^2

Match with required format

The question writes the answer as

\underline{\hspace{1cm}}\times 10^{-1}\,\text{kg m}^2

Since

0.8=8\times 10^{-1}

So the required integer is

\boxed{8}

Comparison with stored answer

Derived answer: $8$

Stored correct answer: $8$

They match.

More from Rotational Motion

View written solutionFree

Correct answer: 8

Understand the geometry

A thin uniform bar of total length $2.4\,\text{m}$ is bent into an equilateral hexagon.

So each side of the hexagon is

a=\frac{2.4}{6}=0.4\,\text{m}

Total mass is $6\,\text{kg}$ , so mass per side is

m_s=\frac{6}{6}=1\,\text{kg}

Required moment of inertia

We need the moment of inertia of the hexagonal wire about an axis through its centre and perpendicular to its plane.

The hexagon consists of 6 identical thin rods.

So,

I=6\times I_{\text{one side about centre of hexagon}}

Moment of inertia of one side

For one side, use the parallel axis theorem.

(a) MOI of one rod about an axis through its own centre and perpendicular to plane

For a thin rod of length $a$ :

I_{\text{cm, rod}}=\frac{1}{12}m_s a^2

(b) Distance of side from centre of regular hexagon

For a regular hexagon of side $a$ , the apothem is

d=\frac{\sqrt{3}}{2}a

Thus,

d=\frac{\sqrt{3}}{2}(0.4)

So,

d^2=\frac{3}{4}a^2

(c) By parallel axis theorem

I_{\text{one side}}=\frac{1}{12}m_s a^2+m_s d^2

=\frac{1}{12}m_s a^2+m_s\cdot \frac{3}{4}a^2

=m_s a^2\left(\frac{1}{12}+\frac{3}{4}\right)

=m_s a^2\left(\frac{1}{12}+\frac{9}{12}\right)

=m_s a^2\cdot \frac{10}{12} =\frac{5}{6}m_s a^2

Total MOI of hexagon

I=6\times \frac{5}{6}m_s a^2=5m_s a^2

Since $m_s=1\,\text{kg}$ and $a=0.4\,\text{m}$ ,

I=5(1)(0.4)^2

I=5\times 0.16=0.8\,\text{kg m}^2

Match with required format

The question writes the answer as

\underline{\hspace{1cm}}\times 10^{-1}\,\text{kg m}^2

Since

0.8=8\times 10^{-1}

So the required integer is

\boxed{8}

Comparison with stored answer

Derived answer: $8$

Stored correct answer: $8$

They match.

Rotational Motion question

Rotational Motion question

(a) MOI of one rod about an axis through its own centre and perpendicular to plane

(b) Distance of side from centre of regular hexagon

(c) By parallel axis theorem

More from Rotational Motion

Rotational Motion question

Rotational Motion question

(a) MOI of one rod about an axis through its own centre and perpendicular to plane

(b) Distance of side from centre of regular hexagon

(c) By parallel axis theorem

More from Rotational Motion