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PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365956 In Column I, shape of different bodies are given and in column II their moment of inertia are given. Match column I wit column II.
supporting img

1 A - S, B - P, C - Q, D - P

2 A - Q, B - P, C - S, D - S

3 A - S, B - R, C - Q, D - P

4 A - Q, B - S, C - P, D - P

Explanation:

$I_{ring} = \frac{m R^{2}}{2} + m R^{2} = \frac{3 m R^{2}}{2}$
$I_{disc} = \frac{m R^{2}}{4}$
$I_{rod} = \frac{m (2 R)^{2}}{12} = \frac{m R^{2}}{3}$
$I_{L shape} = \frac{m R^{2}}{3} + m R^{2} = \frac{4 m R^{2}}{3}$
Correct Option is (3).

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365957 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 and it passes through the centre of the plate making an angle $θ$ with $AB$ . The moment of inertia of the plate about the axis $CD$ is equal to:
supporting img

1

I \cos^{2} θ

2 I

3

I \sin^{2} θ

4

I \cos^{2} (\frac{θ}{2})

Explanation:

As shown in the figure
supporting img
$I_{C D} = I_{C^{'} D^{'}}$
So, $2 I_{C D} = I \Rightarrow I_{C D} = \frac{I}{2}$
Also, $2 I_{A B} = I = I_{A B} = \frac{I}{2}$
So $I_{C D} = I_{A B}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365958 Consider a thin uniform square sheet made of a rigid material. If its side is ' $a$ ' mass $m$ and moment of inertia $I$ about one of its diagonals, then:

1

I = \frac{m a^{2}}{12}

2

\frac{m a^{2}}{24} < I < \frac{m a^{2}}{12}

3

I > \frac{m a^{2}}{12}

4

I = \frac{m a^{2}}{24}

Explanation:

Using perpendicular axis theorem.
$\begin{aligned} I_{0} = I_{x} + I_{y} \\ I_{0} = 2 I (I_{x} = I_{y} = I) \end{aligned}$
supporting img
$\Rightarrow I = \frac{I_{0}}{2} = \frac{m a^{2}}{12}$

JEE - 2015

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365956 In Column I, shape of different bodies are given and in column II their moment of inertia are given. Match column I wit column II.
supporting img

1 A - S, B - P, C - Q, D - P

2 A - Q, B - P, C - S, D - S

3 A - S, B - R, C - Q, D - P

4 A - Q, B - S, C - P, D - P

Explanation:

$I_{ring} = \frac{m R^{2}}{2} + m R^{2} = \frac{3 m R^{2}}{2}$
$I_{disc} = \frac{m R^{2}}{4}$
$I_{rod} = \frac{m (2 R)^{2}}{12} = \frac{m R^{2}}{3}$
$I_{L shape} = \frac{m R^{2}}{3} + m R^{2} = \frac{4 m R^{2}}{3}$
Correct Option is (3).

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365957 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 and it passes through the centre of the plate making an angle $θ$ with $AB$ . The moment of inertia of the plate about the axis $CD$ is equal to:
supporting img

1

I \cos^{2} θ

2 I

3

I \sin^{2} θ

4

I \cos^{2} (\frac{θ}{2})

Explanation:

As shown in the figure
supporting img
$I_{C D} = I_{C^{'} D^{'}}$
So, $2 I_{C D} = I \Rightarrow I_{C D} = \frac{I}{2}$
Also, $2 I_{A B} = I = I_{A B} = \frac{I}{2}$
So $I_{C D} = I_{A B}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365958 Consider a thin uniform square sheet made of a rigid material. If its side is ' $a$ ' mass $m$ and moment of inertia $I$ about one of its diagonals, then:

1

I = \frac{m a^{2}}{12}

2

\frac{m a^{2}}{24} < I < \frac{m a^{2}}{12}

3

I > \frac{m a^{2}}{12}

4

I = \frac{m a^{2}}{24}

Explanation:

Using perpendicular axis theorem.
$\begin{aligned} I_{0} = I_{x} + I_{y} \\ I_{0} = 2 I (I_{x} = I_{y} = I) \end{aligned}$
supporting img
$\Rightarrow I = \frac{I_{0}}{2} = \frac{m a^{2}}{12}$

JEE - 2015

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365956 In Column I, shape of different bodies are given and in column II their moment of inertia are given. Match column I wit column II.
supporting img

1 A - S, B - P, C - Q, D - P

2 A - Q, B - P, C - S, D - S

3 A - S, B - R, C - Q, D - P

4 A - Q, B - S, C - P, D - P

Explanation:

$I_{ring} = \frac{m R^{2}}{2} + m R^{2} = \frac{3 m R^{2}}{2}$
$I_{disc} = \frac{m R^{2}}{4}$
$I_{rod} = \frac{m (2 R)^{2}}{12} = \frac{m R^{2}}{3}$
$I_{L shape} = \frac{m R^{2}}{3} + m R^{2} = \frac{4 m R^{2}}{3}$
Correct Option is (3).

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365957 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 and it passes through the centre of the plate making an angle $θ$ with $AB$ . The moment of inertia of the plate about the axis $CD$ is equal to:
supporting img

1

I \cos^{2} θ

2 I

3

I \sin^{2} θ

4

I \cos^{2} (\frac{θ}{2})

Explanation:

As shown in the figure
supporting img
$I_{C D} = I_{C^{'} D^{'}}$
So, $2 I_{C D} = I \Rightarrow I_{C D} = \frac{I}{2}$
Also, $2 I_{A B} = I = I_{A B} = \frac{I}{2}$
So $I_{C D} = I_{A B}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365958 Consider a thin uniform square sheet made of a rigid material. If its side is ' $a$ ' mass $m$ and moment of inertia $I$ about one of its diagonals, then:

1

I = \frac{m a^{2}}{12}

2

\frac{m a^{2}}{24} < I < \frac{m a^{2}}{12}

3

I > \frac{m a^{2}}{12}

4

I = \frac{m a^{2}}{24}

Explanation:

Using perpendicular axis theorem.
$\begin{aligned} I_{0} = I_{x} + I_{y} \\ I_{0} = 2 I (I_{x} = I_{y} = I) \end{aligned}$
supporting img
$\Rightarrow I = \frac{I_{0}}{2} = \frac{m a^{2}}{12}$

JEE - 2015

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