QBManager

Rotational Motion

150093 Five particles of mass $2 kg$ are attached to the rim of a circular disc of radius $0.1 m$ and negligible mass. Moment of inertia of the system about the axis passing through the centre of the disc and perpendicular to its plane is

1

1 kg m^{2}

2

0.1 kg m^{2}

3

2 kg m^{2}

4

0.2 kg m^{2}

Explanation:

B Given,
Mass of particle $m = 2 kg$
Radius of circular disc $= 0.1 m$
According to question,
original image
As we know that,
Moment of inertia of a circular disc about an axis through its center of mass and perpendicular to the disc
$I_{c . m} = {MR}^{2}$
$I_{net} = 5 \times I_{each}$
$= 5 {mr}^{2}$
$= 5 {mr}^{2} = 5 \times 2 \times (0.1)^{2}$
$= 10 \times 1 / 100 = 0.1 kg \cdot m^{2}$
$I_{net} = 0.1 Kg \cdot m^{2}$

RPMT-2006

Rotational Motion

150094 The ratio of the radii of gyration of a circular disc to that of a circular ring, each of same mass and radius, around their respective axis is

1

\sqrt{3} : \sqrt{2}

2

1 : \sqrt{2}

3

\sqrt{3} : 1

4

\sqrt{5} : \sqrt{3}

Explanation:

B Radius of gyration,
$∵$ Moment of inertia, $I = {MK}^{2}$
$K^{2} = \frac{I}{M}$
$Radius of gyration, K = \sqrt{\frac{I}{M}}$
Moment of inertia of disc, $I = \frac{1}{2} {MR}^{2}$
$K_{disc} = \sqrt{\frac{I}{M}} = \sqrt{\frac{\frac{1}{2} {MR}^{2}}{M}}$
$K_{disc} = \sqrt{\frac{R^{2}}{2}}$
Moment of inertia of ring, $I = {MR}^{2}$
$K_{ring} = \sqrt{\frac{M R^{2}}{M}}$
$K_{ring} = \sqrt{R^{2}}$
Formeq.(i) $\div$ (ii),
$∵ \frac{K_{disc}}{K_{ring}} = \frac{\sqrt{\frac{1}{2} R^{2}}}{\sqrt{R^{2}}} = \frac{1}{\sqrt{2}}$

JCECE-2014

Rotational Motion

150095 The moment of inertia of a uniform thin rod of length $L$ and mass $M$ about an axis passing through a point at a distance $\frac{L}{3}$ from one of its ends and perpendicular to the rod is

1

\frac{7 {ML}^{2}}{48}

2

\frac{{ML}^{2}}{9}

3

\frac{{ML}^{2}}{12}

4

\frac{{ML}^{2}}{3}

Explanation:

B
According to parallel axis theorem-
$I = I_{cm} + M {(\frac{L}{6})}^{2}$
$I = \frac{M L^{2}}{12} + M {(\frac{L}{6})}^{2}$
$= \frac{{ML}^{2}}{12} + \frac{{ML}^{2}}{36}$
$= \frac{4 {ML}^{2}}{36}$
$= \frac{{ML}^{2}}{}$

AMU-2010

Rotational Motion

150096 From a disc of mass ' $M$ ' and radius ' $R$ ', a circular hole of a diameter ' $R$ ' is cut whose rim passes through the centre. The moment of inertia of the remaining part of the disc about perpendicular axis passing through the centre is

1

\frac{9 {MR}^{2}}{32}

2

\frac{13 {MR}^{2}}{32}

3

\frac{11 {MR}^{2}}{32}

4

\frac{7 {MR}^{2}}{32}

Explanation:

B Given that, mass of disc $=$ M, radius of disc $=$ $R$
Moment of inertia $(I_{big disc}) = \frac{1}{2} {MR}^{2}$
Let, moment of inertia of small disc is $I_{1}$ which is cut from big disc.
So, $I_{remaining} = I_{big disc} - I_{1}$
From parallel axis theorem, moment of inertia of the small disc $(I_{1}) = I_{c} + m_{1} R_{1}^{2}$
$I_{1} = \frac{1}{2} \cdot \frac{M}{4} \cdot {(\frac{R}{2})}^{2} + \frac{M}{4} \times {(\frac{R}{2})}^{2} [\begin{array}{l} ∵ m_{1} = M / 4 \\ R_{1} = R / 2 \end{array} & I_{c} = \frac{1}{2} m_{1} R_{1}^{2}]$
$= \frac{M^{2}}{32} + \frac{M R^{2}}{16}$
$I_{1} = \frac{3 M^{2}}{32}$
Putting the value of these in equation (ii), we get-
$∴ I_{remaining} = I_{big disc} - I_{1}$
$I_{remaining} = \frac{1}{2} {MR}^{2} - \frac{3 {MR}^{2}}{32}$
$I_{remaining} = \frac{13 {MR}^{2}}{32}$

MHT CET-2020

Rotational Motion

150093 Five particles of mass $2 kg$ are attached to the rim of a circular disc of radius $0.1 m$ and negligible mass. Moment of inertia of the system about the axis passing through the centre of the disc and perpendicular to its plane is

1

1 kg m^{2}

2

0.1 kg m^{2}

3

2 kg m^{2}

4

0.2 kg m^{2}

Explanation:

B Given,
Mass of particle $m = 2 kg$
Radius of circular disc $= 0.1 m$
According to question,
original image
As we know that,
Moment of inertia of a circular disc about an axis through its center of mass and perpendicular to the disc
$I_{c . m} = {MR}^{2}$
$I_{net} = 5 \times I_{each}$
$= 5 {mr}^{2}$
$= 5 {mr}^{2} = 5 \times 2 \times (0.1)^{2}$
$= 10 \times 1 / 100 = 0.1 kg \cdot m^{2}$
$I_{net} = 0.1 Kg \cdot m^{2}$

RPMT-2006

Rotational Motion

150094 The ratio of the radii of gyration of a circular disc to that of a circular ring, each of same mass and radius, around their respective axis is

1

\sqrt{3} : \sqrt{2}

2

1 : \sqrt{2}

3

\sqrt{3} : 1

4

\sqrt{5} : \sqrt{3}

Explanation:

B Radius of gyration,
$∵$ Moment of inertia, $I = {MK}^{2}$
$K^{2} = \frac{I}{M}$
$Radius of gyration, K = \sqrt{\frac{I}{M}}$
Moment of inertia of disc, $I = \frac{1}{2} {MR}^{2}$
$K_{disc} = \sqrt{\frac{I}{M}} = \sqrt{\frac{\frac{1}{2} {MR}^{2}}{M}}$
$K_{disc} = \sqrt{\frac{R^{2}}{2}}$
Moment of inertia of ring, $I = {MR}^{2}$
$K_{ring} = \sqrt{\frac{M R^{2}}{M}}$
$K_{ring} = \sqrt{R^{2}}$
Formeq.(i) $\div$ (ii),
$∵ \frac{K_{disc}}{K_{ring}} = \frac{\sqrt{\frac{1}{2} R^{2}}}{\sqrt{R^{2}}} = \frac{1}{\sqrt{2}}$

JCECE-2014

Rotational Motion

150095 The moment of inertia of a uniform thin rod of length $L$ and mass $M$ about an axis passing through a point at a distance $\frac{L}{3}$ from one of its ends and perpendicular to the rod is

1

\frac{7 {ML}^{2}}{48}

2

\frac{{ML}^{2}}{9}

3

\frac{{ML}^{2}}{12}

4

\frac{{ML}^{2}}{3}

Explanation:

B
According to parallel axis theorem-
$I = I_{cm} + M {(\frac{L}{6})}^{2}$
$I = \frac{M L^{2}}{12} + M {(\frac{L}{6})}^{2}$
$= \frac{{ML}^{2}}{12} + \frac{{ML}^{2}}{36}$
$= \frac{4 {ML}^{2}}{36}$
$= \frac{{ML}^{2}}{}$

AMU-2010

Rotational Motion

150096 From a disc of mass ' $M$ ' and radius ' $R$ ', a circular hole of a diameter ' $R$ ' is cut whose rim passes through the centre. The moment of inertia of the remaining part of the disc about perpendicular axis passing through the centre is

1

\frac{9 {MR}^{2}}{32}

2

\frac{13 {MR}^{2}}{32}

3

\frac{11 {MR}^{2}}{32}

4

\frac{7 {MR}^{2}}{32}

Explanation:

B Given that, mass of disc $=$ M, radius of disc $=$ $R$
Moment of inertia $(I_{big disc}) = \frac{1}{2} {MR}^{2}$
Let, moment of inertia of small disc is $I_{1}$ which is cut from big disc.
So, $I_{remaining} = I_{big disc} - I_{1}$
From parallel axis theorem, moment of inertia of the small disc $(I_{1}) = I_{c} + m_{1} R_{1}^{2}$
$I_{1} = \frac{1}{2} \cdot \frac{M}{4} \cdot {(\frac{R}{2})}^{2} + \frac{M}{4} \times {(\frac{R}{2})}^{2} [\begin{array}{l} ∵ m_{1} = M / 4 \\ R_{1} = R / 2 \end{array} & I_{c} = \frac{1}{2} m_{1} R_{1}^{2}]$
$= \frac{M^{2}}{32} + \frac{M R^{2}}{16}$
$I_{1} = \frac{3 M^{2}}{32}$
Putting the value of these in equation (ii), we get-
$∴ I_{remaining} = I_{big disc} - I_{1}$
$I_{remaining} = \frac{1}{2} {MR}^{2} - \frac{3 {MR}^{2}}{32}$
$I_{remaining} = \frac{13 {MR}^{2}}{32}$

MHT CET-2020

Rotational Motion

150093 Five particles of mass $2 kg$ are attached to the rim of a circular disc of radius $0.1 m$ and negligible mass. Moment of inertia of the system about the axis passing through the centre of the disc and perpendicular to its plane is

1

1 kg m^{2}

2

0.1 kg m^{2}

3

2 kg m^{2}

4

0.2 kg m^{2}

Explanation:

B Given,
Mass of particle $m = 2 kg$
Radius of circular disc $= 0.1 m$
According to question,
original image
As we know that,
Moment of inertia of a circular disc about an axis through its center of mass and perpendicular to the disc
$I_{c . m} = {MR}^{2}$
$I_{net} = 5 \times I_{each}$
$= 5 {mr}^{2}$
$= 5 {mr}^{2} = 5 \times 2 \times (0.1)^{2}$
$= 10 \times 1 / 100 = 0.1 kg \cdot m^{2}$
$I_{net} = 0.1 Kg \cdot m^{2}$

RPMT-2006

Rotational Motion

150094 The ratio of the radii of gyration of a circular disc to that of a circular ring, each of same mass and radius, around their respective axis is

1

\sqrt{3} : \sqrt{2}

2

1 : \sqrt{2}

3

\sqrt{3} : 1

4

\sqrt{5} : \sqrt{3}

Explanation:

B Radius of gyration,
$∵$ Moment of inertia, $I = {MK}^{2}$
$K^{2} = \frac{I}{M}$
$Radius of gyration, K = \sqrt{\frac{I}{M}}$
Moment of inertia of disc, $I = \frac{1}{2} {MR}^{2}$
$K_{disc} = \sqrt{\frac{I}{M}} = \sqrt{\frac{\frac{1}{2} {MR}^{2}}{M}}$
$K_{disc} = \sqrt{\frac{R^{2}}{2}}$
Moment of inertia of ring, $I = {MR}^{2}$
$K_{ring} = \sqrt{\frac{M R^{2}}{M}}$
$K_{ring} = \sqrt{R^{2}}$
Formeq.(i) $\div$ (ii),
$∵ \frac{K_{disc}}{K_{ring}} = \frac{\sqrt{\frac{1}{2} R^{2}}}{\sqrt{R^{2}}} = \frac{1}{\sqrt{2}}$

JCECE-2014

Rotational Motion

150095 The moment of inertia of a uniform thin rod of length $L$ and mass $M$ about an axis passing through a point at a distance $\frac{L}{3}$ from one of its ends and perpendicular to the rod is

1

\frac{7 {ML}^{2}}{48}

2

\frac{{ML}^{2}}{9}

3

\frac{{ML}^{2}}{12}

4

\frac{{ML}^{2}}{3}

Explanation:

B
According to parallel axis theorem-
$I = I_{cm} + M {(\frac{L}{6})}^{2}$
$I = \frac{M L^{2}}{12} + M {(\frac{L}{6})}^{2}$
$= \frac{{ML}^{2}}{12} + \frac{{ML}^{2}}{36}$
$= \frac{4 {ML}^{2}}{36}$
$= \frac{{ML}^{2}}{}$

AMU-2010

Rotational Motion

150096 From a disc of mass ' $M$ ' and radius ' $R$ ', a circular hole of a diameter ' $R$ ' is cut whose rim passes through the centre. The moment of inertia of the remaining part of the disc about perpendicular axis passing through the centre is

1

\frac{9 {MR}^{2}}{32}

2

\frac{13 {MR}^{2}}{32}

3

\frac{11 {MR}^{2}}{32}

4

\frac{7 {MR}^{2}}{32}

Explanation:

B Given that, mass of disc $=$ M, radius of disc $=$ $R$
Moment of inertia $(I_{big disc}) = \frac{1}{2} {MR}^{2}$
Let, moment of inertia of small disc is $I_{1}$ which is cut from big disc.
So, $I_{remaining} = I_{big disc} - I_{1}$
From parallel axis theorem, moment of inertia of the small disc $(I_{1}) = I_{c} + m_{1} R_{1}^{2}$
$I_{1} = \frac{1}{2} \cdot \frac{M}{4} \cdot {(\frac{R}{2})}^{2} + \frac{M}{4} \times {(\frac{R}{2})}^{2} [\begin{array}{l} ∵ m_{1} = M / 4 \\ R_{1} = R / 2 \end{array} & I_{c} = \frac{1}{2} m_{1} R_{1}^{2}]$
$= \frac{M^{2}}{32} + \frac{M R^{2}}{16}$
$I_{1} = \frac{3 M^{2}}{32}$
Putting the value of these in equation (ii), we get-
$∴ I_{remaining} = I_{big disc} - I_{1}$
$I_{remaining} = \frac{1}{2} {MR}^{2} - \frac{3 {MR}^{2}}{32}$
$I_{remaining} = \frac{13 {MR}^{2}}{32}$

MHT CET-2020

Rotational Motion

150093 Five particles of mass $2 kg$ are attached to the rim of a circular disc of radius $0.1 m$ and negligible mass. Moment of inertia of the system about the axis passing through the centre of the disc and perpendicular to its plane is

1

1 kg m^{2}

2

0.1 kg m^{2}

3

2 kg m^{2}

4

0.2 kg m^{2}

Explanation:

B Given,
Mass of particle $m = 2 kg$
Radius of circular disc $= 0.1 m$
According to question,
original image
As we know that,
Moment of inertia of a circular disc about an axis through its center of mass and perpendicular to the disc
$I_{c . m} = {MR}^{2}$
$I_{net} = 5 \times I_{each}$
$= 5 {mr}^{2}$
$= 5 {mr}^{2} = 5 \times 2 \times (0.1)^{2}$
$= 10 \times 1 / 100 = 0.1 kg \cdot m^{2}$
$I_{net} = 0.1 Kg \cdot m^{2}$

RPMT-2006

Rotational Motion

150094 The ratio of the radii of gyration of a circular disc to that of a circular ring, each of same mass and radius, around their respective axis is

1

\sqrt{3} : \sqrt{2}

2

1 : \sqrt{2}

3

\sqrt{3} : 1

4

\sqrt{5} : \sqrt{3}

Explanation:

B Radius of gyration,
$∵$ Moment of inertia, $I = {MK}^{2}$
$K^{2} = \frac{I}{M}$
$Radius of gyration, K = \sqrt{\frac{I}{M}}$
Moment of inertia of disc, $I = \frac{1}{2} {MR}^{2}$
$K_{disc} = \sqrt{\frac{I}{M}} = \sqrt{\frac{\frac{1}{2} {MR}^{2}}{M}}$
$K_{disc} = \sqrt{\frac{R^{2}}{2}}$
Moment of inertia of ring, $I = {MR}^{2}$
$K_{ring} = \sqrt{\frac{M R^{2}}{M}}$
$K_{ring} = \sqrt{R^{2}}$
Formeq.(i) $\div$ (ii),
$∵ \frac{K_{disc}}{K_{ring}} = \frac{\sqrt{\frac{1}{2} R^{2}}}{\sqrt{R^{2}}} = \frac{1}{\sqrt{2}}$

JCECE-2014

Rotational Motion

150095 The moment of inertia of a uniform thin rod of length $L$ and mass $M$ about an axis passing through a point at a distance $\frac{L}{3}$ from one of its ends and perpendicular to the rod is

1

\frac{7 {ML}^{2}}{48}

2

\frac{{ML}^{2}}{9}

3

\frac{{ML}^{2}}{12}

4

\frac{{ML}^{2}}{3}

Explanation:

B
According to parallel axis theorem-
$I = I_{cm} + M {(\frac{L}{6})}^{2}$
$I = \frac{M L^{2}}{12} + M {(\frac{L}{6})}^{2}$
$= \frac{{ML}^{2}}{12} + \frac{{ML}^{2}}{36}$
$= \frac{4 {ML}^{2}}{36}$
$= \frac{{ML}^{2}}{}$

AMU-2010

Rotational Motion

150096 From a disc of mass ' $M$ ' and radius ' $R$ ', a circular hole of a diameter ' $R$ ' is cut whose rim passes through the centre. The moment of inertia of the remaining part of the disc about perpendicular axis passing through the centre is

1

\frac{9 {MR}^{2}}{32}

2

\frac{13 {MR}^{2}}{32}

3

\frac{11 {MR}^{2}}{32}

4

\frac{7 {MR}^{2}}{32}

Explanation:

B Given that, mass of disc $=$ M, radius of disc $=$ $R$
Moment of inertia $(I_{big disc}) = \frac{1}{2} {MR}^{2}$
Let, moment of inertia of small disc is $I_{1}$ which is cut from big disc.
So, $I_{remaining} = I_{big disc} - I_{1}$
From parallel axis theorem, moment of inertia of the small disc $(I_{1}) = I_{c} + m_{1} R_{1}^{2}$
$I_{1} = \frac{1}{2} \cdot \frac{M}{4} \cdot {(\frac{R}{2})}^{2} + \frac{M}{4} \times {(\frac{R}{2})}^{2} [\begin{array}{l} ∵ m_{1} = M / 4 \\ R_{1} = R / 2 \end{array} & I_{c} = \frac{1}{2} m_{1} R_{1}^{2}]$
$= \frac{M^{2}}{32} + \frac{M R^{2}}{16}$
$I_{1} = \frac{3 M^{2}}{32}$
Putting the value of these in equation (ii), we get-
$∴ I_{remaining} = I_{big disc} - I_{1}$
$I_{remaining} = \frac{1}{2} {MR}^{2} - \frac{3 {MR}^{2}}{32}$
$I_{remaining} = \frac{13 {MR}^{2}}{32}$

MHT CET-2020

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