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Rotational Motion

150112 Let $M$ and $L$ be the mass and length of thin uniform rod respectively. In $1^{st}$ case, axis of rotation is passing through centre and perpendicular to its length. In $2^{nd}$ case, axis of rotation is passing through one end and perpendicular to its length. The ratio of radius of gyration in first case to second case is

1

1 : 2

2

2 : 1

3

3 : 1

4

1 : 3

Explanation:

A Given, mass of thin $rod = M$
Length of thin $rod = L$
From $1^{st}$ case -
$I_{1} = \frac{{ML}^{2}}{12}$
And $I_{1} = {MK}^{2}$
Equating equation (i) and (ii), we get
$K_{1} = \frac{L}{\sqrt{12}}$
From II $^{nd}$ case -
$I_{2} = \frac{{ML}^{2}}{3}$
And $I_{2} = {MK}_{2}^{2}$
Equating equation (iii) and (iv), we get
$K_{2} = \frac{L}{\sqrt{3}}$
Taking ratios of $K_{1}$ and $K_{2}$ , we get
$\frac{K_{1}}{K_{2}} = \frac{L}{\sqrt{12}} \times \frac{\sqrt{3}}{L}$
$∴ \frac{K_{1}}{K_{2}} = \frac{1}{2}$

MHT-CET- 2020

Rotational Motion

150113 A thin uniform rod of length ' $L$ ' and mass ' $M$ ' is bent at the middle point ' $O$ ' at an angle of $45^{0}$ as shown in the figure. The moment of inertia of the system about an axis passing through ' $O$ ' and perpendicular to the plane of the bent rod, is

1

\frac{{ML}^{2}}{6}

2

\frac{{ML}^{2}}{24}

3

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

4

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

Explanation:

D Given, length of rod = L
$mass of rod = M$
When rod is bent at the middle so each part of it will have the same length $(\frac{L}{2})$ and mass $(\frac{M}{2})$ then-
Moment of inertia of each part-
$I_{1} = \frac{1}{3} (\frac{M}{2}) {(\frac{L}{2})}^{2} = \frac{{ML}^{2}}{24}$
$∴$ Total moment of inertia -
$I = 2 I_{1}$
Putting the value of $I_{1}$ from equation (i), we get-
$I = 2 \times \frac{{ML}^{2}}{24}$
$I = \frac{{ML}^{2}}{12}$

MHT-CET 2020

Rotational Motion

150114 A solid sphere of mass ' $M$ ' and radius ' $R$ ' has moment of inertia ' $I$ ' about its diameter. It recast into a disc of thickness ' $t$ ' whose moment of inertia about an axis passing through its edge and perpendicular to its plane, remains ' $I$ '. Radius of the disc will be

1

\frac{R}{\sqrt{19}}

2

\frac{2 R}{\sqrt{15}}

3

\frac{2 R}{\sqrt{19}}

4

\frac{R}{\sqrt{15}}

Explanation:

B Given, mass of solid sphere $= M$
$Radius of sphere = R$
$Radius of disk = r$
Moment of inertia of solid sphere $I) = \frac{2}{5} {MR}^{2}$
Moment of inertia of $disc = \frac{1}{2} {Mr}^{2}$
From parallel axis theorem -
$I = I_{CM} + {Mr}^{2}$
$\frac{2}{5} {MR}^{2} = \frac{1}{2} {Mr}^{2} + {Mr}^{2} = \frac{3}{2} {Mr}^{2}$
$\frac{2}{5} {MR}^{2} = \frac{3}{2} {Mr}^{2}$
$r = \frac{2}{\sqrt{15}} R$

MHT-CET 2020

Rotational Motion

150115 A ring and a disc have same mass and same radius. The ratio of moment of inertia of a ring about a tangent in its plane to that of the disc about its diameter is

1

6 : 1

2

4 : 1

3

2 : 1

4

8 : 1

Explanation:

A Let, mass $= m$ , radius $= r$
For ring -
By Perpendicular axis theorem
original image
$\begin{array}{ll} I_{z} = I_{x} + I_{y} \\ I_{z} = 2 I_{y} \\ I_{y} = \frac{I_{z}}{2} = \frac{{mr}^{2}}{2} \end{array}$
By using parallel axis theorem -
$I_{R} = I_{com} + {mx}^{2}$
$I_{R} = \frac{{mr}^{2}}{2} + {mr}^{2}$
$I_{R} = \frac{3}{2} {mr}^{2}$
Disc -

$I_{y} = \frac{I_{z}}{2}$
$I_{y} = \frac{m^{2}}{4} = I_{D}$
$\frac{I_{R}}{I_{D}} = \frac{3}{2} \frac{m^{2}}{m^{2}} \times 4$
$I_{R} : I_{D} = 6 : 1$

MHT-CET 2020

Rotational Motion

150112 Let $M$ and $L$ be the mass and length of thin uniform rod respectively. In $1^{st}$ case, axis of rotation is passing through centre and perpendicular to its length. In $2^{nd}$ case, axis of rotation is passing through one end and perpendicular to its length. The ratio of radius of gyration in first case to second case is

1

1 : 2

2

2 : 1

3

3 : 1

4

1 : 3

Explanation:

A Given, mass of thin $rod = M$
Length of thin $rod = L$
From $1^{st}$ case -
$I_{1} = \frac{{ML}^{2}}{12}$
And $I_{1} = {MK}^{2}$
Equating equation (i) and (ii), we get
$K_{1} = \frac{L}{\sqrt{12}}$
From II $^{nd}$ case -
$I_{2} = \frac{{ML}^{2}}{3}$
And $I_{2} = {MK}_{2}^{2}$
Equating equation (iii) and (iv), we get
$K_{2} = \frac{L}{\sqrt{3}}$
Taking ratios of $K_{1}$ and $K_{2}$ , we get
$\frac{K_{1}}{K_{2}} = \frac{L}{\sqrt{12}} \times \frac{\sqrt{3}}{L}$
$∴ \frac{K_{1}}{K_{2}} = \frac{1}{2}$

MHT-CET- 2020

Rotational Motion

150113 A thin uniform rod of length ' $L$ ' and mass ' $M$ ' is bent at the middle point ' $O$ ' at an angle of $45^{0}$ as shown in the figure. The moment of inertia of the system about an axis passing through ' $O$ ' and perpendicular to the plane of the bent rod, is

1

\frac{{ML}^{2}}{6}

2

\frac{{ML}^{2}}{24}

3

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

4

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

Explanation:

D Given, length of rod = L
$mass of rod = M$
When rod is bent at the middle so each part of it will have the same length $(\frac{L}{2})$ and mass $(\frac{M}{2})$ then-
Moment of inertia of each part-
$I_{1} = \frac{1}{3} (\frac{M}{2}) {(\frac{L}{2})}^{2} = \frac{{ML}^{2}}{24}$
$∴$ Total moment of inertia -
$I = 2 I_{1}$
Putting the value of $I_{1}$ from equation (i), we get-
$I = 2 \times \frac{{ML}^{2}}{24}$
$I = \frac{{ML}^{2}}{12}$

MHT-CET 2020

Rotational Motion

150114 A solid sphere of mass ' $M$ ' and radius ' $R$ ' has moment of inertia ' $I$ ' about its diameter. It recast into a disc of thickness ' $t$ ' whose moment of inertia about an axis passing through its edge and perpendicular to its plane, remains ' $I$ '. Radius of the disc will be

1

\frac{R}{\sqrt{19}}

2

\frac{2 R}{\sqrt{15}}

3

\frac{2 R}{\sqrt{19}}

4

\frac{R}{\sqrt{15}}

Explanation:

B Given, mass of solid sphere $= M$
$Radius of sphere = R$
$Radius of disk = r$
Moment of inertia of solid sphere $I) = \frac{2}{5} {MR}^{2}$
Moment of inertia of $disc = \frac{1}{2} {Mr}^{2}$
From parallel axis theorem -
$I = I_{CM} + {Mr}^{2}$
$\frac{2}{5} {MR}^{2} = \frac{1}{2} {Mr}^{2} + {Mr}^{2} = \frac{3}{2} {Mr}^{2}$
$\frac{2}{5} {MR}^{2} = \frac{3}{2} {Mr}^{2}$
$r = \frac{2}{\sqrt{15}} R$

MHT-CET 2020

Rotational Motion

150115 A ring and a disc have same mass and same radius. The ratio of moment of inertia of a ring about a tangent in its plane to that of the disc about its diameter is

1

6 : 1

2

4 : 1

3

2 : 1

4

8 : 1

Explanation:

A Let, mass $= m$ , radius $= r$
For ring -
By Perpendicular axis theorem
original image
$\begin{array}{ll} I_{z} = I_{x} + I_{y} \\ I_{z} = 2 I_{y} \\ I_{y} = \frac{I_{z}}{2} = \frac{{mr}^{2}}{2} \end{array}$
By using parallel axis theorem -
$I_{R} = I_{com} + {mx}^{2}$
$I_{R} = \frac{{mr}^{2}}{2} + {mr}^{2}$
$I_{R} = \frac{3}{2} {mr}^{2}$
Disc -

$I_{y} = \frac{I_{z}}{2}$
$I_{y} = \frac{m^{2}}{4} = I_{D}$
$\frac{I_{R}}{I_{D}} = \frac{3}{2} \frac{m^{2}}{m^{2}} \times 4$
$I_{R} : I_{D} = 6 : 1$

MHT-CET 2020

Rotational Motion

150112 Let $M$ and $L$ be the mass and length of thin uniform rod respectively. In $1^{st}$ case, axis of rotation is passing through centre and perpendicular to its length. In $2^{nd}$ case, axis of rotation is passing through one end and perpendicular to its length. The ratio of radius of gyration in first case to second case is

1

1 : 2

2

2 : 1

3

3 : 1

4

1 : 3

Explanation:

A Given, mass of thin $rod = M$
Length of thin $rod = L$
From $1^{st}$ case -
$I_{1} = \frac{{ML}^{2}}{12}$
And $I_{1} = {MK}^{2}$
Equating equation (i) and (ii), we get
$K_{1} = \frac{L}{\sqrt{12}}$
From II $^{nd}$ case -
$I_{2} = \frac{{ML}^{2}}{3}$
And $I_{2} = {MK}_{2}^{2}$
Equating equation (iii) and (iv), we get
$K_{2} = \frac{L}{\sqrt{3}}$
Taking ratios of $K_{1}$ and $K_{2}$ , we get
$\frac{K_{1}}{K_{2}} = \frac{L}{\sqrt{12}} \times \frac{\sqrt{3}}{L}$
$∴ \frac{K_{1}}{K_{2}} = \frac{1}{2}$

MHT-CET- 2020

Rotational Motion

150113 A thin uniform rod of length ' $L$ ' and mass ' $M$ ' is bent at the middle point ' $O$ ' at an angle of $45^{0}$ as shown in the figure. The moment of inertia of the system about an axis passing through ' $O$ ' and perpendicular to the plane of the bent rod, is

1

\frac{{ML}^{2}}{6}

2

\frac{{ML}^{2}}{24}

3

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

4

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

Explanation:

D Given, length of rod = L
$mass of rod = M$
When rod is bent at the middle so each part of it will have the same length $(\frac{L}{2})$ and mass $(\frac{M}{2})$ then-
Moment of inertia of each part-
$I_{1} = \frac{1}{3} (\frac{M}{2}) {(\frac{L}{2})}^{2} = \frac{{ML}^{2}}{24}$
$∴$ Total moment of inertia -
$I = 2 I_{1}$
Putting the value of $I_{1}$ from equation (i), we get-
$I = 2 \times \frac{{ML}^{2}}{24}$
$I = \frac{{ML}^{2}}{12}$

MHT-CET 2020

Rotational Motion

150114 A solid sphere of mass ' $M$ ' and radius ' $R$ ' has moment of inertia ' $I$ ' about its diameter. It recast into a disc of thickness ' $t$ ' whose moment of inertia about an axis passing through its edge and perpendicular to its plane, remains ' $I$ '. Radius of the disc will be

1

\frac{R}{\sqrt{19}}

2

\frac{2 R}{\sqrt{15}}

3

\frac{2 R}{\sqrt{19}}

4

\frac{R}{\sqrt{15}}

Explanation:

B Given, mass of solid sphere $= M$
$Radius of sphere = R$
$Radius of disk = r$
Moment of inertia of solid sphere $I) = \frac{2}{5} {MR}^{2}$
Moment of inertia of $disc = \frac{1}{2} {Mr}^{2}$
From parallel axis theorem -
$I = I_{CM} + {Mr}^{2}$
$\frac{2}{5} {MR}^{2} = \frac{1}{2} {Mr}^{2} + {Mr}^{2} = \frac{3}{2} {Mr}^{2}$
$\frac{2}{5} {MR}^{2} = \frac{3}{2} {Mr}^{2}$
$r = \frac{2}{\sqrt{15}} R$

MHT-CET 2020

Rotational Motion

150115 A ring and a disc have same mass and same radius. The ratio of moment of inertia of a ring about a tangent in its plane to that of the disc about its diameter is

1

6 : 1

2

4 : 1

3

2 : 1

4

8 : 1

Explanation:

A Let, mass $= m$ , radius $= r$
For ring -
By Perpendicular axis theorem
original image
$\begin{array}{ll} I_{z} = I_{x} + I_{y} \\ I_{z} = 2 I_{y} \\ I_{y} = \frac{I_{z}}{2} = \frac{{mr}^{2}}{2} \end{array}$
By using parallel axis theorem -
$I_{R} = I_{com} + {mx}^{2}$
$I_{R} = \frac{{mr}^{2}}{2} + {mr}^{2}$
$I_{R} = \frac{3}{2} {mr}^{2}$
Disc -

$I_{y} = \frac{I_{z}}{2}$
$I_{y} = \frac{m^{2}}{4} = I_{D}$
$\frac{I_{R}}{I_{D}} = \frac{3}{2} \frac{m^{2}}{m^{2}} \times 4$
$I_{R} : I_{D} = 6 : 1$

MHT-CET 2020

Rotational Motion

150112 Let $M$ and $L$ be the mass and length of thin uniform rod respectively. In $1^{st}$ case, axis of rotation is passing through centre and perpendicular to its length. In $2^{nd}$ case, axis of rotation is passing through one end and perpendicular to its length. The ratio of radius of gyration in first case to second case is

1

1 : 2

2

2 : 1

3

3 : 1

4

1 : 3

Explanation:

A Given, mass of thin $rod = M$
Length of thin $rod = L$
From $1^{st}$ case -
$I_{1} = \frac{{ML}^{2}}{12}$
And $I_{1} = {MK}^{2}$
Equating equation (i) and (ii), we get
$K_{1} = \frac{L}{\sqrt{12}}$
From II $^{nd}$ case -
$I_{2} = \frac{{ML}^{2}}{3}$
And $I_{2} = {MK}_{2}^{2}$
Equating equation (iii) and (iv), we get
$K_{2} = \frac{L}{\sqrt{3}}$
Taking ratios of $K_{1}$ and $K_{2}$ , we get
$\frac{K_{1}}{K_{2}} = \frac{L}{\sqrt{12}} \times \frac{\sqrt{3}}{L}$
$∴ \frac{K_{1}}{K_{2}} = \frac{1}{2}$

MHT-CET- 2020

Rotational Motion

150113 A thin uniform rod of length ' $L$ ' and mass ' $M$ ' is bent at the middle point ' $O$ ' at an angle of $45^{0}$ as shown in the figure. The moment of inertia of the system about an axis passing through ' $O$ ' and perpendicular to the plane of the bent rod, is

1

\frac{{ML}^{2}}{6}

2

\frac{{ML}^{2}}{24}

3

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

4

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

Explanation:

D Given, length of rod = L
$mass of rod = M$
When rod is bent at the middle so each part of it will have the same length $(\frac{L}{2})$ and mass $(\frac{M}{2})$ then-
Moment of inertia of each part-
$I_{1} = \frac{1}{3} (\frac{M}{2}) {(\frac{L}{2})}^{2} = \frac{{ML}^{2}}{24}$
$∴$ Total moment of inertia -
$I = 2 I_{1}$
Putting the value of $I_{1}$ from equation (i), we get-
$I = 2 \times \frac{{ML}^{2}}{24}$
$I = \frac{{ML}^{2}}{12}$

MHT-CET 2020

Rotational Motion

150114 A solid sphere of mass ' $M$ ' and radius ' $R$ ' has moment of inertia ' $I$ ' about its diameter. It recast into a disc of thickness ' $t$ ' whose moment of inertia about an axis passing through its edge and perpendicular to its plane, remains ' $I$ '. Radius of the disc will be

1

\frac{R}{\sqrt{19}}

2

\frac{2 R}{\sqrt{15}}

3

\frac{2 R}{\sqrt{19}}

4

\frac{R}{\sqrt{15}}

Explanation:

B Given, mass of solid sphere $= M$
$Radius of sphere = R$
$Radius of disk = r$
Moment of inertia of solid sphere $I) = \frac{2}{5} {MR}^{2}$
Moment of inertia of $disc = \frac{1}{2} {Mr}^{2}$
From parallel axis theorem -
$I = I_{CM} + {Mr}^{2}$
$\frac{2}{5} {MR}^{2} = \frac{1}{2} {Mr}^{2} + {Mr}^{2} = \frac{3}{2} {Mr}^{2}$
$\frac{2}{5} {MR}^{2} = \frac{3}{2} {Mr}^{2}$
$r = \frac{2}{\sqrt{15}} R$

MHT-CET 2020

Rotational Motion

150115 A ring and a disc have same mass and same radius. The ratio of moment of inertia of a ring about a tangent in its plane to that of the disc about its diameter is

1

6 : 1

2

4 : 1

3

2 : 1

4

8 : 1

Explanation:

A Let, mass $= m$ , radius $= r$
For ring -
By Perpendicular axis theorem
original image
$\begin{array}{ll} I_{z} = I_{x} + I_{y} \\ I_{z} = 2 I_{y} \\ I_{y} = \frac{I_{z}}{2} = \frac{{mr}^{2}}{2} \end{array}$
By using parallel axis theorem -
$I_{R} = I_{com} + {mx}^{2}$
$I_{R} = \frac{{mr}^{2}}{2} + {mr}^{2}$
$I_{R} = \frac{3}{2} {mr}^{2}$
Disc -

$I_{y} = \frac{I_{z}}{2}$
$I_{y} = \frac{m^{2}}{4} = I_{D}$
$\frac{I_{R}}{I_{D}} = \frac{3}{2} \frac{m^{2}}{m^{2}} \times 4$
$I_{R} : I_{D} = 6 : 1$

MHT-CET 2020