QBManager

Rotational Motion

150139 A circular disc ' $X$ ' of radius ' $R$ ' made from iron plate of thickness ' $t$ ' has moment of inertia ' $I_{x}$ ' about an axis passing through the centre of disc and perpendicular to its plane. Another disc ' $Y$ ' of radius ' $3 R$ ' made from an iron plate of thickness $(\frac{t}{3})$ has moment of inertia ' $I_{y}$ ', about the axis same as that of disc $X$ . The relation between $I_{x}$ and $I_{y}$ is

1

I_{y} = 9 I_{x}

2

I_{y} = I_{x}

3

I_{y} = 3 I_{x}

4

I_{y} = 27 I_{x}

Explanation:

D Mass $=$ volume $\times$ density
Radius of disc $X = R$
So, mass of $disc X (M_{x}) = (π R^{2} . t) ρ$
Radius of disc $Y = 3 R$ and thickness of disc $Y = t / 3$
So, mass of $disc Y (M_{y}) = [π (3 R)^{2} \frac{t}{3}] ρ$
$= (3 π R^{2} t) ρ$
Moment of inertia of disc about an axis passing through center and perpendicular to the plane $I = \frac{{MR}^{2}}{2}$
So, $I_{x} = \frac{M_{x} R^{2}}{2}$
$= \frac{(π R^{2} \cdot t) ρ \cdot R^{2}}{2} = \frac{1}{2} π R^{4} t \cdot ρ$
$I_{y} = \frac{M_{y} (3 R)^{2}}{2}$
$= \frac{(3 π R^{2} \cdot t) ρ \cdot 9 R^{2}}{2} = \frac{27}{2} π R^{4} t \cdot ρ$
$\frac{I_{x}}{I_{y}} = \frac{\frac{1}{2} π R^{4} t . ρ}{\frac{27}{2} π R^{4} t . ρ}$
$I_{y} = 27 I_{x}$

MHT-CET 2020

Rotational Motion

150140 Two rings of radii $R$ and $n R$ made from the same wire have the ratio of moments of inertia about an axis passing through their centre and perpendicular to the plane of the rings is $1 : 8$ . The value of $n$ is

1

2 \sqrt{2}

2 2

3 4

4

\frac{1}{2}

Explanation:

B Given,
$R_{1} = R, R_{2} = nR, I_{1} : I_{2} = 1 : 8$
$m_{1} = (2 π$ R.t. $ρ$
$m_{2} = (2 π$ nR.t. $ρ$
$∵ I = {MR}^{2}$
$∴ \frac{I_{1}}{I_{2}} = \frac{m_{1} R_{1}^{2}}{m_{2} R_{2}^{2}}$ $\frac{1}{8} = \frac{(2 π R \cdot t \cdot ρ) \cdot R^{2}}{(2 π n R \cdot t \cdot ρ) \cdot (n R)^{2}}$
$\frac{1}{8} = \frac{1}{n^{3}}$
$\frac{1}{2^{3}} = \frac{1}{n^{3}}$
$n = 2$

MHT-CET 2020

Rotational Motion

150141 The torque necessary to produce an angular acceleration of 25 rad. $s^{- 2}$ in a flywheel of mass $50 kg$ and radius of gyration $50 cm$ about its axis is

1 312.5 dyne.

cm

2

31.25 N . m

3 31.25 dyne.

cm

4

312.5 N . m

Explanation:

D Given that,
Radius $R) = 50 cm = 1 / 2 m$
Angular acceleration $α) = 25 rad / \sec^{2}$
Mass (M) $= 50 kg$
$Moment of inertia = {MR}^{2}$
$= 50 \times (1 / 2)^{2} = 50 / 4 {kgm}^{2}$
We know that,
$Torque = I α$
$Torque = 50 / 4 \times 25 = 1250 / 4 = 312.5 N.m$

AP EAMCET-25.09.2020

Rotational Motion

150142 The moment of inertia of a slab about a perpendicular axis passing through in centre is
$M (\frac{a^{2} + b^{2}}{12})$ . The value of radius of Gyration,
$K$ is

1

\frac{{(a^{2} + b^{2})}^{1 / 2}}{2 \sqrt{3}}

2

\frac{a^{2} + b^{2}}{12}

3

\frac{\sqrt{a + b}}{5}

4

\frac{2}{5} \sqrt{a^{2} + b^{2}}

Explanation:

A Given that,
Moment of inertia of a slab about a perpendicular axis passing through in centre is -
$I = M (\frac{a^{2} + b^{2}}{12})$
Then, we know that,
$I = {MK}^{2}$
Where $K$ is radius of gyration
From equation (i) and (ii), we get
${MK}^{2} = M (\frac{a^{2} + b^{2}}{12})$
$K = \frac{{(a^{2} + b^{2})}^{1 / 2}}{2 \sqrt{3}}$

Assam CEE-2020

Rotational Motion

150139 A circular disc ' $X$ ' of radius ' $R$ ' made from iron plate of thickness ' $t$ ' has moment of inertia ' $I_{x}$ ' about an axis passing through the centre of disc and perpendicular to its plane. Another disc ' $Y$ ' of radius ' $3 R$ ' made from an iron plate of thickness $(\frac{t}{3})$ has moment of inertia ' $I_{y}$ ', about the axis same as that of disc $X$ . The relation between $I_{x}$ and $I_{y}$ is

1

I_{y} = 9 I_{x}

2

I_{y} = I_{x}

3

I_{y} = 3 I_{x}

4

I_{y} = 27 I_{x}

Explanation:

D Mass $=$ volume $\times$ density
Radius of disc $X = R$
So, mass of $disc X (M_{x}) = (π R^{2} . t) ρ$
Radius of disc $Y = 3 R$ and thickness of disc $Y = t / 3$
So, mass of $disc Y (M_{y}) = [π (3 R)^{2} \frac{t}{3}] ρ$
$= (3 π R^{2} t) ρ$
Moment of inertia of disc about an axis passing through center and perpendicular to the plane $I = \frac{{MR}^{2}}{2}$
So, $I_{x} = \frac{M_{x} R^{2}}{2}$
$= \frac{(π R^{2} \cdot t) ρ \cdot R^{2}}{2} = \frac{1}{2} π R^{4} t \cdot ρ$
$I_{y} = \frac{M_{y} (3 R)^{2}}{2}$
$= \frac{(3 π R^{2} \cdot t) ρ \cdot 9 R^{2}}{2} = \frac{27}{2} π R^{4} t \cdot ρ$
$\frac{I_{x}}{I_{y}} = \frac{\frac{1}{2} π R^{4} t . ρ}{\frac{27}{2} π R^{4} t . ρ}$
$I_{y} = 27 I_{x}$

MHT-CET 2020

Rotational Motion

150140 Two rings of radii $R$ and $n R$ made from the same wire have the ratio of moments of inertia about an axis passing through their centre and perpendicular to the plane of the rings is $1 : 8$ . The value of $n$ is

1

2 \sqrt{2}

2 2

3 4

4

\frac{1}{2}

Explanation:

B Given,
$R_{1} = R, R_{2} = nR, I_{1} : I_{2} = 1 : 8$
$m_{1} = (2 π$ R.t. $ρ$
$m_{2} = (2 π$ nR.t. $ρ$
$∵ I = {MR}^{2}$
$∴ \frac{I_{1}}{I_{2}} = \frac{m_{1} R_{1}^{2}}{m_{2} R_{2}^{2}}$ $\frac{1}{8} = \frac{(2 π R \cdot t \cdot ρ) \cdot R^{2}}{(2 π n R \cdot t \cdot ρ) \cdot (n R)^{2}}$
$\frac{1}{8} = \frac{1}{n^{3}}$
$\frac{1}{2^{3}} = \frac{1}{n^{3}}$
$n = 2$

MHT-CET 2020

Rotational Motion

150141 The torque necessary to produce an angular acceleration of 25 rad. $s^{- 2}$ in a flywheel of mass $50 kg$ and radius of gyration $50 cm$ about its axis is

1 312.5 dyne.

cm

2

31.25 N . m

3 31.25 dyne.

cm

4

312.5 N . m

Explanation:

D Given that,
Radius $R) = 50 cm = 1 / 2 m$
Angular acceleration $α) = 25 rad / \sec^{2}$
Mass (M) $= 50 kg$
$Moment of inertia = {MR}^{2}$
$= 50 \times (1 / 2)^{2} = 50 / 4 {kgm}^{2}$
We know that,
$Torque = I α$
$Torque = 50 / 4 \times 25 = 1250 / 4 = 312.5 N.m$

AP EAMCET-25.09.2020

Rotational Motion

150142 The moment of inertia of a slab about a perpendicular axis passing through in centre is
$M (\frac{a^{2} + b^{2}}{12})$ . The value of radius of Gyration,
$K$ is

1

\frac{{(a^{2} + b^{2})}^{1 / 2}}{2 \sqrt{3}}

2

\frac{a^{2} + b^{2}}{12}

3

\frac{\sqrt{a + b}}{5}

4

\frac{2}{5} \sqrt{a^{2} + b^{2}}

Explanation:

A Given that,
Moment of inertia of a slab about a perpendicular axis passing through in centre is -
$I = M (\frac{a^{2} + b^{2}}{12})$
Then, we know that,
$I = {MK}^{2}$
Where $K$ is radius of gyration
From equation (i) and (ii), we get
${MK}^{2} = M (\frac{a^{2} + b^{2}}{12})$
$K = \frac{{(a^{2} + b^{2})}^{1 / 2}}{2 \sqrt{3}}$

Assam CEE-2020

Rotational Motion

150139 A circular disc ' $X$ ' of radius ' $R$ ' made from iron plate of thickness ' $t$ ' has moment of inertia ' $I_{x}$ ' about an axis passing through the centre of disc and perpendicular to its plane. Another disc ' $Y$ ' of radius ' $3 R$ ' made from an iron plate of thickness $(\frac{t}{3})$ has moment of inertia ' $I_{y}$ ', about the axis same as that of disc $X$ . The relation between $I_{x}$ and $I_{y}$ is

1

I_{y} = 9 I_{x}

2

I_{y} = I_{x}

3

I_{y} = 3 I_{x}

4

I_{y} = 27 I_{x}

Explanation:

D Mass $=$ volume $\times$ density
Radius of disc $X = R$
So, mass of $disc X (M_{x}) = (π R^{2} . t) ρ$
Radius of disc $Y = 3 R$ and thickness of disc $Y = t / 3$
So, mass of $disc Y (M_{y}) = [π (3 R)^{2} \frac{t}{3}] ρ$
$= (3 π R^{2} t) ρ$
Moment of inertia of disc about an axis passing through center and perpendicular to the plane $I = \frac{{MR}^{2}}{2}$
So, $I_{x} = \frac{M_{x} R^{2}}{2}$
$= \frac{(π R^{2} \cdot t) ρ \cdot R^{2}}{2} = \frac{1}{2} π R^{4} t \cdot ρ$
$I_{y} = \frac{M_{y} (3 R)^{2}}{2}$
$= \frac{(3 π R^{2} \cdot t) ρ \cdot 9 R^{2}}{2} = \frac{27}{2} π R^{4} t \cdot ρ$
$\frac{I_{x}}{I_{y}} = \frac{\frac{1}{2} π R^{4} t . ρ}{\frac{27}{2} π R^{4} t . ρ}$
$I_{y} = 27 I_{x}$

MHT-CET 2020

Rotational Motion

150140 Two rings of radii $R$ and $n R$ made from the same wire have the ratio of moments of inertia about an axis passing through their centre and perpendicular to the plane of the rings is $1 : 8$ . The value of $n$ is

1

2 \sqrt{2}

2 2

3 4

4

\frac{1}{2}

Explanation:

B Given,
$R_{1} = R, R_{2} = nR, I_{1} : I_{2} = 1 : 8$
$m_{1} = (2 π$ R.t. $ρ$
$m_{2} = (2 π$ nR.t. $ρ$
$∵ I = {MR}^{2}$
$∴ \frac{I_{1}}{I_{2}} = \frac{m_{1} R_{1}^{2}}{m_{2} R_{2}^{2}}$ $\frac{1}{8} = \frac{(2 π R \cdot t \cdot ρ) \cdot R^{2}}{(2 π n R \cdot t \cdot ρ) \cdot (n R)^{2}}$
$\frac{1}{8} = \frac{1}{n^{3}}$
$\frac{1}{2^{3}} = \frac{1}{n^{3}}$
$n = 2$

MHT-CET 2020

Rotational Motion

150141 The torque necessary to produce an angular acceleration of 25 rad. $s^{- 2}$ in a flywheel of mass $50 kg$ and radius of gyration $50 cm$ about its axis is

1 312.5 dyne.

cm

2

31.25 N . m

3 31.25 dyne.

cm

4

312.5 N . m

Explanation:

D Given that,
Radius $R) = 50 cm = 1 / 2 m$
Angular acceleration $α) = 25 rad / \sec^{2}$
Mass (M) $= 50 kg$
$Moment of inertia = {MR}^{2}$
$= 50 \times (1 / 2)^{2} = 50 / 4 {kgm}^{2}$
We know that,
$Torque = I α$
$Torque = 50 / 4 \times 25 = 1250 / 4 = 312.5 N.m$

AP EAMCET-25.09.2020

Rotational Motion

150142 The moment of inertia of a slab about a perpendicular axis passing through in centre is
$M (\frac{a^{2} + b^{2}}{12})$ . The value of radius of Gyration,
$K$ is

1

\frac{{(a^{2} + b^{2})}^{1 / 2}}{2 \sqrt{3}}

2

\frac{a^{2} + b^{2}}{12}

3

\frac{\sqrt{a + b}}{5}

4

\frac{2}{5} \sqrt{a^{2} + b^{2}}

Explanation:

A Given that,
Moment of inertia of a slab about a perpendicular axis passing through in centre is -
$I = M (\frac{a^{2} + b^{2}}{12})$
Then, we know that,
$I = {MK}^{2}$
Where $K$ is radius of gyration
From equation (i) and (ii), we get
${MK}^{2} = M (\frac{a^{2} + b^{2}}{12})$
$K = \frac{{(a^{2} + b^{2})}^{1 / 2}}{2 \sqrt{3}}$

Assam CEE-2020

Rotational Motion

150139 A circular disc ' $X$ ' of radius ' $R$ ' made from iron plate of thickness ' $t$ ' has moment of inertia ' $I_{x}$ ' about an axis passing through the centre of disc and perpendicular to its plane. Another disc ' $Y$ ' of radius ' $3 R$ ' made from an iron plate of thickness $(\frac{t}{3})$ has moment of inertia ' $I_{y}$ ', about the axis same as that of disc $X$ . The relation between $I_{x}$ and $I_{y}$ is

1

I_{y} = 9 I_{x}

2

I_{y} = I_{x}

3

I_{y} = 3 I_{x}

4

I_{y} = 27 I_{x}

Explanation:

D Mass $=$ volume $\times$ density
Radius of disc $X = R$
So, mass of $disc X (M_{x}) = (π R^{2} . t) ρ$
Radius of disc $Y = 3 R$ and thickness of disc $Y = t / 3$
So, mass of $disc Y (M_{y}) = [π (3 R)^{2} \frac{t}{3}] ρ$
$= (3 π R^{2} t) ρ$
Moment of inertia of disc about an axis passing through center and perpendicular to the plane $I = \frac{{MR}^{2}}{2}$
So, $I_{x} = \frac{M_{x} R^{2}}{2}$
$= \frac{(π R^{2} \cdot t) ρ \cdot R^{2}}{2} = \frac{1}{2} π R^{4} t \cdot ρ$
$I_{y} = \frac{M_{y} (3 R)^{2}}{2}$
$= \frac{(3 π R^{2} \cdot t) ρ \cdot 9 R^{2}}{2} = \frac{27}{2} π R^{4} t \cdot ρ$
$\frac{I_{x}}{I_{y}} = \frac{\frac{1}{2} π R^{4} t . ρ}{\frac{27}{2} π R^{4} t . ρ}$
$I_{y} = 27 I_{x}$

MHT-CET 2020

Rotational Motion

150140 Two rings of radii $R$ and $n R$ made from the same wire have the ratio of moments of inertia about an axis passing through their centre and perpendicular to the plane of the rings is $1 : 8$ . The value of $n$ is

1

2 \sqrt{2}

2 2

3 4

4

\frac{1}{2}

Explanation:

B Given,
$R_{1} = R, R_{2} = nR, I_{1} : I_{2} = 1 : 8$
$m_{1} = (2 π$ R.t. $ρ$
$m_{2} = (2 π$ nR.t. $ρ$
$∵ I = {MR}^{2}$
$∴ \frac{I_{1}}{I_{2}} = \frac{m_{1} R_{1}^{2}}{m_{2} R_{2}^{2}}$ $\frac{1}{8} = \frac{(2 π R \cdot t \cdot ρ) \cdot R^{2}}{(2 π n R \cdot t \cdot ρ) \cdot (n R)^{2}}$
$\frac{1}{8} = \frac{1}{n^{3}}$
$\frac{1}{2^{3}} = \frac{1}{n^{3}}$
$n = 2$

MHT-CET 2020

Rotational Motion

150141 The torque necessary to produce an angular acceleration of 25 rad. $s^{- 2}$ in a flywheel of mass $50 kg$ and radius of gyration $50 cm$ about its axis is

1 312.5 dyne.

cm

2

31.25 N . m

3 31.25 dyne.

cm

4

312.5 N . m

Explanation:

D Given that,
Radius $R) = 50 cm = 1 / 2 m$
Angular acceleration $α) = 25 rad / \sec^{2}$
Mass (M) $= 50 kg$
$Moment of inertia = {MR}^{2}$
$= 50 \times (1 / 2)^{2} = 50 / 4 {kgm}^{2}$
We know that,
$Torque = I α$
$Torque = 50 / 4 \times 25 = 1250 / 4 = 312.5 N.m$

AP EAMCET-25.09.2020

Rotational Motion

150142 The moment of inertia of a slab about a perpendicular axis passing through in centre is
$M (\frac{a^{2} + b^{2}}{12})$ . The value of radius of Gyration,
$K$ is

1

\frac{{(a^{2} + b^{2})}^{1 / 2}}{2 \sqrt{3}}

2

\frac{a^{2} + b^{2}}{12}

3

\frac{\sqrt{a + b}}{5}

4

\frac{2}{5} \sqrt{a^{2} + b^{2}}

Explanation:

A Given that,
Moment of inertia of a slab about a perpendicular axis passing through in centre is -
$I = M (\frac{a^{2} + b^{2}}{12})$
Then, we know that,
$I = {MK}^{2}$
Where $K$ is radius of gyration
From equation (i) and (ii), we get
${MK}^{2} = M (\frac{a^{2} + b^{2}}{12})$
$K = \frac{{(a^{2} + b^{2})}^{1 / 2}}{2 \sqrt{3}}$

Assam CEE-2020

Question Bank Series

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