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Rotational Motion

269368 A thin uniform rod of length" $L$ " is bent at its mid point as shown in the figure. The distance of the centre of mass from the point " $O$ " is

1

\frac{L}{2} \sin \frac{θ}{2}

2

\frac{L}{2} \cos \frac{θ}{2}

3

\frac{L}{4} \sin \frac{θ}{2}

4

\frac{L}{4} \cos \frac{θ}{2}

Explanation:

$x_{c m} = \frac{- \frac{m}{2} (L / 4) - \frac{m}{2} (\cos θ) L / 4}{m}$ $y_{c m} = \frac{\frac{m}{2} \sin θ (L / 4)}{m}; r_{c m} = \sqrt{x_{c m}^{2} + y_{c m}^{2}}$

Rotational Motion

269369 Three identical spheres each of mass ' $m$ ' and radius ' $R$ ' are placed touching each other so that theircentres $A, B$ and $C$ lie on a straight line. The position of their centre of mass from centre of $A$ is

1

\frac{2 R}{3}

2

2 R

3

\frac{5 R}{3}

4

\frac{4 R}{3}

Explanation:

$x_{c m} = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3}}{m_{1} + m_{2} + m_{3}}$

Rotational Motion

269370 A boy of mass $50 kg$ is standing at one end of a boat of length $9 m$ and mass $400 kg$ . He runs to the other end. The distance through which the centre of mass of the boat boy system moves is

1 0

2

1 m

3

2 m

4

3 m

Rotational Motion

269371 A dog weighing $5 kg$ is standing on a flat boat so that it is 10 metres from the shore. It walks $4 m$ on the boat towards the shore and then halts. The boat weighs $20 kg$ and one can assume that there is no friction between it and water. The dog from the shore at the end of this time is

1

3.4 m

2

6.8 m

3

12.6 m

4

10 m

Explanation:

Distance from shore $= (10 - l + d)$

Rotational Motion

269368 A thin uniform rod of length" $L$ " is bent at its mid point as shown in the figure. The distance of the centre of mass from the point " $O$ " is

1

\frac{L}{2} \sin \frac{θ}{2}

2

\frac{L}{2} \cos \frac{θ}{2}

3

\frac{L}{4} \sin \frac{θ}{2}

4

\frac{L}{4} \cos \frac{θ}{2}

Explanation:

$x_{c m} = \frac{- \frac{m}{2} (L / 4) - \frac{m}{2} (\cos θ) L / 4}{m}$ $y_{c m} = \frac{\frac{m}{2} \sin θ (L / 4)}{m}; r_{c m} = \sqrt{x_{c m}^{2} + y_{c m}^{2}}$

Rotational Motion

269369 Three identical spheres each of mass ' $m$ ' and radius ' $R$ ' are placed touching each other so that theircentres $A, B$ and $C$ lie on a straight line. The position of their centre of mass from centre of $A$ is

1

\frac{2 R}{3}

2

2 R

3

\frac{5 R}{3}

4

\frac{4 R}{3}

Explanation:

$x_{c m} = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3}}{m_{1} + m_{2} + m_{3}}$

Rotational Motion

269370 A boy of mass $50 kg$ is standing at one end of a boat of length $9 m$ and mass $400 kg$ . He runs to the other end. The distance through which the centre of mass of the boat boy system moves is

1 0

2

1 m

3

2 m

4

3 m

Rotational Motion

269371 A dog weighing $5 kg$ is standing on a flat boat so that it is 10 metres from the shore. It walks $4 m$ on the boat towards the shore and then halts. The boat weighs $20 kg$ and one can assume that there is no friction between it and water. The dog from the shore at the end of this time is

1

3.4 m

2

6.8 m

3

12.6 m

4

10 m

Explanation:

Distance from shore $= (10 - l + d)$

Rotational Motion

269368 A thin uniform rod of length" $L$ " is bent at its mid point as shown in the figure. The distance of the centre of mass from the point " $O$ " is

1

\frac{L}{2} \sin \frac{θ}{2}

2

\frac{L}{2} \cos \frac{θ}{2}

3

\frac{L}{4} \sin \frac{θ}{2}

4

\frac{L}{4} \cos \frac{θ}{2}

Explanation:

$x_{c m} = \frac{- \frac{m}{2} (L / 4) - \frac{m}{2} (\cos θ) L / 4}{m}$ $y_{c m} = \frac{\frac{m}{2} \sin θ (L / 4)}{m}; r_{c m} = \sqrt{x_{c m}^{2} + y_{c m}^{2}}$

Rotational Motion

269369 Three identical spheres each of mass ' $m$ ' and radius ' $R$ ' are placed touching each other so that theircentres $A, B$ and $C$ lie on a straight line. The position of their centre of mass from centre of $A$ is

1

\frac{2 R}{3}

2

2 R

3

\frac{5 R}{3}

4

\frac{4 R}{3}

Explanation:

$x_{c m} = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3}}{m_{1} + m_{2} + m_{3}}$

Rotational Motion

269370 A boy of mass $50 kg$ is standing at one end of a boat of length $9 m$ and mass $400 kg$ . He runs to the other end. The distance through which the centre of mass of the boat boy system moves is

1 0

2

1 m

3

2 m

4

3 m

Rotational Motion

269371 A dog weighing $5 kg$ is standing on a flat boat so that it is 10 metres from the shore. It walks $4 m$ on the boat towards the shore and then halts. The boat weighs $20 kg$ and one can assume that there is no friction between it and water. The dog from the shore at the end of this time is

1

3.4 m

2

6.8 m

3

12.6 m

4

10 m

Explanation:

Distance from shore $= (10 - l + d)$

Rotational Motion

269368 A thin uniform rod of length" $L$ " is bent at its mid point as shown in the figure. The distance of the centre of mass from the point " $O$ " is

1

\frac{L}{2} \sin \frac{θ}{2}

2

\frac{L}{2} \cos \frac{θ}{2}

3

\frac{L}{4} \sin \frac{θ}{2}

4

\frac{L}{4} \cos \frac{θ}{2}

Explanation:

$x_{c m} = \frac{- \frac{m}{2} (L / 4) - \frac{m}{2} (\cos θ) L / 4}{m}$ $y_{c m} = \frac{\frac{m}{2} \sin θ (L / 4)}{m}; r_{c m} = \sqrt{x_{c m}^{2} + y_{c m}^{2}}$

Rotational Motion

269369 Three identical spheres each of mass ' $m$ ' and radius ' $R$ ' are placed touching each other so that theircentres $A, B$ and $C$ lie on a straight line. The position of their centre of mass from centre of $A$ is

1

\frac{2 R}{3}

2

2 R

3

\frac{5 R}{3}

4

\frac{4 R}{3}

Explanation:

$x_{c m} = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3}}{m_{1} + m_{2} + m_{3}}$

Rotational Motion

269370 A boy of mass $50 kg$ is standing at one end of a boat of length $9 m$ and mass $400 kg$ . He runs to the other end. The distance through which the centre of mass of the boat boy system moves is

1 0

2

1 m

3

2 m

4

3 m

Rotational Motion

269371 A dog weighing $5 kg$ is standing on a flat boat so that it is 10 metres from the shore. It walks $4 m$ on the boat towards the shore and then halts. The boat weighs $20 kg$ and one can assume that there is no friction between it and water. The dog from the shore at the end of this time is

1

3.4 m

2

6.8 m

3

12.6 m

4

10 m

Explanation:

Distance from shore $= (10 - l + d)$

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