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Rotational Motion

149719 The centre of mass of a system of two bodies of masses $M$ and $m, (M > m)$ , separated by a distance $d$ is

1 Midway between the bodies

2 Closer to the heavier body

3 Closer to the lighter body

4 At the centre of the heavier body

Explanation:

B
Let $x_{1}$ and $x_{2}$ be distance of centre of mass from the two bodies of masses $M$ and $m (M > m)$ respectively.
${Mx}_{1} = {mx}_{2}$
$\frac{x_{1}}{x_{2}} = \frac{m}{M}$
$∵ m < M$
$∴ \frac{x_{1}}{x_{2}} < 1 \Rightarrow x_{1} < x_{2}$
Hence, the centre of mass is closer to the heavier body.

COMEDK 2014

Rotational Motion

149720 Assertion: If no external force acts on a system of particles, then the centre of mass will not move in any direction.
Reason: If net external force is zero, then the linear momentum of the system changes.

1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

Explanation:

D We know that,
${\vec{a}}_{cm} = \frac{{\vec{F}}_{ext}}{M}$
If ${\vec{F}}_{ext} = 0$
Thus, ${\vec{a}}_{cm} = 0$
Thus, when no external force acts on a system of particles, it will move continuously with its initial velocity.
Now,
$\frac{d \vec{P}}{dt} = {\vec{F}}_{ext}$
$∴ \frac{d \vec{P}}{dt} = 0 (∵ {\vec{F}}_{ext} = 0)$
Hence, if no external force acts on the system, then linear momentum of the system is conserved.
So, the both Assertion and Reason is incorrect

AIIMS-2011

Rotational Motion

149721 Assertion : The position of centre of mass of a body depends upon shape and size of the body.
Reason: Centre of mass of a body lies always at the centre of the body.

1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

AIIMS-2009

Rotational Motion

149724 If linear density of a rod of length $3 m$ varies as $λ = 2 + x$ , then the position of the centre of gravity of the rod is :

1

\frac{7}{3} m

2

\frac{12}{7} m

3

\frac{10}{7} m

4

\frac{9}{7} m

Explanation:

B Given, length of $rod = 3 m, λ = 2 + x$
$(0, 0)$

Linear density of rod varies with distance
$\frac{dm}{dx} = λ$
$dm = λ dx$
Position of center of mass,
$x_{CM} = \frac{\int dm \times x}{\int dm}$
Putting the value of $dm$ from equation (i), we get
$= \frac{\int_{0}^{3} (λ d x) x}{\int_{0}^{3} λ d x} = \frac{\int_{0}^{3} (2 + x) x d x}{\int_{0}^{3} (2 + x) d x}$
$= \frac{{[x^{2} + \frac{x^{3}}{3}]}_{0}^{3}}{{[2 x + \frac{x^{2}}{2}]}_{0}^{3}}$
$= \frac{9 + \frac{27}{3}}{6 + \frac{9}{2}} = \frac{12}{7}$
$x_{C M} = \frac{12}{7} m$

BCECE-2005

Rotational Motion

149719 The centre of mass of a system of two bodies of masses $M$ and $m, (M > m)$ , separated by a distance $d$ is

1 Midway between the bodies

2 Closer to the heavier body

3 Closer to the lighter body

4 At the centre of the heavier body

Explanation:

B
Let $x_{1}$ and $x_{2}$ be distance of centre of mass from the two bodies of masses $M$ and $m (M > m)$ respectively.
${Mx}_{1} = {mx}_{2}$
$\frac{x_{1}}{x_{2}} = \frac{m}{M}$
$∵ m < M$
$∴ \frac{x_{1}}{x_{2}} < 1 \Rightarrow x_{1} < x_{2}$
Hence, the centre of mass is closer to the heavier body.

COMEDK 2014

Rotational Motion

149720 Assertion: If no external force acts on a system of particles, then the centre of mass will not move in any direction.
Reason: If net external force is zero, then the linear momentum of the system changes.

1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

Explanation:

D We know that,
${\vec{a}}_{cm} = \frac{{\vec{F}}_{ext}}{M}$
If ${\vec{F}}_{ext} = 0$
Thus, ${\vec{a}}_{cm} = 0$
Thus, when no external force acts on a system of particles, it will move continuously with its initial velocity.
Now,
$\frac{d \vec{P}}{dt} = {\vec{F}}_{ext}$
$∴ \frac{d \vec{P}}{dt} = 0 (∵ {\vec{F}}_{ext} = 0)$
Hence, if no external force acts on the system, then linear momentum of the system is conserved.
So, the both Assertion and Reason is incorrect

AIIMS-2011

Rotational Motion

149721 Assertion : The position of centre of mass of a body depends upon shape and size of the body.
Reason: Centre of mass of a body lies always at the centre of the body.

1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

AIIMS-2009

Rotational Motion

149724 If linear density of a rod of length $3 m$ varies as $λ = 2 + x$ , then the position of the centre of gravity of the rod is :

1

\frac{7}{3} m

2

\frac{12}{7} m

3

\frac{10}{7} m

4

\frac{9}{7} m

Explanation:

B Given, length of $rod = 3 m, λ = 2 + x$
$(0, 0)$

Linear density of rod varies with distance
$\frac{dm}{dx} = λ$
$dm = λ dx$
Position of center of mass,
$x_{CM} = \frac{\int dm \times x}{\int dm}$
Putting the value of $dm$ from equation (i), we get
$= \frac{\int_{0}^{3} (λ d x) x}{\int_{0}^{3} λ d x} = \frac{\int_{0}^{3} (2 + x) x d x}{\int_{0}^{3} (2 + x) d x}$
$= \frac{{[x^{2} + \frac{x^{3}}{3}]}_{0}^{3}}{{[2 x + \frac{x^{2}}{2}]}_{0}^{3}}$
$= \frac{9 + \frac{27}{3}}{6 + \frac{9}{2}} = \frac{12}{7}$
$x_{C M} = \frac{12}{7} m$

BCECE-2005

Rotational Motion

149719 The centre of mass of a system of two bodies of masses $M$ and $m, (M > m)$ , separated by a distance $d$ is

1 Midway between the bodies

2 Closer to the heavier body

3 Closer to the lighter body

4 At the centre of the heavier body

Explanation:

B
Let $x_{1}$ and $x_{2}$ be distance of centre of mass from the two bodies of masses $M$ and $m (M > m)$ respectively.
${Mx}_{1} = {mx}_{2}$
$\frac{x_{1}}{x_{2}} = \frac{m}{M}$
$∵ m < M$
$∴ \frac{x_{1}}{x_{2}} < 1 \Rightarrow x_{1} < x_{2}$
Hence, the centre of mass is closer to the heavier body.

COMEDK 2014

Rotational Motion

149720 Assertion: If no external force acts on a system of particles, then the centre of mass will not move in any direction.
Reason: If net external force is zero, then the linear momentum of the system changes.

1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

Explanation:

D We know that,
${\vec{a}}_{cm} = \frac{{\vec{F}}_{ext}}{M}$
If ${\vec{F}}_{ext} = 0$
Thus, ${\vec{a}}_{cm} = 0$
Thus, when no external force acts on a system of particles, it will move continuously with its initial velocity.
Now,
$\frac{d \vec{P}}{dt} = {\vec{F}}_{ext}$
$∴ \frac{d \vec{P}}{dt} = 0 (∵ {\vec{F}}_{ext} = 0)$
Hence, if no external force acts on the system, then linear momentum of the system is conserved.
So, the both Assertion and Reason is incorrect

AIIMS-2011

Rotational Motion

149721 Assertion : The position of centre of mass of a body depends upon shape and size of the body.
Reason: Centre of mass of a body lies always at the centre of the body.

1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

AIIMS-2009

Rotational Motion

149724 If linear density of a rod of length $3 m$ varies as $λ = 2 + x$ , then the position of the centre of gravity of the rod is :

1

\frac{7}{3} m

2

\frac{12}{7} m

3

\frac{10}{7} m

4

\frac{9}{7} m

Explanation:

B Given, length of $rod = 3 m, λ = 2 + x$
$(0, 0)$

Linear density of rod varies with distance
$\frac{dm}{dx} = λ$
$dm = λ dx$
Position of center of mass,
$x_{CM} = \frac{\int dm \times x}{\int dm}$
Putting the value of $dm$ from equation (i), we get
$= \frac{\int_{0}^{3} (λ d x) x}{\int_{0}^{3} λ d x} = \frac{\int_{0}^{3} (2 + x) x d x}{\int_{0}^{3} (2 + x) d x}$
$= \frac{{[x^{2} + \frac{x^{3}}{3}]}_{0}^{3}}{{[2 x + \frac{x^{2}}{2}]}_{0}^{3}}$
$= \frac{9 + \frac{27}{3}}{6 + \frac{9}{2}} = \frac{12}{7}$
$x_{C M} = \frac{12}{7} m$

BCECE-2005

Rotational Motion

149719 The centre of mass of a system of two bodies of masses $M$ and $m, (M > m)$ , separated by a distance $d$ is

1 Midway between the bodies

2 Closer to the heavier body

3 Closer to the lighter body

4 At the centre of the heavier body

Explanation:

B
Let $x_{1}$ and $x_{2}$ be distance of centre of mass from the two bodies of masses $M$ and $m (M > m)$ respectively.
${Mx}_{1} = {mx}_{2}$
$\frac{x_{1}}{x_{2}} = \frac{m}{M}$
$∵ m < M$
$∴ \frac{x_{1}}{x_{2}} < 1 \Rightarrow x_{1} < x_{2}$
Hence, the centre of mass is closer to the heavier body.

COMEDK 2014

Rotational Motion

149720 Assertion: If no external force acts on a system of particles, then the centre of mass will not move in any direction.
Reason: If net external force is zero, then the linear momentum of the system changes.

1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

Explanation:

D We know that,
${\vec{a}}_{cm} = \frac{{\vec{F}}_{ext}}{M}$
If ${\vec{F}}_{ext} = 0$
Thus, ${\vec{a}}_{cm} = 0$
Thus, when no external force acts on a system of particles, it will move continuously with its initial velocity.
Now,
$\frac{d \vec{P}}{dt} = {\vec{F}}_{ext}$
$∴ \frac{d \vec{P}}{dt} = 0 (∵ {\vec{F}}_{ext} = 0)$
Hence, if no external force acts on the system, then linear momentum of the system is conserved.
So, the both Assertion and Reason is incorrect

AIIMS-2011

Rotational Motion

149721 Assertion : The position of centre of mass of a body depends upon shape and size of the body.
Reason: Centre of mass of a body lies always at the centre of the body.

1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

AIIMS-2009

Rotational Motion

149724 If linear density of a rod of length $3 m$ varies as $λ = 2 + x$ , then the position of the centre of gravity of the rod is :

1

\frac{7}{3} m

2

\frac{12}{7} m

3

\frac{10}{7} m

4

\frac{9}{7} m

Explanation:

B Given, length of $rod = 3 m, λ = 2 + x$
$(0, 0)$

Linear density of rod varies with distance
$\frac{dm}{dx} = λ$
$dm = λ dx$
Position of center of mass,
$x_{CM} = \frac{\int dm \times x}{\int dm}$
Putting the value of $dm$ from equation (i), we get
$= \frac{\int_{0}^{3} (λ d x) x}{\int_{0}^{3} λ d x} = \frac{\int_{0}^{3} (2 + x) x d x}{\int_{0}^{3} (2 + x) d x}$
$= \frac{{[x^{2} + \frac{x^{3}}{3}]}_{0}^{3}}{{[2 x + \frac{x^{2}}{2}]}_{0}^{3}}$
$= \frac{9 + \frac{27}{3}}{6 + \frac{9}{2}} = \frac{12}{7}$
$x_{C M} = \frac{12}{7} m$

BCECE-2005

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