QBManager

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365701 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{12}{7} m

2

\frac{7}{3} m

3

\frac{9}{7} m

4

\frac{10}{7} m

Explanation:

Linear density of the rod varies with distance as
$\begin{aligned} λ = 2 + x \\ \frac{d m}{d x} = λ [given] ∴ d m = λ d x \end{aligned}$
supporting img
Position of centre of mass
$x_{c m} = \frac{\int d m \times x}{\int d m}$
$= \frac{\int_{0}^{3} (λ d x) \times x}{\int_{0}^{3} λ d x} = \frac{\int_{0}^{3} (2 + x) \times x d x}{\int_{0}^{3} (2 + x) \times d x}$
$= \frac{{[x^{2} + \frac{x^{3}}{3}]}_{0}^{3}}{{[2 x + \frac{x^{2}}{2}]}_{0}^{3}} = \frac{9 + 9}{6 + \frac{9}{2}} = \frac{12}{7} m .$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365702 From a uniform disc of radius $R$ , a circular section of radius $\frac{R}{2}$ is cut out. The centre of the hole is at $\frac{R}{2}$ from the centre of the original disc. Locate the centre of gravity of the resulting flat body.

1

\frac{R}{3}

to the right of centre

O

2

\frac{R}{6}

to the right of centre

O

3

\frac{R}{6}

to the left of centre

O

4

\frac{R}{3}

to the left of centre

O

Explanation:

Let mass per unit area of the disc be $σ$ .
$∴$ Mass of the disc $(M) = π R^{2} σ$ .
Mass of the portion removed from the disc $(M^{'})$
$= π {(\frac{R}{2})}^{2} σ = \frac{M}{4}$
The position of centre of mass of the remaining portion is
$x = \frac{- \frac{M}{4} \times \frac{R}{2}}{M - \frac{M}{4}}$
$= - \frac{M R}{8} \times \frac{4}{3 M}$
$x = \frac{- R}{6}$

NCERT Exemplar

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365703 Which of the following statements are correct?

1 Centre of mass of a body always coincides with the centre of gravity of the body.

2 Centre of mass of a body is the point at which the total gravitational torque on the body is zero.

3 A couple on a body produces both translational and rotational motion in a body.

4 Mechanical advantage greater than one means that small effort can be used to lift a large load.

NEET - 2017

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365704 A small card board is balanced on the tip of a pencil. The centre of mass coincides with centre of gravity $G$ . When the card board is in equilibrium (translational and rotational) then predict the correct option.
supporting img

1

\vec{R} = M \vec{g}

2

{\vec{τ}}_{G} = \sum {\vec{r}}_{i} \times m_{i} \vec{g} = 0

3

\sum m_{i} {\vec{r}}_{i} = 0

4 All the above

Explanation:

As the card board is in translational equilibrium the net vertical force should be zero.
$\vec{R} = M \vec{g}$
For rotational equilibrium the net torque should be zero about any point. The torque about $G$ is
${\vec{τ}}_{G} = \sum {\vec{r}}_{i} \times {\vec{F}}_{g} = \sum {\vec{r}}_{i} \times {\vec{m}}_{i} \vec{g} = 0$
Where ${\vec{r}}_{i}$ is the position vector of the $p$ article $m_{i}$ is the mass of the $i^{th}$ particle.
$\vec{g}$ is the acceleration vector.
For a small object $\vec{g}$ is same for all the particles
so $\vec{g}$ can be taken out of summation.
$\vec{g} \sum {\vec{r}}_{i} m_{i} = 0 \Rightarrow \sum m_{i} {\vec{r}}_{i} = \frac{{\vec{r}}_{C M}}{\sum m_{i}} = 0$
So the $C M$ coincides with centre of gravity
So option (4) is correct.

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365705 An irregular shaped body like a cardboard is suspended by a string. The body is in equilibrium about the lines $A A_{1}, B B_{1} & C C_{1}$ . Then the centre of gravity of the body lies
supporting img

1 on line

A A_{1}

2 on line

B B_{1}

3 on line

C C_{1}

4 All the above

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365701 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{12}{7} m

2

\frac{7}{3} m

3

\frac{9}{7} m

4

\frac{10}{7} m

Explanation:

Linear density of the rod varies with distance as
$\begin{aligned} λ = 2 + x \\ \frac{d m}{d x} = λ [given] ∴ d m = λ d x \end{aligned}$
supporting img
Position of centre of mass
$x_{c m} = \frac{\int d m \times x}{\int d m}$
$= \frac{\int_{0}^{3} (λ d x) \times x}{\int_{0}^{3} λ d x} = \frac{\int_{0}^{3} (2 + x) \times x d x}{\int_{0}^{3} (2 + x) \times d x}$
$= \frac{{[x^{2} + \frac{x^{3}}{3}]}_{0}^{3}}{{[2 x + \frac{x^{2}}{2}]}_{0}^{3}} = \frac{9 + 9}{6 + \frac{9}{2}} = \frac{12}{7} m .$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365702 From a uniform disc of radius $R$ , a circular section of radius $\frac{R}{2}$ is cut out. The centre of the hole is at $\frac{R}{2}$ from the centre of the original disc. Locate the centre of gravity of the resulting flat body.

1

\frac{R}{3}

to the right of centre

O

2

\frac{R}{6}

to the right of centre

O

3

\frac{R}{6}

to the left of centre

O

4

\frac{R}{3}

to the left of centre

O

Explanation:

Let mass per unit area of the disc be $σ$ .
$∴$ Mass of the disc $(M) = π R^{2} σ$ .
Mass of the portion removed from the disc $(M^{'})$
$= π {(\frac{R}{2})}^{2} σ = \frac{M}{4}$
The position of centre of mass of the remaining portion is
$x = \frac{- \frac{M}{4} \times \frac{R}{2}}{M - \frac{M}{4}}$
$= - \frac{M R}{8} \times \frac{4}{3 M}$
$x = \frac{- R}{6}$

NCERT Exemplar

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365703 Which of the following statements are correct?

1 Centre of mass of a body always coincides with the centre of gravity of the body.

2 Centre of mass of a body is the point at which the total gravitational torque on the body is zero.

3 A couple on a body produces both translational and rotational motion in a body.

4 Mechanical advantage greater than one means that small effort can be used to lift a large load.

NEET - 2017

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365704 A small card board is balanced on the tip of a pencil. The centre of mass coincides with centre of gravity $G$ . When the card board is in equilibrium (translational and rotational) then predict the correct option.
supporting img

1

\vec{R} = M \vec{g}

2

{\vec{τ}}_{G} = \sum {\vec{r}}_{i} \times m_{i} \vec{g} = 0

3

\sum m_{i} {\vec{r}}_{i} = 0

4 All the above

Explanation:

As the card board is in translational equilibrium the net vertical force should be zero.
$\vec{R} = M \vec{g}$
For rotational equilibrium the net torque should be zero about any point. The torque about $G$ is
${\vec{τ}}_{G} = \sum {\vec{r}}_{i} \times {\vec{F}}_{g} = \sum {\vec{r}}_{i} \times {\vec{m}}_{i} \vec{g} = 0$
Where ${\vec{r}}_{i}$ is the position vector of the $p$ article $m_{i}$ is the mass of the $i^{th}$ particle.
$\vec{g}$ is the acceleration vector.
For a small object $\vec{g}$ is same for all the particles
so $\vec{g}$ can be taken out of summation.
$\vec{g} \sum {\vec{r}}_{i} m_{i} = 0 \Rightarrow \sum m_{i} {\vec{r}}_{i} = \frac{{\vec{r}}_{C M}}{\sum m_{i}} = 0$
So the $C M$ coincides with centre of gravity
So option (4) is correct.

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365705 An irregular shaped body like a cardboard is suspended by a string. The body is in equilibrium about the lines $A A_{1}, B B_{1} & C C_{1}$ . Then the centre of gravity of the body lies
supporting img

1 on line

A A_{1}

2 on line

B B_{1}

3 on line

C C_{1}

4 All the above

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365701 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{12}{7} m

2

\frac{7}{3} m

3

\frac{9}{7} m

4

\frac{10}{7} m

Explanation:

Linear density of the rod varies with distance as
$\begin{aligned} λ = 2 + x \\ \frac{d m}{d x} = λ [given] ∴ d m = λ d x \end{aligned}$
supporting img
Position of centre of mass
$x_{c m} = \frac{\int d m \times x}{\int d m}$
$= \frac{\int_{0}^{3} (λ d x) \times x}{\int_{0}^{3} λ d x} = \frac{\int_{0}^{3} (2 + x) \times x d x}{\int_{0}^{3} (2 + x) \times d x}$
$= \frac{{[x^{2} + \frac{x^{3}}{3}]}_{0}^{3}}{{[2 x + \frac{x^{2}}{2}]}_{0}^{3}} = \frac{9 + 9}{6 + \frac{9}{2}} = \frac{12}{7} m .$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365702 From a uniform disc of radius $R$ , a circular section of radius $\frac{R}{2}$ is cut out. The centre of the hole is at $\frac{R}{2}$ from the centre of the original disc. Locate the centre of gravity of the resulting flat body.

1

\frac{R}{3}

to the right of centre

O

2

\frac{R}{6}

to the right of centre

O

3

\frac{R}{6}

to the left of centre

O

4

\frac{R}{3}

to the left of centre

O

Explanation:

Let mass per unit area of the disc be $σ$ .
$∴$ Mass of the disc $(M) = π R^{2} σ$ .
Mass of the portion removed from the disc $(M^{'})$
$= π {(\frac{R}{2})}^{2} σ = \frac{M}{4}$
The position of centre of mass of the remaining portion is
$x = \frac{- \frac{M}{4} \times \frac{R}{2}}{M - \frac{M}{4}}$
$= - \frac{M R}{8} \times \frac{4}{3 M}$
$x = \frac{- R}{6}$

NCERT Exemplar

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365703 Which of the following statements are correct?

1 Centre of mass of a body always coincides with the centre of gravity of the body.

2 Centre of mass of a body is the point at which the total gravitational torque on the body is zero.

3 A couple on a body produces both translational and rotational motion in a body.

4 Mechanical advantage greater than one means that small effort can be used to lift a large load.

NEET - 2017

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365704 A small card board is balanced on the tip of a pencil. The centre of mass coincides with centre of gravity $G$ . When the card board is in equilibrium (translational and rotational) then predict the correct option.
supporting img

1

\vec{R} = M \vec{g}

2

{\vec{τ}}_{G} = \sum {\vec{r}}_{i} \times m_{i} \vec{g} = 0

3

\sum m_{i} {\vec{r}}_{i} = 0

4 All the above

Explanation:

As the card board is in translational equilibrium the net vertical force should be zero.
$\vec{R} = M \vec{g}$
For rotational equilibrium the net torque should be zero about any point. The torque about $G$ is
${\vec{τ}}_{G} = \sum {\vec{r}}_{i} \times {\vec{F}}_{g} = \sum {\vec{r}}_{i} \times {\vec{m}}_{i} \vec{g} = 0$
Where ${\vec{r}}_{i}$ is the position vector of the $p$ article $m_{i}$ is the mass of the $i^{th}$ particle.
$\vec{g}$ is the acceleration vector.
For a small object $\vec{g}$ is same for all the particles
so $\vec{g}$ can be taken out of summation.
$\vec{g} \sum {\vec{r}}_{i} m_{i} = 0 \Rightarrow \sum m_{i} {\vec{r}}_{i} = \frac{{\vec{r}}_{C M}}{\sum m_{i}} = 0$
So the $C M$ coincides with centre of gravity
So option (4) is correct.

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365705 An irregular shaped body like a cardboard is suspended by a string. The body is in equilibrium about the lines $A A_{1}, B B_{1} & C C_{1}$ . Then the centre of gravity of the body lies
supporting img

1 on line

A A_{1}

2 on line

B B_{1}

3 on line

C C_{1}

4 All the above

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365701 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{12}{7} m

2

\frac{7}{3} m

3

\frac{9}{7} m

4

\frac{10}{7} m

Explanation:

Linear density of the rod varies with distance as
$\begin{aligned} λ = 2 + x \\ \frac{d m}{d x} = λ [given] ∴ d m = λ d x \end{aligned}$
supporting img
Position of centre of mass
$x_{c m} = \frac{\int d m \times x}{\int d m}$
$= \frac{\int_{0}^{3} (λ d x) \times x}{\int_{0}^{3} λ d x} = \frac{\int_{0}^{3} (2 + x) \times x d x}{\int_{0}^{3} (2 + x) \times d x}$
$= \frac{{[x^{2} + \frac{x^{3}}{3}]}_{0}^{3}}{{[2 x + \frac{x^{2}}{2}]}_{0}^{3}} = \frac{9 + 9}{6 + \frac{9}{2}} = \frac{12}{7} m .$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365702 From a uniform disc of radius $R$ , a circular section of radius $\frac{R}{2}$ is cut out. The centre of the hole is at $\frac{R}{2}$ from the centre of the original disc. Locate the centre of gravity of the resulting flat body.

1

\frac{R}{3}

to the right of centre

O

2

\frac{R}{6}

to the right of centre

O

3

\frac{R}{6}

to the left of centre

O

4

\frac{R}{3}

to the left of centre

O

Explanation:

Let mass per unit area of the disc be $σ$ .
$∴$ Mass of the disc $(M) = π R^{2} σ$ .
Mass of the portion removed from the disc $(M^{'})$
$= π {(\frac{R}{2})}^{2} σ = \frac{M}{4}$
The position of centre of mass of the remaining portion is
$x = \frac{- \frac{M}{4} \times \frac{R}{2}}{M - \frac{M}{4}}$
$= - \frac{M R}{8} \times \frac{4}{3 M}$
$x = \frac{- R}{6}$

NCERT Exemplar

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365703 Which of the following statements are correct?

1 Centre of mass of a body always coincides with the centre of gravity of the body.

2 Centre of mass of a body is the point at which the total gravitational torque on the body is zero.

3 A couple on a body produces both translational and rotational motion in a body.

4 Mechanical advantage greater than one means that small effort can be used to lift a large load.

NEET - 2017

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365704 A small card board is balanced on the tip of a pencil. The centre of mass coincides with centre of gravity $G$ . When the card board is in equilibrium (translational and rotational) then predict the correct option.
supporting img

1

\vec{R} = M \vec{g}

2

{\vec{τ}}_{G} = \sum {\vec{r}}_{i} \times m_{i} \vec{g} = 0

3

\sum m_{i} {\vec{r}}_{i} = 0

4 All the above

Explanation:

As the card board is in translational equilibrium the net vertical force should be zero.
$\vec{R} = M \vec{g}$
For rotational equilibrium the net torque should be zero about any point. The torque about $G$ is
${\vec{τ}}_{G} = \sum {\vec{r}}_{i} \times {\vec{F}}_{g} = \sum {\vec{r}}_{i} \times {\vec{m}}_{i} \vec{g} = 0$
Where ${\vec{r}}_{i}$ is the position vector of the $p$ article $m_{i}$ is the mass of the $i^{th}$ particle.
$\vec{g}$ is the acceleration vector.
For a small object $\vec{g}$ is same for all the particles
so $\vec{g}$ can be taken out of summation.
$\vec{g} \sum {\vec{r}}_{i} m_{i} = 0 \Rightarrow \sum m_{i} {\vec{r}}_{i} = \frac{{\vec{r}}_{C M}}{\sum m_{i}} = 0$
So the $C M$ coincides with centre of gravity
So option (4) is correct.

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365705 An irregular shaped body like a cardboard is suspended by a string. The body is in equilibrium about the lines $A A_{1}, B B_{1} & C C_{1}$ . Then the centre of gravity of the body lies
supporting img

1 on line

A A_{1}

2 on line

B B_{1}

3 on line

C C_{1}

4 All the above

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365701 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{12}{7} m

2

\frac{7}{3} m

3

\frac{9}{7} m

4

\frac{10}{7} m

Explanation:

Linear density of the rod varies with distance as
$\begin{aligned} λ = 2 + x \\ \frac{d m}{d x} = λ [given] ∴ d m = λ d x \end{aligned}$
supporting img
Position of centre of mass
$x_{c m} = \frac{\int d m \times x}{\int d m}$
$= \frac{\int_{0}^{3} (λ d x) \times x}{\int_{0}^{3} λ d x} = \frac{\int_{0}^{3} (2 + x) \times x d x}{\int_{0}^{3} (2 + x) \times d x}$
$= \frac{{[x^{2} + \frac{x^{3}}{3}]}_{0}^{3}}{{[2 x + \frac{x^{2}}{2}]}_{0}^{3}} = \frac{9 + 9}{6 + \frac{9}{2}} = \frac{12}{7} m .$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365702 From a uniform disc of radius $R$ , a circular section of radius $\frac{R}{2}$ is cut out. The centre of the hole is at $\frac{R}{2}$ from the centre of the original disc. Locate the centre of gravity of the resulting flat body.

1

\frac{R}{3}

to the right of centre

O

2

\frac{R}{6}

to the right of centre

O

3

\frac{R}{6}

to the left of centre

O

4

\frac{R}{3}

to the left of centre

O

Explanation:

Let mass per unit area of the disc be $σ$ .
$∴$ Mass of the disc $(M) = π R^{2} σ$ .
Mass of the portion removed from the disc $(M^{'})$
$= π {(\frac{R}{2})}^{2} σ = \frac{M}{4}$
The position of centre of mass of the remaining portion is
$x = \frac{- \frac{M}{4} \times \frac{R}{2}}{M - \frac{M}{4}}$
$= - \frac{M R}{8} \times \frac{4}{3 M}$
$x = \frac{- R}{6}$

NCERT Exemplar

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365703 Which of the following statements are correct?

1 Centre of mass of a body always coincides with the centre of gravity of the body.

2 Centre of mass of a body is the point at which the total gravitational torque on the body is zero.

3 A couple on a body produces both translational and rotational motion in a body.

4 Mechanical advantage greater than one means that small effort can be used to lift a large load.

NEET - 2017

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365704 A small card board is balanced on the tip of a pencil. The centre of mass coincides with centre of gravity $G$ . When the card board is in equilibrium (translational and rotational) then predict the correct option.
supporting img

1

\vec{R} = M \vec{g}

2

{\vec{τ}}_{G} = \sum {\vec{r}}_{i} \times m_{i} \vec{g} = 0

3

\sum m_{i} {\vec{r}}_{i} = 0

4 All the above

Explanation:

As the card board is in translational equilibrium the net vertical force should be zero.
$\vec{R} = M \vec{g}$
For rotational equilibrium the net torque should be zero about any point. The torque about $G$ is
${\vec{τ}}_{G} = \sum {\vec{r}}_{i} \times {\vec{F}}_{g} = \sum {\vec{r}}_{i} \times {\vec{m}}_{i} \vec{g} = 0$
Where ${\vec{r}}_{i}$ is the position vector of the $p$ article $m_{i}$ is the mass of the $i^{th}$ particle.
$\vec{g}$ is the acceleration vector.
For a small object $\vec{g}$ is same for all the particles
so $\vec{g}$ can be taken out of summation.
$\vec{g} \sum {\vec{r}}_{i} m_{i} = 0 \Rightarrow \sum m_{i} {\vec{r}}_{i} = \frac{{\vec{r}}_{C M}}{\sum m_{i}} = 0$
So the $C M$ coincides with centre of gravity
So option (4) is correct.

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365705 An irregular shaped body like a cardboard is suspended by a string. The body is in equilibrium about the lines $A A_{1}, B B_{1} & C C_{1}$ . Then the centre of gravity of the body lies
supporting img

1 on line

A A_{1}

2 on line

B B_{1}

3 on line

C C_{1}

4 All the above