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4 RBTS PAPER

163783 A circular plate of diameter $d$ is kept in contact with a square plate of edge $d$ as shown in fig. The density of the material and the thickness are same everywhere. The centre of mass of the composite system will be

1 inside the circular plate

2 inside the square plate

3 at the point of contact

4 outside the system

Explanation:

$Area of square = a \times a$
$= a^{2}$

Area of circular plate $= π \times {(\frac{^{'}}{2})}^{2} = π \times \frac{.2}{4}$
It is clear that Area of square is more than area of circular plate hence centre of mass lie within square plate.

NCERT-XI- I -115

4 RBTS PAPER

163784 The moment of inertia of a rod about an axis through its centre and perpendicular to it's length is $\frac{1}{12} M L^{2} (m =$ mass $& L$ is the length of rod). The rod is bent in the middle so that the two halves make an angle of $60^{\circ}$ . The moment of inertia of the bent rod about the same axis perpendicular to length of each halves is :

1

\frac{1}{24} M L^{2}

2

\frac{1}{48} M L^{2}

3

\frac{1}{12} M L^{2}

4

\frac{1}{8} M L^{2}

Explanation:

$I = \frac{\frac{M}{2} {(\frac{L}{2})}^{2}}{3} + \frac{\frac{M}{2} {(\frac{L}{2})}^{2}}{3}$
$= \frac{M L^{2}}{12}$

NCERT-XI- I -114

4 RBTS PAPER

163785 Two bodies of mass $1 kg & 3 kg$ have position vectors $\hat{i} + 2 \hat{j} + \hat{k}$ and $- 3 \hat{i} - 2 \hat{j} + \hat{k}$ , respectively. The centre of mass of this system has a position vector :

1

- 2 \hat{i} + \hat{k}

2

- 2 \hat{i} - \hat{j} + \hat{k}

3

2 \hat{i} - \hat{j} - \hat{k}

4

- \hat{i} + \hat{j} + \hat{k}

Explanation:

${\vec{r}}_{c m} = \frac{m_{1} {\vec{r}}_{1} + m_{2} {\vec{r}}_{2}}{m_{1} + m_{2}}$
$= \frac{1 (\hat{i} + 2 \hat{j} + \hat{k}) + 3 (- 3 \hat{i} - 2 \hat{j} + \hat{k})}{1 + 3}$
$\Rightarrow {\vec{r}}_{c m} = - 2 \hat{i} - \hat{j} + \hat{k}$

NCERT-XI- I -97

4 RBTS PAPER

163786 A constant torque of $1000 N - m$ turns a wheel of moment of inertia $200 kg - m^{2}$ about an axis through its centre. Its angular velocity after 3 sec is (starts from rest):

1

1 rad / \sec

2

5 rad / \sec

3

10 rad / \sec

4

15 rad / \sec

Explanation:

$ω = ω_{0} + at \Rightarrow ω = 0 + (\frac{τ}{l}) t$
$ω = 0 + \frac{1000}{200} \times 3 = 15 rad / s$

NCERT-XI- I -119

4 RBTS PAPER

163783 A circular plate of diameter $d$ is kept in contact with a square plate of edge $d$ as shown in fig. The density of the material and the thickness are same everywhere. The centre of mass of the composite system will be

1 inside the circular plate

2 inside the square plate

3 at the point of contact

4 outside the system

Explanation:

$Area of square = a \times a$
$= a^{2}$

Area of circular plate $= π \times {(\frac{^{'}}{2})}^{2} = π \times \frac{.2}{4}$
It is clear that Area of square is more than area of circular plate hence centre of mass lie within square plate.

NCERT-XI- I -115

4 RBTS PAPER

163784 The moment of inertia of a rod about an axis through its centre and perpendicular to it's length is $\frac{1}{12} M L^{2} (m =$ mass $& L$ is the length of rod). The rod is bent in the middle so that the two halves make an angle of $60^{\circ}$ . The moment of inertia of the bent rod about the same axis perpendicular to length of each halves is :

1

\frac{1}{24} M L^{2}

2

\frac{1}{48} M L^{2}

3

\frac{1}{12} M L^{2}

4

\frac{1}{8} M L^{2}

Explanation:

$I = \frac{\frac{M}{2} {(\frac{L}{2})}^{2}}{3} + \frac{\frac{M}{2} {(\frac{L}{2})}^{2}}{3}$
$= \frac{M L^{2}}{12}$

NCERT-XI- I -114

4 RBTS PAPER

163785 Two bodies of mass $1 kg & 3 kg$ have position vectors $\hat{i} + 2 \hat{j} + \hat{k}$ and $- 3 \hat{i} - 2 \hat{j} + \hat{k}$ , respectively. The centre of mass of this system has a position vector :

1

- 2 \hat{i} + \hat{k}

2

- 2 \hat{i} - \hat{j} + \hat{k}

3

2 \hat{i} - \hat{j} - \hat{k}

4

- \hat{i} + \hat{j} + \hat{k}

Explanation:

${\vec{r}}_{c m} = \frac{m_{1} {\vec{r}}_{1} + m_{2} {\vec{r}}_{2}}{m_{1} + m_{2}}$
$= \frac{1 (\hat{i} + 2 \hat{j} + \hat{k}) + 3 (- 3 \hat{i} - 2 \hat{j} + \hat{k})}{1 + 3}$
$\Rightarrow {\vec{r}}_{c m} = - 2 \hat{i} - \hat{j} + \hat{k}$

NCERT-XI- I -97

4 RBTS PAPER

163786 A constant torque of $1000 N - m$ turns a wheel of moment of inertia $200 kg - m^{2}$ about an axis through its centre. Its angular velocity after 3 sec is (starts from rest):

1

1 rad / \sec

2

5 rad / \sec

3

10 rad / \sec

4

15 rad / \sec

Explanation:

$ω = ω_{0} + at \Rightarrow ω = 0 + (\frac{τ}{l}) t$
$ω = 0 + \frac{1000}{200} \times 3 = 15 rad / s$

NCERT-XI- I -119

4 RBTS PAPER

163783 A circular plate of diameter $d$ is kept in contact with a square plate of edge $d$ as shown in fig. The density of the material and the thickness are same everywhere. The centre of mass of the composite system will be

1 inside the circular plate

2 inside the square plate

3 at the point of contact

4 outside the system

Explanation:

$Area of square = a \times a$
$= a^{2}$

Area of circular plate $= π \times {(\frac{^{'}}{2})}^{2} = π \times \frac{.2}{4}$
It is clear that Area of square is more than area of circular plate hence centre of mass lie within square plate.

NCERT-XI- I -115

4 RBTS PAPER

163784 The moment of inertia of a rod about an axis through its centre and perpendicular to it's length is $\frac{1}{12} M L^{2} (m =$ mass $& L$ is the length of rod). The rod is bent in the middle so that the two halves make an angle of $60^{\circ}$ . The moment of inertia of the bent rod about the same axis perpendicular to length of each halves is :

1

\frac{1}{24} M L^{2}

2

\frac{1}{48} M L^{2}

3

\frac{1}{12} M L^{2}

4

\frac{1}{8} M L^{2}

Explanation:

$I = \frac{\frac{M}{2} {(\frac{L}{2})}^{2}}{3} + \frac{\frac{M}{2} {(\frac{L}{2})}^{2}}{3}$
$= \frac{M L^{2}}{12}$

NCERT-XI- I -114

4 RBTS PAPER

163785 Two bodies of mass $1 kg & 3 kg$ have position vectors $\hat{i} + 2 \hat{j} + \hat{k}$ and $- 3 \hat{i} - 2 \hat{j} + \hat{k}$ , respectively. The centre of mass of this system has a position vector :

1

- 2 \hat{i} + \hat{k}

2

- 2 \hat{i} - \hat{j} + \hat{k}

3

2 \hat{i} - \hat{j} - \hat{k}

4

- \hat{i} + \hat{j} + \hat{k}

Explanation:

${\vec{r}}_{c m} = \frac{m_{1} {\vec{r}}_{1} + m_{2} {\vec{r}}_{2}}{m_{1} + m_{2}}$
$= \frac{1 (\hat{i} + 2 \hat{j} + \hat{k}) + 3 (- 3 \hat{i} - 2 \hat{j} + \hat{k})}{1 + 3}$
$\Rightarrow {\vec{r}}_{c m} = - 2 \hat{i} - \hat{j} + \hat{k}$

NCERT-XI- I -97

4 RBTS PAPER

163786 A constant torque of $1000 N - m$ turns a wheel of moment of inertia $200 kg - m^{2}$ about an axis through its centre. Its angular velocity after 3 sec is (starts from rest):

1

1 rad / \sec

2

5 rad / \sec

3

10 rad / \sec

4

15 rad / \sec

Explanation:

$ω = ω_{0} + at \Rightarrow ω = 0 + (\frac{τ}{l}) t$
$ω = 0 + \frac{1000}{200} \times 3 = 15 rad / s$

NCERT-XI- I -119

4 RBTS PAPER

163783 A circular plate of diameter $d$ is kept in contact with a square plate of edge $d$ as shown in fig. The density of the material and the thickness are same everywhere. The centre of mass of the composite system will be

1 inside the circular plate

2 inside the square plate

3 at the point of contact

4 outside the system

Explanation:

$Area of square = a \times a$
$= a^{2}$

Area of circular plate $= π \times {(\frac{^{'}}{2})}^{2} = π \times \frac{.2}{4}$
It is clear that Area of square is more than area of circular plate hence centre of mass lie within square plate.

NCERT-XI- I -115

4 RBTS PAPER

163784 The moment of inertia of a rod about an axis through its centre and perpendicular to it's length is $\frac{1}{12} M L^{2} (m =$ mass $& L$ is the length of rod). The rod is bent in the middle so that the two halves make an angle of $60^{\circ}$ . The moment of inertia of the bent rod about the same axis perpendicular to length of each halves is :

1

\frac{1}{24} M L^{2}

2

\frac{1}{48} M L^{2}

3

\frac{1}{12} M L^{2}

4

\frac{1}{8} M L^{2}

Explanation:

$I = \frac{\frac{M}{2} {(\frac{L}{2})}^{2}}{3} + \frac{\frac{M}{2} {(\frac{L}{2})}^{2}}{3}$
$= \frac{M L^{2}}{12}$

NCERT-XI- I -114

4 RBTS PAPER

163785 Two bodies of mass $1 kg & 3 kg$ have position vectors $\hat{i} + 2 \hat{j} + \hat{k}$ and $- 3 \hat{i} - 2 \hat{j} + \hat{k}$ , respectively. The centre of mass of this system has a position vector :

1

- 2 \hat{i} + \hat{k}

2

- 2 \hat{i} - \hat{j} + \hat{k}

3

2 \hat{i} - \hat{j} - \hat{k}

4

- \hat{i} + \hat{j} + \hat{k}

Explanation:

${\vec{r}}_{c m} = \frac{m_{1} {\vec{r}}_{1} + m_{2} {\vec{r}}_{2}}{m_{1} + m_{2}}$
$= \frac{1 (\hat{i} + 2 \hat{j} + \hat{k}) + 3 (- 3 \hat{i} - 2 \hat{j} + \hat{k})}{1 + 3}$
$\Rightarrow {\vec{r}}_{c m} = - 2 \hat{i} - \hat{j} + \hat{k}$

NCERT-XI- I -97

4 RBTS PAPER

163786 A constant torque of $1000 N - m$ turns a wheel of moment of inertia $200 kg - m^{2}$ about an axis through its centre. Its angular velocity after 3 sec is (starts from rest):

1

1 rad / \sec

2

5 rad / \sec

3

10 rad / \sec

4

15 rad / \sec

Explanation:

$ω = ω_{0} + at \Rightarrow ω = 0 + (\frac{τ}{l}) t$
$ω = 0 + \frac{1000}{200} \times 3 = 15 rad / s$

NCERT-XI- I -119

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