163778
The position vector of the centre of mass \(\overrightarrow{\mathrm{r}} \mathrm{cm}\) of an symmetric uniform bar of negligible area of cross section as shown in figure is:
1 \(\overrightarrow{\mathrm{rcm}}=\frac{13}{8} L \hat{\mathrm{x}}+\frac{5}{8} L \hat{y}\)
2 \(\overrightarrow{\mathrm{rcm}}=\frac{11}{8} L \hat{x}+\frac{3}{8} L \hat{y}\)
3 \(\overrightarrow{\mathrm{rcm}}=\frac{3}{8} L \hat{x}+\frac{11}{8} L \hat{y}\)
4 \(\overrightarrow{\mathrm{rcm}}=\frac{5}{8} L \hat{x}+\frac{13}{8} L \hat{y}\)
Explanation:
\( X_{c m}=\frac{2 m L+2 m L+\frac{5 m L}{2}}{4 m}=\frac{13}{8} L \hat{x} \) \( Y_{c m}=\frac{2 m \times L+m \times\left(\frac{L}{2}\right)+m \times 0}{4 m}=\frac{5 L}{8} \hat{y} \)
NCERT-XI- I -98
4 RBTS PAPER
163779
The moment of inertia of a solid sphere about an axis parallel to its diameter and at a distance of \(x\) from it, is \(I(x)^{\prime}\). Which one of the graphs represents the variation of \(I(x)\) with \(x\) correctly :
163780
A uniform rectangular thin sheet \(A B C D\) of mass \(M\) has length \(a\) and breadth \(b\), as shown in the figure. If the shaded portion HBGO is cut-off, the coordinates of the centre of mass of the remaining portion will be :
163781
The time dependence of the position of a particle of mass \(m=2\) is given by \(\vec{r}(t)=2 t \hat{i}-3 t^2 \hat{j}\). Its angular momentum, with respect to the origin at time \(\mathbf{t}=\mathbf{2}\) is :
1 \(36 \hat{k}\)
2 \(-34(\hat{\mathrm{k}}-\hat{\mathrm{i}})\)
3 \(48(\hat{i}+\hat{j})\)
4 \(-48 \hat{k}\)
Explanation:
\( \vec{L} =m \) \( m =2 k g \) \( \vec{r} =2 t \hat{i}-3 t^2 \hat{j} \) \( =4 \hat{i}-12 \hat{j} \quad(A t t=2 \mathrm{sec}) \) \( \bar{v} =\frac{d \vec{r}}{d t}=2 \hat{i}-6 t \hat{j}=2 \hat{i}-12 \hat{j} \) \( \vec{r} \times \vec{v}=(4 \hat{i}-12 \hat{j}) \times(2 \hat{i}-12 \hat{j})=-24 \hat{k} \) \( \vec{L} =m(\vec{r} \times \vec{v})=-48 \hat{k} \)
NCERT-XI- I -107
4 RBTS PAPER
163782
Three point masses \(m\) are placed at the vertices of an equilateral triangle of side a. Moment of inertia of the system about an axis COD passing through a mass \(m\) at \(O\) and lying in the plane of \(A O B\) and perpendicular to \(O A\) is :
1 \(2 \mathrm{ma}^2\)
2 \(2 / 3 \mathrm{ma}^2\)
3 \(5 / 4 \mathrm{ma}^2\)
4 \(7 / 4 \mathrm{ma}^2\)
Explanation:
\( \mathrm{l}_{C O D} =\sum m r^2 \) \( =m_A r_A^2+m_B r_B^2 \) \( =m a^2+m\left(a \cos 60^{\circ}\right)^2 \) \( =m a^2+m \frac{a^2}{4}=\frac{5}{4} m a^2 \)
163778
The position vector of the centre of mass \(\overrightarrow{\mathrm{r}} \mathrm{cm}\) of an symmetric uniform bar of negligible area of cross section as shown in figure is:
1 \(\overrightarrow{\mathrm{rcm}}=\frac{13}{8} L \hat{\mathrm{x}}+\frac{5}{8} L \hat{y}\)
2 \(\overrightarrow{\mathrm{rcm}}=\frac{11}{8} L \hat{x}+\frac{3}{8} L \hat{y}\)
3 \(\overrightarrow{\mathrm{rcm}}=\frac{3}{8} L \hat{x}+\frac{11}{8} L \hat{y}\)
4 \(\overrightarrow{\mathrm{rcm}}=\frac{5}{8} L \hat{x}+\frac{13}{8} L \hat{y}\)
Explanation:
\( X_{c m}=\frac{2 m L+2 m L+\frac{5 m L}{2}}{4 m}=\frac{13}{8} L \hat{x} \) \( Y_{c m}=\frac{2 m \times L+m \times\left(\frac{L}{2}\right)+m \times 0}{4 m}=\frac{5 L}{8} \hat{y} \)
NCERT-XI- I -98
4 RBTS PAPER
163779
The moment of inertia of a solid sphere about an axis parallel to its diameter and at a distance of \(x\) from it, is \(I(x)^{\prime}\). Which one of the graphs represents the variation of \(I(x)\) with \(x\) correctly :
163780
A uniform rectangular thin sheet \(A B C D\) of mass \(M\) has length \(a\) and breadth \(b\), as shown in the figure. If the shaded portion HBGO is cut-off, the coordinates of the centre of mass of the remaining portion will be :
163781
The time dependence of the position of a particle of mass \(m=2\) is given by \(\vec{r}(t)=2 t \hat{i}-3 t^2 \hat{j}\). Its angular momentum, with respect to the origin at time \(\mathbf{t}=\mathbf{2}\) is :
1 \(36 \hat{k}\)
2 \(-34(\hat{\mathrm{k}}-\hat{\mathrm{i}})\)
3 \(48(\hat{i}+\hat{j})\)
4 \(-48 \hat{k}\)
Explanation:
\( \vec{L} =m \) \( m =2 k g \) \( \vec{r} =2 t \hat{i}-3 t^2 \hat{j} \) \( =4 \hat{i}-12 \hat{j} \quad(A t t=2 \mathrm{sec}) \) \( \bar{v} =\frac{d \vec{r}}{d t}=2 \hat{i}-6 t \hat{j}=2 \hat{i}-12 \hat{j} \) \( \vec{r} \times \vec{v}=(4 \hat{i}-12 \hat{j}) \times(2 \hat{i}-12 \hat{j})=-24 \hat{k} \) \( \vec{L} =m(\vec{r} \times \vec{v})=-48 \hat{k} \)
NCERT-XI- I -107
4 RBTS PAPER
163782
Three point masses \(m\) are placed at the vertices of an equilateral triangle of side a. Moment of inertia of the system about an axis COD passing through a mass \(m\) at \(O\) and lying in the plane of \(A O B\) and perpendicular to \(O A\) is :
1 \(2 \mathrm{ma}^2\)
2 \(2 / 3 \mathrm{ma}^2\)
3 \(5 / 4 \mathrm{ma}^2\)
4 \(7 / 4 \mathrm{ma}^2\)
Explanation:
\( \mathrm{l}_{C O D} =\sum m r^2 \) \( =m_A r_A^2+m_B r_B^2 \) \( =m a^2+m\left(a \cos 60^{\circ}\right)^2 \) \( =m a^2+m \frac{a^2}{4}=\frac{5}{4} m a^2 \)
163778
The position vector of the centre of mass \(\overrightarrow{\mathrm{r}} \mathrm{cm}\) of an symmetric uniform bar of negligible area of cross section as shown in figure is:
1 \(\overrightarrow{\mathrm{rcm}}=\frac{13}{8} L \hat{\mathrm{x}}+\frac{5}{8} L \hat{y}\)
2 \(\overrightarrow{\mathrm{rcm}}=\frac{11}{8} L \hat{x}+\frac{3}{8} L \hat{y}\)
3 \(\overrightarrow{\mathrm{rcm}}=\frac{3}{8} L \hat{x}+\frac{11}{8} L \hat{y}\)
4 \(\overrightarrow{\mathrm{rcm}}=\frac{5}{8} L \hat{x}+\frac{13}{8} L \hat{y}\)
Explanation:
\( X_{c m}=\frac{2 m L+2 m L+\frac{5 m L}{2}}{4 m}=\frac{13}{8} L \hat{x} \) \( Y_{c m}=\frac{2 m \times L+m \times\left(\frac{L}{2}\right)+m \times 0}{4 m}=\frac{5 L}{8} \hat{y} \)
NCERT-XI- I -98
4 RBTS PAPER
163779
The moment of inertia of a solid sphere about an axis parallel to its diameter and at a distance of \(x\) from it, is \(I(x)^{\prime}\). Which one of the graphs represents the variation of \(I(x)\) with \(x\) correctly :
163780
A uniform rectangular thin sheet \(A B C D\) of mass \(M\) has length \(a\) and breadth \(b\), as shown in the figure. If the shaded portion HBGO is cut-off, the coordinates of the centre of mass of the remaining portion will be :
163781
The time dependence of the position of a particle of mass \(m=2\) is given by \(\vec{r}(t)=2 t \hat{i}-3 t^2 \hat{j}\). Its angular momentum, with respect to the origin at time \(\mathbf{t}=\mathbf{2}\) is :
1 \(36 \hat{k}\)
2 \(-34(\hat{\mathrm{k}}-\hat{\mathrm{i}})\)
3 \(48(\hat{i}+\hat{j})\)
4 \(-48 \hat{k}\)
Explanation:
\( \vec{L} =m \) \( m =2 k g \) \( \vec{r} =2 t \hat{i}-3 t^2 \hat{j} \) \( =4 \hat{i}-12 \hat{j} \quad(A t t=2 \mathrm{sec}) \) \( \bar{v} =\frac{d \vec{r}}{d t}=2 \hat{i}-6 t \hat{j}=2 \hat{i}-12 \hat{j} \) \( \vec{r} \times \vec{v}=(4 \hat{i}-12 \hat{j}) \times(2 \hat{i}-12 \hat{j})=-24 \hat{k} \) \( \vec{L} =m(\vec{r} \times \vec{v})=-48 \hat{k} \)
NCERT-XI- I -107
4 RBTS PAPER
163782
Three point masses \(m\) are placed at the vertices of an equilateral triangle of side a. Moment of inertia of the system about an axis COD passing through a mass \(m\) at \(O\) and lying in the plane of \(A O B\) and perpendicular to \(O A\) is :
1 \(2 \mathrm{ma}^2\)
2 \(2 / 3 \mathrm{ma}^2\)
3 \(5 / 4 \mathrm{ma}^2\)
4 \(7 / 4 \mathrm{ma}^2\)
Explanation:
\( \mathrm{l}_{C O D} =\sum m r^2 \) \( =m_A r_A^2+m_B r_B^2 \) \( =m a^2+m\left(a \cos 60^{\circ}\right)^2 \) \( =m a^2+m \frac{a^2}{4}=\frac{5}{4} m a^2 \)
163778
The position vector of the centre of mass \(\overrightarrow{\mathrm{r}} \mathrm{cm}\) of an symmetric uniform bar of negligible area of cross section as shown in figure is:
1 \(\overrightarrow{\mathrm{rcm}}=\frac{13}{8} L \hat{\mathrm{x}}+\frac{5}{8} L \hat{y}\)
2 \(\overrightarrow{\mathrm{rcm}}=\frac{11}{8} L \hat{x}+\frac{3}{8} L \hat{y}\)
3 \(\overrightarrow{\mathrm{rcm}}=\frac{3}{8} L \hat{x}+\frac{11}{8} L \hat{y}\)
4 \(\overrightarrow{\mathrm{rcm}}=\frac{5}{8} L \hat{x}+\frac{13}{8} L \hat{y}\)
Explanation:
\( X_{c m}=\frac{2 m L+2 m L+\frac{5 m L}{2}}{4 m}=\frac{13}{8} L \hat{x} \) \( Y_{c m}=\frac{2 m \times L+m \times\left(\frac{L}{2}\right)+m \times 0}{4 m}=\frac{5 L}{8} \hat{y} \)
NCERT-XI- I -98
4 RBTS PAPER
163779
The moment of inertia of a solid sphere about an axis parallel to its diameter and at a distance of \(x\) from it, is \(I(x)^{\prime}\). Which one of the graphs represents the variation of \(I(x)\) with \(x\) correctly :
163780
A uniform rectangular thin sheet \(A B C D\) of mass \(M\) has length \(a\) and breadth \(b\), as shown in the figure. If the shaded portion HBGO is cut-off, the coordinates of the centre of mass of the remaining portion will be :
163781
The time dependence of the position of a particle of mass \(m=2\) is given by \(\vec{r}(t)=2 t \hat{i}-3 t^2 \hat{j}\). Its angular momentum, with respect to the origin at time \(\mathbf{t}=\mathbf{2}\) is :
1 \(36 \hat{k}\)
2 \(-34(\hat{\mathrm{k}}-\hat{\mathrm{i}})\)
3 \(48(\hat{i}+\hat{j})\)
4 \(-48 \hat{k}\)
Explanation:
\( \vec{L} =m \) \( m =2 k g \) \( \vec{r} =2 t \hat{i}-3 t^2 \hat{j} \) \( =4 \hat{i}-12 \hat{j} \quad(A t t=2 \mathrm{sec}) \) \( \bar{v} =\frac{d \vec{r}}{d t}=2 \hat{i}-6 t \hat{j}=2 \hat{i}-12 \hat{j} \) \( \vec{r} \times \vec{v}=(4 \hat{i}-12 \hat{j}) \times(2 \hat{i}-12 \hat{j})=-24 \hat{k} \) \( \vec{L} =m(\vec{r} \times \vec{v})=-48 \hat{k} \)
NCERT-XI- I -107
4 RBTS PAPER
163782
Three point masses \(m\) are placed at the vertices of an equilateral triangle of side a. Moment of inertia of the system about an axis COD passing through a mass \(m\) at \(O\) and lying in the plane of \(A O B\) and perpendicular to \(O A\) is :
1 \(2 \mathrm{ma}^2\)
2 \(2 / 3 \mathrm{ma}^2\)
3 \(5 / 4 \mathrm{ma}^2\)
4 \(7 / 4 \mathrm{ma}^2\)
Explanation:
\( \mathrm{l}_{C O D} =\sum m r^2 \) \( =m_A r_A^2+m_B r_B^2 \) \( =m a^2+m\left(a \cos 60^{\circ}\right)^2 \) \( =m a^2+m \frac{a^2}{4}=\frac{5}{4} m a^2 \)
163778
The position vector of the centre of mass \(\overrightarrow{\mathrm{r}} \mathrm{cm}\) of an symmetric uniform bar of negligible area of cross section as shown in figure is:
1 \(\overrightarrow{\mathrm{rcm}}=\frac{13}{8} L \hat{\mathrm{x}}+\frac{5}{8} L \hat{y}\)
2 \(\overrightarrow{\mathrm{rcm}}=\frac{11}{8} L \hat{x}+\frac{3}{8} L \hat{y}\)
3 \(\overrightarrow{\mathrm{rcm}}=\frac{3}{8} L \hat{x}+\frac{11}{8} L \hat{y}\)
4 \(\overrightarrow{\mathrm{rcm}}=\frac{5}{8} L \hat{x}+\frac{13}{8} L \hat{y}\)
Explanation:
\( X_{c m}=\frac{2 m L+2 m L+\frac{5 m L}{2}}{4 m}=\frac{13}{8} L \hat{x} \) \( Y_{c m}=\frac{2 m \times L+m \times\left(\frac{L}{2}\right)+m \times 0}{4 m}=\frac{5 L}{8} \hat{y} \)
NCERT-XI- I -98
4 RBTS PAPER
163779
The moment of inertia of a solid sphere about an axis parallel to its diameter and at a distance of \(x\) from it, is \(I(x)^{\prime}\). Which one of the graphs represents the variation of \(I(x)\) with \(x\) correctly :
163780
A uniform rectangular thin sheet \(A B C D\) of mass \(M\) has length \(a\) and breadth \(b\), as shown in the figure. If the shaded portion HBGO is cut-off, the coordinates of the centre of mass of the remaining portion will be :
163781
The time dependence of the position of a particle of mass \(m=2\) is given by \(\vec{r}(t)=2 t \hat{i}-3 t^2 \hat{j}\). Its angular momentum, with respect to the origin at time \(\mathbf{t}=\mathbf{2}\) is :
1 \(36 \hat{k}\)
2 \(-34(\hat{\mathrm{k}}-\hat{\mathrm{i}})\)
3 \(48(\hat{i}+\hat{j})\)
4 \(-48 \hat{k}\)
Explanation:
\( \vec{L} =m \) \( m =2 k g \) \( \vec{r} =2 t \hat{i}-3 t^2 \hat{j} \) \( =4 \hat{i}-12 \hat{j} \quad(A t t=2 \mathrm{sec}) \) \( \bar{v} =\frac{d \vec{r}}{d t}=2 \hat{i}-6 t \hat{j}=2 \hat{i}-12 \hat{j} \) \( \vec{r} \times \vec{v}=(4 \hat{i}-12 \hat{j}) \times(2 \hat{i}-12 \hat{j})=-24 \hat{k} \) \( \vec{L} =m(\vec{r} \times \vec{v})=-48 \hat{k} \)
NCERT-XI- I -107
4 RBTS PAPER
163782
Three point masses \(m\) are placed at the vertices of an equilateral triangle of side a. Moment of inertia of the system about an axis COD passing through a mass \(m\) at \(O\) and lying in the plane of \(A O B\) and perpendicular to \(O A\) is :
1 \(2 \mathrm{ma}^2\)
2 \(2 / 3 \mathrm{ma}^2\)
3 \(5 / 4 \mathrm{ma}^2\)
4 \(7 / 4 \mathrm{ma}^2\)
Explanation:
\( \mathrm{l}_{C O D} =\sum m r^2 \) \( =m_A r_A^2+m_B r_B^2 \) \( =m a^2+m\left(a \cos 60^{\circ}\right)^2 \) \( =m a^2+m \frac{a^2}{4}=\frac{5}{4} m a^2 \)
