NEET Test Series from KOTA - 10 Papers In MS WORD
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Rotational Motion
149709
In a two-particle system with particle masses \(m_{1}\) and \(m_{2}\), the first particle is pushed towards the centre of mass through a distance \(d\), the distance through which second particle must be moved to keep the centre of mass at the same position is
E \(\mathrm{x}_{\mathrm{cm}}=\frac{\mathrm{m}_{1}(0)+\mathrm{m}_{2}(l)}{\mathrm{m}_{1}+\mathrm{m}_{2}}\) \(\mathrm{x}_{\mathrm{cm}}=\frac{\mathrm{m}_{2} l}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\) \(\mathrm{x}_{\mathrm{cm}}=\frac{\mathrm{m}_{1}(\mathrm{~d})+\mathrm{m}_{2}(l-\mathrm{x})}{\mathrm{m}_{1}+\mathrm{m}_{2}}\) On equating equation (i) \& (ii), we get \(\frac{\mathrm{m}_{1} \mathrm{~d}+\mathrm{m}_{2}(l-\mathrm{x})}{\mathrm{m}_{1}+\mathrm{m}_{2}}=\frac{\mathrm{m}_{2} l}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\) \(\mathrm{~m}_{1} \mathrm{~d}+\mathrm{m}_{2} l-\mathrm{m}_{2} \mathrm{x}=\mathrm{m}_{2} l\) \(\mathrm{~m}_{1} \mathrm{~d}=\mathrm{m}_{2} \mathrm{x} \Rightarrow \mathrm{x}=\frac{\mathrm{m}_{1} \mathrm{~d}}{\mathrm{~m}_{2}}\)
Kerala CEE- 2013
Rotational Motion
149710
A system consists of 3 particles each of mass \(m\) located at points \((1,1)(2,2)\) and \((3,3)\). The coordinates of the centre of mass are
1 \((6,6)\)
2 \((3,3)\)
3 \((1,1)\)
4 \((2,2)\)
5 \((5,5)\)
Explanation:
D Given, \(\mathrm{m}_{1}=\mathrm{m}_{2}=\mathrm{m}_{3}=\mathrm{m}\) \(\mathrm{x}_{1}=1, \mathrm{x}_{2}=2\) and \(\mathrm{x}_{3}=3\) \(\mathrm{y}_{1}=1, \mathrm{y}_{2}=2\), and \(\mathrm{y}_{3}=3\) \(\mathrm{x}_{\mathrm{cm}}=\frac{\mathrm{m}_{1} \mathrm{x}_{1}+\mathrm{m}_{2} \mathrm{x}_{2}+\mathrm{m}_{3} \mathrm{x}_{3}}{\mathrm{~m}_{1}+\mathrm{m}_{2}+\mathrm{m}_{3}}\) \(\mathrm{x}_{\mathrm{cm}}=\frac{\mathrm{mx}_{1}+\mathrm{mx}_{2}+\mathrm{mx}_{3}}{\mathrm{~m}+\mathrm{m}+\mathrm{m}}=\frac{\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}}{3}\) \(=\frac{1+2+3}{3}=\frac{6}{3}=2\) \(\mathrm{y}_{\mathrm{cm}}=\frac{\mathrm{my}_{1}+\mathrm{my}_{2}+\mathrm{my}_{3}}{\mathrm{~m}+\mathrm{m}+\mathrm{m}}=\frac{\mathrm{y}_{1}+\mathrm{y}_{2}+\mathrm{y}_{3}}{3}\) \(=\frac{1+2+3}{3}=\frac{6}{3}=2\) Hence, the coordinates of the centre of mass are \((2,2)\).
BITSAT-2010
Rotational Motion
149711
Three identical spheres, each of mass \(3 \mathrm{~kg}\) are placed touching each other with their centres lying on a straight line. The centres of the sphere are marked at \(P, Q\) and \(R\) respectively. The distance of centre of mass of system from \(P\) is
B \lt smiles>[R]c1cccc2cccc([R])c12 \lt /smiles> The centre of mass of the given system consisting of three identical spheres of mass \(3 \mathrm{~kg}\) each is \(x_{c m}=\frac{M_{P}\left(x_{P}\right)+M_{Q}\left(x_{Q}\right)+M_{R}\left(x_{R}\right)}{M_{P}+M_{Q}+M_{R}}\) \(\mathrm{x}_{\mathrm{cm}}=\frac{\mathrm{M}(0)+\mathrm{M}(\mathrm{PQ})+\mathrm{M}(\mathrm{PR})}{\mathrm{M}+\mathrm{M}+\mathrm{M}}\) \(=\frac{\mathrm{M}(\mathrm{PQ}+\mathrm{PR})}{3 \mathrm{M}}\) \(\mathrm{x}_{\mathrm{cm}}=\frac{\mathrm{PQ}+\mathrm{PR}}{3}\) Hence, COM of the given system is \(\frac{P Q+P R}{3}\)
Kerala CEE - 2008
Rotational Motion
149712
The distance between the centers of carbon and oxygen atoms in the carbon monoxide molecule is \(1.130 \AA\). Locate the centre of mass of the molecule relative to the carbon atom :
1 \(5.428 \AA\)
2 \(1.130 \AA\)
3 \(0.6457 \AA\)
4 \(0.3260 \AA\)
5 none of these
Explanation:
C \(\mathrm{m}_{1}=16\) (Oxygen mass number) \(\mathrm{m}_{2}=12\) (carbon mass number) \(\mathrm{m}_{2} \mathrm{x}_{2}=\mathrm{m}_{1} \mathrm{x}_{1}\) \(12 \mathrm{x}_{2}=16\left(1.130 \times 10^{-10}-\mathrm{x}_{2}\right)\) \(12 \mathrm{x}_{2}=16 \times 1.130 \times 10^{-10}-16 \mathrm{x}_{2}\) \(\mathrm{x}_{2}=\frac{16 \times 1.13 \times 10^{-10}}{28}\) \(\mathrm{x}_{2}=0.6457 \AA\) Hence, the centre of mass (COM) of molecule relative to the carbon atom is \(0.6457{ }^{\circ} \mathrm{A}\).
149709
In a two-particle system with particle masses \(m_{1}\) and \(m_{2}\), the first particle is pushed towards the centre of mass through a distance \(d\), the distance through which second particle must be moved to keep the centre of mass at the same position is
E \(\mathrm{x}_{\mathrm{cm}}=\frac{\mathrm{m}_{1}(0)+\mathrm{m}_{2}(l)}{\mathrm{m}_{1}+\mathrm{m}_{2}}\) \(\mathrm{x}_{\mathrm{cm}}=\frac{\mathrm{m}_{2} l}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\) \(\mathrm{x}_{\mathrm{cm}}=\frac{\mathrm{m}_{1}(\mathrm{~d})+\mathrm{m}_{2}(l-\mathrm{x})}{\mathrm{m}_{1}+\mathrm{m}_{2}}\) On equating equation (i) \& (ii), we get \(\frac{\mathrm{m}_{1} \mathrm{~d}+\mathrm{m}_{2}(l-\mathrm{x})}{\mathrm{m}_{1}+\mathrm{m}_{2}}=\frac{\mathrm{m}_{2} l}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\) \(\mathrm{~m}_{1} \mathrm{~d}+\mathrm{m}_{2} l-\mathrm{m}_{2} \mathrm{x}=\mathrm{m}_{2} l\) \(\mathrm{~m}_{1} \mathrm{~d}=\mathrm{m}_{2} \mathrm{x} \Rightarrow \mathrm{x}=\frac{\mathrm{m}_{1} \mathrm{~d}}{\mathrm{~m}_{2}}\)
Kerala CEE- 2013
Rotational Motion
149710
A system consists of 3 particles each of mass \(m\) located at points \((1,1)(2,2)\) and \((3,3)\). The coordinates of the centre of mass are
1 \((6,6)\)
2 \((3,3)\)
3 \((1,1)\)
4 \((2,2)\)
5 \((5,5)\)
Explanation:
D Given, \(\mathrm{m}_{1}=\mathrm{m}_{2}=\mathrm{m}_{3}=\mathrm{m}\) \(\mathrm{x}_{1}=1, \mathrm{x}_{2}=2\) and \(\mathrm{x}_{3}=3\) \(\mathrm{y}_{1}=1, \mathrm{y}_{2}=2\), and \(\mathrm{y}_{3}=3\) \(\mathrm{x}_{\mathrm{cm}}=\frac{\mathrm{m}_{1} \mathrm{x}_{1}+\mathrm{m}_{2} \mathrm{x}_{2}+\mathrm{m}_{3} \mathrm{x}_{3}}{\mathrm{~m}_{1}+\mathrm{m}_{2}+\mathrm{m}_{3}}\) \(\mathrm{x}_{\mathrm{cm}}=\frac{\mathrm{mx}_{1}+\mathrm{mx}_{2}+\mathrm{mx}_{3}}{\mathrm{~m}+\mathrm{m}+\mathrm{m}}=\frac{\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}}{3}\) \(=\frac{1+2+3}{3}=\frac{6}{3}=2\) \(\mathrm{y}_{\mathrm{cm}}=\frac{\mathrm{my}_{1}+\mathrm{my}_{2}+\mathrm{my}_{3}}{\mathrm{~m}+\mathrm{m}+\mathrm{m}}=\frac{\mathrm{y}_{1}+\mathrm{y}_{2}+\mathrm{y}_{3}}{3}\) \(=\frac{1+2+3}{3}=\frac{6}{3}=2\) Hence, the coordinates of the centre of mass are \((2,2)\).
BITSAT-2010
Rotational Motion
149711
Three identical spheres, each of mass \(3 \mathrm{~kg}\) are placed touching each other with their centres lying on a straight line. The centres of the sphere are marked at \(P, Q\) and \(R\) respectively. The distance of centre of mass of system from \(P\) is
B \lt smiles>[R]c1cccc2cccc([R])c12 \lt /smiles> The centre of mass of the given system consisting of three identical spheres of mass \(3 \mathrm{~kg}\) each is \(x_{c m}=\frac{M_{P}\left(x_{P}\right)+M_{Q}\left(x_{Q}\right)+M_{R}\left(x_{R}\right)}{M_{P}+M_{Q}+M_{R}}\) \(\mathrm{x}_{\mathrm{cm}}=\frac{\mathrm{M}(0)+\mathrm{M}(\mathrm{PQ})+\mathrm{M}(\mathrm{PR})}{\mathrm{M}+\mathrm{M}+\mathrm{M}}\) \(=\frac{\mathrm{M}(\mathrm{PQ}+\mathrm{PR})}{3 \mathrm{M}}\) \(\mathrm{x}_{\mathrm{cm}}=\frac{\mathrm{PQ}+\mathrm{PR}}{3}\) Hence, COM of the given system is \(\frac{P Q+P R}{3}\)
Kerala CEE - 2008
Rotational Motion
149712
The distance between the centers of carbon and oxygen atoms in the carbon monoxide molecule is \(1.130 \AA\). Locate the centre of mass of the molecule relative to the carbon atom :
1 \(5.428 \AA\)
2 \(1.130 \AA\)
3 \(0.6457 \AA\)
4 \(0.3260 \AA\)
5 none of these
Explanation:
C \(\mathrm{m}_{1}=16\) (Oxygen mass number) \(\mathrm{m}_{2}=12\) (carbon mass number) \(\mathrm{m}_{2} \mathrm{x}_{2}=\mathrm{m}_{1} \mathrm{x}_{1}\) \(12 \mathrm{x}_{2}=16\left(1.130 \times 10^{-10}-\mathrm{x}_{2}\right)\) \(12 \mathrm{x}_{2}=16 \times 1.130 \times 10^{-10}-16 \mathrm{x}_{2}\) \(\mathrm{x}_{2}=\frac{16 \times 1.13 \times 10^{-10}}{28}\) \(\mathrm{x}_{2}=0.6457 \AA\) Hence, the centre of mass (COM) of molecule relative to the carbon atom is \(0.6457{ }^{\circ} \mathrm{A}\).
149709
In a two-particle system with particle masses \(m_{1}\) and \(m_{2}\), the first particle is pushed towards the centre of mass through a distance \(d\), the distance through which second particle must be moved to keep the centre of mass at the same position is
E \(\mathrm{x}_{\mathrm{cm}}=\frac{\mathrm{m}_{1}(0)+\mathrm{m}_{2}(l)}{\mathrm{m}_{1}+\mathrm{m}_{2}}\) \(\mathrm{x}_{\mathrm{cm}}=\frac{\mathrm{m}_{2} l}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\) \(\mathrm{x}_{\mathrm{cm}}=\frac{\mathrm{m}_{1}(\mathrm{~d})+\mathrm{m}_{2}(l-\mathrm{x})}{\mathrm{m}_{1}+\mathrm{m}_{2}}\) On equating equation (i) \& (ii), we get \(\frac{\mathrm{m}_{1} \mathrm{~d}+\mathrm{m}_{2}(l-\mathrm{x})}{\mathrm{m}_{1}+\mathrm{m}_{2}}=\frac{\mathrm{m}_{2} l}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\) \(\mathrm{~m}_{1} \mathrm{~d}+\mathrm{m}_{2} l-\mathrm{m}_{2} \mathrm{x}=\mathrm{m}_{2} l\) \(\mathrm{~m}_{1} \mathrm{~d}=\mathrm{m}_{2} \mathrm{x} \Rightarrow \mathrm{x}=\frac{\mathrm{m}_{1} \mathrm{~d}}{\mathrm{~m}_{2}}\)
Kerala CEE- 2013
Rotational Motion
149710
A system consists of 3 particles each of mass \(m\) located at points \((1,1)(2,2)\) and \((3,3)\). The coordinates of the centre of mass are
1 \((6,6)\)
2 \((3,3)\)
3 \((1,1)\)
4 \((2,2)\)
5 \((5,5)\)
Explanation:
D Given, \(\mathrm{m}_{1}=\mathrm{m}_{2}=\mathrm{m}_{3}=\mathrm{m}\) \(\mathrm{x}_{1}=1, \mathrm{x}_{2}=2\) and \(\mathrm{x}_{3}=3\) \(\mathrm{y}_{1}=1, \mathrm{y}_{2}=2\), and \(\mathrm{y}_{3}=3\) \(\mathrm{x}_{\mathrm{cm}}=\frac{\mathrm{m}_{1} \mathrm{x}_{1}+\mathrm{m}_{2} \mathrm{x}_{2}+\mathrm{m}_{3} \mathrm{x}_{3}}{\mathrm{~m}_{1}+\mathrm{m}_{2}+\mathrm{m}_{3}}\) \(\mathrm{x}_{\mathrm{cm}}=\frac{\mathrm{mx}_{1}+\mathrm{mx}_{2}+\mathrm{mx}_{3}}{\mathrm{~m}+\mathrm{m}+\mathrm{m}}=\frac{\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}}{3}\) \(=\frac{1+2+3}{3}=\frac{6}{3}=2\) \(\mathrm{y}_{\mathrm{cm}}=\frac{\mathrm{my}_{1}+\mathrm{my}_{2}+\mathrm{my}_{3}}{\mathrm{~m}+\mathrm{m}+\mathrm{m}}=\frac{\mathrm{y}_{1}+\mathrm{y}_{2}+\mathrm{y}_{3}}{3}\) \(=\frac{1+2+3}{3}=\frac{6}{3}=2\) Hence, the coordinates of the centre of mass are \((2,2)\).
BITSAT-2010
Rotational Motion
149711
Three identical spheres, each of mass \(3 \mathrm{~kg}\) are placed touching each other with their centres lying on a straight line. The centres of the sphere are marked at \(P, Q\) and \(R\) respectively. The distance of centre of mass of system from \(P\) is
B \lt smiles>[R]c1cccc2cccc([R])c12 \lt /smiles> The centre of mass of the given system consisting of three identical spheres of mass \(3 \mathrm{~kg}\) each is \(x_{c m}=\frac{M_{P}\left(x_{P}\right)+M_{Q}\left(x_{Q}\right)+M_{R}\left(x_{R}\right)}{M_{P}+M_{Q}+M_{R}}\) \(\mathrm{x}_{\mathrm{cm}}=\frac{\mathrm{M}(0)+\mathrm{M}(\mathrm{PQ})+\mathrm{M}(\mathrm{PR})}{\mathrm{M}+\mathrm{M}+\mathrm{M}}\) \(=\frac{\mathrm{M}(\mathrm{PQ}+\mathrm{PR})}{3 \mathrm{M}}\) \(\mathrm{x}_{\mathrm{cm}}=\frac{\mathrm{PQ}+\mathrm{PR}}{3}\) Hence, COM of the given system is \(\frac{P Q+P R}{3}\)
Kerala CEE - 2008
Rotational Motion
149712
The distance between the centers of carbon and oxygen atoms in the carbon monoxide molecule is \(1.130 \AA\). Locate the centre of mass of the molecule relative to the carbon atom :
1 \(5.428 \AA\)
2 \(1.130 \AA\)
3 \(0.6457 \AA\)
4 \(0.3260 \AA\)
5 none of these
Explanation:
C \(\mathrm{m}_{1}=16\) (Oxygen mass number) \(\mathrm{m}_{2}=12\) (carbon mass number) \(\mathrm{m}_{2} \mathrm{x}_{2}=\mathrm{m}_{1} \mathrm{x}_{1}\) \(12 \mathrm{x}_{2}=16\left(1.130 \times 10^{-10}-\mathrm{x}_{2}\right)\) \(12 \mathrm{x}_{2}=16 \times 1.130 \times 10^{-10}-16 \mathrm{x}_{2}\) \(\mathrm{x}_{2}=\frac{16 \times 1.13 \times 10^{-10}}{28}\) \(\mathrm{x}_{2}=0.6457 \AA\) Hence, the centre of mass (COM) of molecule relative to the carbon atom is \(0.6457{ }^{\circ} \mathrm{A}\).
149709
In a two-particle system with particle masses \(m_{1}\) and \(m_{2}\), the first particle is pushed towards the centre of mass through a distance \(d\), the distance through which second particle must be moved to keep the centre of mass at the same position is
E \(\mathrm{x}_{\mathrm{cm}}=\frac{\mathrm{m}_{1}(0)+\mathrm{m}_{2}(l)}{\mathrm{m}_{1}+\mathrm{m}_{2}}\) \(\mathrm{x}_{\mathrm{cm}}=\frac{\mathrm{m}_{2} l}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\) \(\mathrm{x}_{\mathrm{cm}}=\frac{\mathrm{m}_{1}(\mathrm{~d})+\mathrm{m}_{2}(l-\mathrm{x})}{\mathrm{m}_{1}+\mathrm{m}_{2}}\) On equating equation (i) \& (ii), we get \(\frac{\mathrm{m}_{1} \mathrm{~d}+\mathrm{m}_{2}(l-\mathrm{x})}{\mathrm{m}_{1}+\mathrm{m}_{2}}=\frac{\mathrm{m}_{2} l}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\) \(\mathrm{~m}_{1} \mathrm{~d}+\mathrm{m}_{2} l-\mathrm{m}_{2} \mathrm{x}=\mathrm{m}_{2} l\) \(\mathrm{~m}_{1} \mathrm{~d}=\mathrm{m}_{2} \mathrm{x} \Rightarrow \mathrm{x}=\frac{\mathrm{m}_{1} \mathrm{~d}}{\mathrm{~m}_{2}}\)
Kerala CEE- 2013
Rotational Motion
149710
A system consists of 3 particles each of mass \(m\) located at points \((1,1)(2,2)\) and \((3,3)\). The coordinates of the centre of mass are
1 \((6,6)\)
2 \((3,3)\)
3 \((1,1)\)
4 \((2,2)\)
5 \((5,5)\)
Explanation:
D Given, \(\mathrm{m}_{1}=\mathrm{m}_{2}=\mathrm{m}_{3}=\mathrm{m}\) \(\mathrm{x}_{1}=1, \mathrm{x}_{2}=2\) and \(\mathrm{x}_{3}=3\) \(\mathrm{y}_{1}=1, \mathrm{y}_{2}=2\), and \(\mathrm{y}_{3}=3\) \(\mathrm{x}_{\mathrm{cm}}=\frac{\mathrm{m}_{1} \mathrm{x}_{1}+\mathrm{m}_{2} \mathrm{x}_{2}+\mathrm{m}_{3} \mathrm{x}_{3}}{\mathrm{~m}_{1}+\mathrm{m}_{2}+\mathrm{m}_{3}}\) \(\mathrm{x}_{\mathrm{cm}}=\frac{\mathrm{mx}_{1}+\mathrm{mx}_{2}+\mathrm{mx}_{3}}{\mathrm{~m}+\mathrm{m}+\mathrm{m}}=\frac{\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}}{3}\) \(=\frac{1+2+3}{3}=\frac{6}{3}=2\) \(\mathrm{y}_{\mathrm{cm}}=\frac{\mathrm{my}_{1}+\mathrm{my}_{2}+\mathrm{my}_{3}}{\mathrm{~m}+\mathrm{m}+\mathrm{m}}=\frac{\mathrm{y}_{1}+\mathrm{y}_{2}+\mathrm{y}_{3}}{3}\) \(=\frac{1+2+3}{3}=\frac{6}{3}=2\) Hence, the coordinates of the centre of mass are \((2,2)\).
BITSAT-2010
Rotational Motion
149711
Three identical spheres, each of mass \(3 \mathrm{~kg}\) are placed touching each other with their centres lying on a straight line. The centres of the sphere are marked at \(P, Q\) and \(R\) respectively. The distance of centre of mass of system from \(P\) is
B \lt smiles>[R]c1cccc2cccc([R])c12 \lt /smiles> The centre of mass of the given system consisting of three identical spheres of mass \(3 \mathrm{~kg}\) each is \(x_{c m}=\frac{M_{P}\left(x_{P}\right)+M_{Q}\left(x_{Q}\right)+M_{R}\left(x_{R}\right)}{M_{P}+M_{Q}+M_{R}}\) \(\mathrm{x}_{\mathrm{cm}}=\frac{\mathrm{M}(0)+\mathrm{M}(\mathrm{PQ})+\mathrm{M}(\mathrm{PR})}{\mathrm{M}+\mathrm{M}+\mathrm{M}}\) \(=\frac{\mathrm{M}(\mathrm{PQ}+\mathrm{PR})}{3 \mathrm{M}}\) \(\mathrm{x}_{\mathrm{cm}}=\frac{\mathrm{PQ}+\mathrm{PR}}{3}\) Hence, COM of the given system is \(\frac{P Q+P R}{3}\)
Kerala CEE - 2008
Rotational Motion
149712
The distance between the centers of carbon and oxygen atoms in the carbon monoxide molecule is \(1.130 \AA\). Locate the centre of mass of the molecule relative to the carbon atom :
1 \(5.428 \AA\)
2 \(1.130 \AA\)
3 \(0.6457 \AA\)
4 \(0.3260 \AA\)
5 none of these
Explanation:
C \(\mathrm{m}_{1}=16\) (Oxygen mass number) \(\mathrm{m}_{2}=12\) (carbon mass number) \(\mathrm{m}_{2} \mathrm{x}_{2}=\mathrm{m}_{1} \mathrm{x}_{1}\) \(12 \mathrm{x}_{2}=16\left(1.130 \times 10^{-10}-\mathrm{x}_{2}\right)\) \(12 \mathrm{x}_{2}=16 \times 1.130 \times 10^{-10}-16 \mathrm{x}_{2}\) \(\mathrm{x}_{2}=\frac{16 \times 1.13 \times 10^{-10}}{28}\) \(\mathrm{x}_{2}=0.6457 \AA\) Hence, the centre of mass (COM) of molecule relative to the carbon atom is \(0.6457{ }^{\circ} \mathrm{A}\).
