NEET Test Series from KOTA - 10 Papers In MS WORD
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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. \(\mathrm{Mx}_{1}=\mathrm{mx}_{2}\) \(\frac{x_{1}}{x_{2}}=\frac{m}{M}\) \(\because \quad \mathrm{m} \lt \mathrm{M}\) \(\therefore \quad \frac{\mathrm{x}_{1}}{\mathrm{x}_{2}} \lt 1 \Rightarrow \mathrm{x}_{1} \lt \mathrm{x}_{2}\) 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, \(\overrightarrow{\mathrm{a}}_{\mathrm{cm}}=\frac{\overrightarrow{\mathrm{F}}_{\mathrm{ext}}}{\mathrm{M}}\) If \(\quad \overrightarrow{\mathrm{F}}_{\mathrm{ext}}=0\) Thus, \(\overrightarrow{\mathrm{a}}_{\mathrm{cm}}=0\) Thus, when no external force acts on a system of particles, it will move continuously with its initial velocity. Now, \(\frac{\mathrm{d} \overrightarrow{\mathrm{P}}}{\mathrm{dt}} =\overrightarrow{\mathrm{F}}_{\mathrm{ext}}\) \(\therefore \quad \frac{\mathrm{d} \overrightarrow{\mathrm{P}}}{\mathrm{dt}} =0 \quad\left(\because \overrightarrow{\mathrm{F}}_{\mathrm{ext}}=0\right)\) 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.
Explanation:
C The centre of mass is the point where mass distribution is uniform in all direction. The position of centre of mass of a body depends on shape, size and distribution of the mass of the body. Centre of mass of a body does not always lie on the centre of body. Only centre of mass of a uniform body lies always at geometric centre of the body. Therefore correct answer is option (c).
AIIMS-2009
Rotational Motion
149724
If linear density of a rod of length \(3 \mathrm{~m}\) varies as \(\lambda=2+x\), then the position of the centre of gravity of the rod is :
1 \(\frac{7}{3} \mathrm{~m}\)
2 \(\frac{12}{7} \mathrm{~m}\)
3 \(\frac{10}{7} \mathrm{~m}\)
4 \(\frac{9}{7} \mathrm{~m}\)
Explanation:
B Given, length of \(\operatorname{rod}=3 \mathrm{~m}, \lambda=2+\mathrm{x}\) \((0,0)\) Linear density of rod varies with distance \(\frac{\mathrm{dm}}{\mathrm{dx}}=\lambda\) \(\mathrm{dm}=\lambda \mathrm{dx}\) Position of center of mass, \(\mathrm{x}_{\mathrm{CM}}=\frac{\int \mathrm{dm} \times \mathrm{x}}{\int \mathrm{dm}}\) Putting the value of \(\mathrm{dm}\) from equation (i), we get \(=\frac{\int_{0}^{3}(\lambda d x) x}{\int_{0}^{3} \lambda d x}=\frac{\int_{0}^{3}(2+x) x d x}{\int_{0}^{3}(2+x) d x}\) \(=\frac{\left[x^{2}+\frac{x^{3}}{3}\right]_{0}^{3}}{\left[2 x+\frac{x^{2}}{2}\right]_{0}^{3}}\) \(=\frac{9+\frac{27}{3}}{6+\frac{9}{2}}=\frac{12}{7}\) \(x_{C M} =\frac{12}{7} \mathrm{~m}\)
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. \(\mathrm{Mx}_{1}=\mathrm{mx}_{2}\) \(\frac{x_{1}}{x_{2}}=\frac{m}{M}\) \(\because \quad \mathrm{m} \lt \mathrm{M}\) \(\therefore \quad \frac{\mathrm{x}_{1}}{\mathrm{x}_{2}} \lt 1 \Rightarrow \mathrm{x}_{1} \lt \mathrm{x}_{2}\) 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, \(\overrightarrow{\mathrm{a}}_{\mathrm{cm}}=\frac{\overrightarrow{\mathrm{F}}_{\mathrm{ext}}}{\mathrm{M}}\) If \(\quad \overrightarrow{\mathrm{F}}_{\mathrm{ext}}=0\) Thus, \(\overrightarrow{\mathrm{a}}_{\mathrm{cm}}=0\) Thus, when no external force acts on a system of particles, it will move continuously with its initial velocity. Now, \(\frac{\mathrm{d} \overrightarrow{\mathrm{P}}}{\mathrm{dt}} =\overrightarrow{\mathrm{F}}_{\mathrm{ext}}\) \(\therefore \quad \frac{\mathrm{d} \overrightarrow{\mathrm{P}}}{\mathrm{dt}} =0 \quad\left(\because \overrightarrow{\mathrm{F}}_{\mathrm{ext}}=0\right)\) 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.
Explanation:
C The centre of mass is the point where mass distribution is uniform in all direction. The position of centre of mass of a body depends on shape, size and distribution of the mass of the body. Centre of mass of a body does not always lie on the centre of body. Only centre of mass of a uniform body lies always at geometric centre of the body. Therefore correct answer is option (c).
AIIMS-2009
Rotational Motion
149724
If linear density of a rod of length \(3 \mathrm{~m}\) varies as \(\lambda=2+x\), then the position of the centre of gravity of the rod is :
1 \(\frac{7}{3} \mathrm{~m}\)
2 \(\frac{12}{7} \mathrm{~m}\)
3 \(\frac{10}{7} \mathrm{~m}\)
4 \(\frac{9}{7} \mathrm{~m}\)
Explanation:
B Given, length of \(\operatorname{rod}=3 \mathrm{~m}, \lambda=2+\mathrm{x}\) \((0,0)\) Linear density of rod varies with distance \(\frac{\mathrm{dm}}{\mathrm{dx}}=\lambda\) \(\mathrm{dm}=\lambda \mathrm{dx}\) Position of center of mass, \(\mathrm{x}_{\mathrm{CM}}=\frac{\int \mathrm{dm} \times \mathrm{x}}{\int \mathrm{dm}}\) Putting the value of \(\mathrm{dm}\) from equation (i), we get \(=\frac{\int_{0}^{3}(\lambda d x) x}{\int_{0}^{3} \lambda d x}=\frac{\int_{0}^{3}(2+x) x d x}{\int_{0}^{3}(2+x) d x}\) \(=\frac{\left[x^{2}+\frac{x^{3}}{3}\right]_{0}^{3}}{\left[2 x+\frac{x^{2}}{2}\right]_{0}^{3}}\) \(=\frac{9+\frac{27}{3}}{6+\frac{9}{2}}=\frac{12}{7}\) \(x_{C M} =\frac{12}{7} \mathrm{~m}\)
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. \(\mathrm{Mx}_{1}=\mathrm{mx}_{2}\) \(\frac{x_{1}}{x_{2}}=\frac{m}{M}\) \(\because \quad \mathrm{m} \lt \mathrm{M}\) \(\therefore \quad \frac{\mathrm{x}_{1}}{\mathrm{x}_{2}} \lt 1 \Rightarrow \mathrm{x}_{1} \lt \mathrm{x}_{2}\) 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, \(\overrightarrow{\mathrm{a}}_{\mathrm{cm}}=\frac{\overrightarrow{\mathrm{F}}_{\mathrm{ext}}}{\mathrm{M}}\) If \(\quad \overrightarrow{\mathrm{F}}_{\mathrm{ext}}=0\) Thus, \(\overrightarrow{\mathrm{a}}_{\mathrm{cm}}=0\) Thus, when no external force acts on a system of particles, it will move continuously with its initial velocity. Now, \(\frac{\mathrm{d} \overrightarrow{\mathrm{P}}}{\mathrm{dt}} =\overrightarrow{\mathrm{F}}_{\mathrm{ext}}\) \(\therefore \quad \frac{\mathrm{d} \overrightarrow{\mathrm{P}}}{\mathrm{dt}} =0 \quad\left(\because \overrightarrow{\mathrm{F}}_{\mathrm{ext}}=0\right)\) 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.
Explanation:
C The centre of mass is the point where mass distribution is uniform in all direction. The position of centre of mass of a body depends on shape, size and distribution of the mass of the body. Centre of mass of a body does not always lie on the centre of body. Only centre of mass of a uniform body lies always at geometric centre of the body. Therefore correct answer is option (c).
AIIMS-2009
Rotational Motion
149724
If linear density of a rod of length \(3 \mathrm{~m}\) varies as \(\lambda=2+x\), then the position of the centre of gravity of the rod is :
1 \(\frac{7}{3} \mathrm{~m}\)
2 \(\frac{12}{7} \mathrm{~m}\)
3 \(\frac{10}{7} \mathrm{~m}\)
4 \(\frac{9}{7} \mathrm{~m}\)
Explanation:
B Given, length of \(\operatorname{rod}=3 \mathrm{~m}, \lambda=2+\mathrm{x}\) \((0,0)\) Linear density of rod varies with distance \(\frac{\mathrm{dm}}{\mathrm{dx}}=\lambda\) \(\mathrm{dm}=\lambda \mathrm{dx}\) Position of center of mass, \(\mathrm{x}_{\mathrm{CM}}=\frac{\int \mathrm{dm} \times \mathrm{x}}{\int \mathrm{dm}}\) Putting the value of \(\mathrm{dm}\) from equation (i), we get \(=\frac{\int_{0}^{3}(\lambda d x) x}{\int_{0}^{3} \lambda d x}=\frac{\int_{0}^{3}(2+x) x d x}{\int_{0}^{3}(2+x) d x}\) \(=\frac{\left[x^{2}+\frac{x^{3}}{3}\right]_{0}^{3}}{\left[2 x+\frac{x^{2}}{2}\right]_{0}^{3}}\) \(=\frac{9+\frac{27}{3}}{6+\frac{9}{2}}=\frac{12}{7}\) \(x_{C M} =\frac{12}{7} \mathrm{~m}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
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. \(\mathrm{Mx}_{1}=\mathrm{mx}_{2}\) \(\frac{x_{1}}{x_{2}}=\frac{m}{M}\) \(\because \quad \mathrm{m} \lt \mathrm{M}\) \(\therefore \quad \frac{\mathrm{x}_{1}}{\mathrm{x}_{2}} \lt 1 \Rightarrow \mathrm{x}_{1} \lt \mathrm{x}_{2}\) 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, \(\overrightarrow{\mathrm{a}}_{\mathrm{cm}}=\frac{\overrightarrow{\mathrm{F}}_{\mathrm{ext}}}{\mathrm{M}}\) If \(\quad \overrightarrow{\mathrm{F}}_{\mathrm{ext}}=0\) Thus, \(\overrightarrow{\mathrm{a}}_{\mathrm{cm}}=0\) Thus, when no external force acts on a system of particles, it will move continuously with its initial velocity. Now, \(\frac{\mathrm{d} \overrightarrow{\mathrm{P}}}{\mathrm{dt}} =\overrightarrow{\mathrm{F}}_{\mathrm{ext}}\) \(\therefore \quad \frac{\mathrm{d} \overrightarrow{\mathrm{P}}}{\mathrm{dt}} =0 \quad\left(\because \overrightarrow{\mathrm{F}}_{\mathrm{ext}}=0\right)\) 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.
Explanation:
C The centre of mass is the point where mass distribution is uniform in all direction. The position of centre of mass of a body depends on shape, size and distribution of the mass of the body. Centre of mass of a body does not always lie on the centre of body. Only centre of mass of a uniform body lies always at geometric centre of the body. Therefore correct answer is option (c).
AIIMS-2009
Rotational Motion
149724
If linear density of a rod of length \(3 \mathrm{~m}\) varies as \(\lambda=2+x\), then the position of the centre of gravity of the rod is :
1 \(\frac{7}{3} \mathrm{~m}\)
2 \(\frac{12}{7} \mathrm{~m}\)
3 \(\frac{10}{7} \mathrm{~m}\)
4 \(\frac{9}{7} \mathrm{~m}\)
Explanation:
B Given, length of \(\operatorname{rod}=3 \mathrm{~m}, \lambda=2+\mathrm{x}\) \((0,0)\) Linear density of rod varies with distance \(\frac{\mathrm{dm}}{\mathrm{dx}}=\lambda\) \(\mathrm{dm}=\lambda \mathrm{dx}\) Position of center of mass, \(\mathrm{x}_{\mathrm{CM}}=\frac{\int \mathrm{dm} \times \mathrm{x}}{\int \mathrm{dm}}\) Putting the value of \(\mathrm{dm}\) from equation (i), we get \(=\frac{\int_{0}^{3}(\lambda d x) x}{\int_{0}^{3} \lambda d x}=\frac{\int_{0}^{3}(2+x) x d x}{\int_{0}^{3}(2+x) d x}\) \(=\frac{\left[x^{2}+\frac{x^{3}}{3}\right]_{0}^{3}}{\left[2 x+\frac{x^{2}}{2}\right]_{0}^{3}}\) \(=\frac{9+\frac{27}{3}}{6+\frac{9}{2}}=\frac{12}{7}\) \(x_{C M} =\frac{12}{7} \mathrm{~m}\)
