149693
In the figure shown, the blocks have equal masses. Friction, mass of the string and the mass of the pulley are negligible. The magnitude of the acceleration of the centre of mass of the two blocks is (Acceleration due to gravity \(=\underline{g}\) ).
D \(\operatorname{mg} \sin 60^{\circ}\) Force equation for both the blocks \(\mathrm{T}-\mathrm{mg} \sin 30^{\circ}=\mathrm{ma}\) \(\mathrm{T}-\frac{\mathrm{mg}}{2}=\mathrm{ma}\) \(\mathrm{mg} \sin 60^{\circ}-\mathrm{T}=\mathrm{ma}\) \(\frac{\sqrt{3} \mathrm{mg}}{2}-\mathrm{T}=\mathrm{ma}\) Adding eq \({ }^{\mathrm{n}}\) (i) and (ii) we get, \(\mathrm{a}=\left(\frac{\sqrt{3}-1}{4}\right) \mathrm{g}\) \(\therefore \quad\) Magnitude of the acceleration of centre of mass \(\mathrm{a}_{\mathrm{CM}}=\left|\frac{\mathrm{ma}_{1}+m \mathrm{a}_{2}}{\mathrm{~m}+\mathrm{m}}\right|\) \(\mathrm{a}_{\mathrm{CM}}=\left|\frac{\frac{(\sqrt{3}-1) \mathrm{g}}{4} \hat{\mathrm{i}}+\frac{(\sqrt{3}-1) \mathrm{g}}{4} \hat{\mathrm{j}}}{2}\right|\) \(=\frac{\mathrm{g}}{2} \sqrt{\left(\frac{\sqrt{3}-1}{4}\right)^{2}+\left(\frac{\sqrt{3}-1}{4}\right)^{2}}\) \(=\frac{\mathrm{g}}{2} \times \sqrt{2} \times \frac{\sqrt{3}-1}{4}\) \(\mathrm{a}_{\mathrm{CM}}=\left(\frac{\sqrt{3}-1}{4 \sqrt{2}) \mathrm{g}}\right.\)
AP EAMCET (23.04.2019) Shift-I
Rotational Motion
149694
Three identical spheres each of diameter \(2 \sqrt{3} \mathrm{~m}\) are kept on a horizontal surface such that each sphere touches the other two spheres. If one of the sphere is removed, then the shift in the position of the centre of mass of the system is
1 \(12 \mathrm{~m}\)
2 \(1 \mathrm{~m}\)
3 \(2 \mathrm{~m}\)
4 \(\frac{3}{2} \mathrm{~m}\)
Explanation:
B The centre of mass of the spheres is given by \(\mathrm{x}_{\mathrm{CM}}=\frac{\mathrm{m}_{1} \mathrm{x}_{1}+\mathrm{m}_{2} \mathrm{x}_{2}+\mathrm{m}_{3} \mathrm{x}_{3}}{\mathrm{~m}_{1}+\mathrm{m}_{2}+\mathrm{m}_{3}}\) As the spheres are identical \(\mathrm{x}_{\mathrm{CM}} =\frac{\mathrm{m}}{3 \mathrm{~m}}(0+2 \sqrt{3}+\sqrt{3})\) \(=\frac{3 \sqrt{3}}{3}=\sqrt{3}\) Similarly \(\mathrm{y}_{\mathrm{CM}} =\frac{\mathrm{m}}{3 \mathrm{~m}}\left(\mathrm{y}_{1}+\mathrm{y}_{2}+\mathrm{y}_{3}\right)\) \(=\frac{0+0+3}{3}=1\) Center of mass \(\mathrm{c}_{\mathrm{CM}}=(\sqrt{3}, 1)\) Hence, the centre of mass shifted by \(1 \mathrm{~m}\) in the \(y\) direction.
AP EAMCET (21.04.2019) Shift-I
Rotational Motion
149695
A semicircular plate of mass \(m\) has radius \(r\) and centre \(C\). The centre of mass of the plate is at a distance \(x\) from its centre \(C\). Its moment of inertia about an axis passing through its centre of mass and perpendicular to its plane is
1 \(\frac{m r^{2}}{2}\)
2 \(\frac{\mathrm{mr}^{2}}{4}\)
3 \(\frac{\mathrm{mr}^{2}}{2}+\mathrm{mx}^{2}\)
4 \(\frac{m^{2}}{2}-m x^{2}\)
Explanation:
D From the parallel axis theorem, \(\mathrm{I}_{\mathrm{c}}=\mathrm{I}_{\mathrm{cm}}+\mathrm{mx}^{2}\) \(\mathrm{I}_{\mathrm{cm}}=\mathrm{I}_{\mathrm{c}}-\mathrm{mx}^{2}\) We know, moment of inertia of semicircle \(\left(I_{c}\right)=\frac{m^{2}}{2}\) Hence, moment of inertia of semicircular plate about an axis passing through its centre of mass and perpendicular to its plane is \(\mathrm{I}_{\mathrm{cm}}=\frac{\mathrm{mr}^{2}}{2}-\mathrm{mx}^{2}\)
AP EAMCET (20.04.2019) Shift-1
Rotational Motion
149692
Assertion: The centre of mass of a proton and an electron, released from their respective positions remains at rest. Reason: The centre of mass remains at rest, if no external force is applied.
1 If both assertion and reason are true and reason is the correct explanation of assertion.
2 If both assertion and reason are true but reason is not the correct explanation of assertion.
3 If assertion is true but reason is false.
4 If both assertion and reason are false.
Explanation:
A Initially the electron and proton were at rest so their centre of mass will be rest. When they move toward each other under mutual attraction then centre of mass remains unaffected because external force on the system is zero.
149693
In the figure shown, the blocks have equal masses. Friction, mass of the string and the mass of the pulley are negligible. The magnitude of the acceleration of the centre of mass of the two blocks is (Acceleration due to gravity \(=\underline{g}\) ).
D \(\operatorname{mg} \sin 60^{\circ}\) Force equation for both the blocks \(\mathrm{T}-\mathrm{mg} \sin 30^{\circ}=\mathrm{ma}\) \(\mathrm{T}-\frac{\mathrm{mg}}{2}=\mathrm{ma}\) \(\mathrm{mg} \sin 60^{\circ}-\mathrm{T}=\mathrm{ma}\) \(\frac{\sqrt{3} \mathrm{mg}}{2}-\mathrm{T}=\mathrm{ma}\) Adding eq \({ }^{\mathrm{n}}\) (i) and (ii) we get, \(\mathrm{a}=\left(\frac{\sqrt{3}-1}{4}\right) \mathrm{g}\) \(\therefore \quad\) Magnitude of the acceleration of centre of mass \(\mathrm{a}_{\mathrm{CM}}=\left|\frac{\mathrm{ma}_{1}+m \mathrm{a}_{2}}{\mathrm{~m}+\mathrm{m}}\right|\) \(\mathrm{a}_{\mathrm{CM}}=\left|\frac{\frac{(\sqrt{3}-1) \mathrm{g}}{4} \hat{\mathrm{i}}+\frac{(\sqrt{3}-1) \mathrm{g}}{4} \hat{\mathrm{j}}}{2}\right|\) \(=\frac{\mathrm{g}}{2} \sqrt{\left(\frac{\sqrt{3}-1}{4}\right)^{2}+\left(\frac{\sqrt{3}-1}{4}\right)^{2}}\) \(=\frac{\mathrm{g}}{2} \times \sqrt{2} \times \frac{\sqrt{3}-1}{4}\) \(\mathrm{a}_{\mathrm{CM}}=\left(\frac{\sqrt{3}-1}{4 \sqrt{2}) \mathrm{g}}\right.\)
AP EAMCET (23.04.2019) Shift-I
Rotational Motion
149694
Three identical spheres each of diameter \(2 \sqrt{3} \mathrm{~m}\) are kept on a horizontal surface such that each sphere touches the other two spheres. If one of the sphere is removed, then the shift in the position of the centre of mass of the system is
1 \(12 \mathrm{~m}\)
2 \(1 \mathrm{~m}\)
3 \(2 \mathrm{~m}\)
4 \(\frac{3}{2} \mathrm{~m}\)
Explanation:
B The centre of mass of the spheres is given by \(\mathrm{x}_{\mathrm{CM}}=\frac{\mathrm{m}_{1} \mathrm{x}_{1}+\mathrm{m}_{2} \mathrm{x}_{2}+\mathrm{m}_{3} \mathrm{x}_{3}}{\mathrm{~m}_{1}+\mathrm{m}_{2}+\mathrm{m}_{3}}\) As the spheres are identical \(\mathrm{x}_{\mathrm{CM}} =\frac{\mathrm{m}}{3 \mathrm{~m}}(0+2 \sqrt{3}+\sqrt{3})\) \(=\frac{3 \sqrt{3}}{3}=\sqrt{3}\) Similarly \(\mathrm{y}_{\mathrm{CM}} =\frac{\mathrm{m}}{3 \mathrm{~m}}\left(\mathrm{y}_{1}+\mathrm{y}_{2}+\mathrm{y}_{3}\right)\) \(=\frac{0+0+3}{3}=1\) Center of mass \(\mathrm{c}_{\mathrm{CM}}=(\sqrt{3}, 1)\) Hence, the centre of mass shifted by \(1 \mathrm{~m}\) in the \(y\) direction.
AP EAMCET (21.04.2019) Shift-I
Rotational Motion
149695
A semicircular plate of mass \(m\) has radius \(r\) and centre \(C\). The centre of mass of the plate is at a distance \(x\) from its centre \(C\). Its moment of inertia about an axis passing through its centre of mass and perpendicular to its plane is
1 \(\frac{m r^{2}}{2}\)
2 \(\frac{\mathrm{mr}^{2}}{4}\)
3 \(\frac{\mathrm{mr}^{2}}{2}+\mathrm{mx}^{2}\)
4 \(\frac{m^{2}}{2}-m x^{2}\)
Explanation:
D From the parallel axis theorem, \(\mathrm{I}_{\mathrm{c}}=\mathrm{I}_{\mathrm{cm}}+\mathrm{mx}^{2}\) \(\mathrm{I}_{\mathrm{cm}}=\mathrm{I}_{\mathrm{c}}-\mathrm{mx}^{2}\) We know, moment of inertia of semicircle \(\left(I_{c}\right)=\frac{m^{2}}{2}\) Hence, moment of inertia of semicircular plate about an axis passing through its centre of mass and perpendicular to its plane is \(\mathrm{I}_{\mathrm{cm}}=\frac{\mathrm{mr}^{2}}{2}-\mathrm{mx}^{2}\)
AP EAMCET (20.04.2019) Shift-1
Rotational Motion
149692
Assertion: The centre of mass of a proton and an electron, released from their respective positions remains at rest. Reason: The centre of mass remains at rest, if no external force is applied.
1 If both assertion and reason are true and reason is the correct explanation of assertion.
2 If both assertion and reason are true but reason is not the correct explanation of assertion.
3 If assertion is true but reason is false.
4 If both assertion and reason are false.
Explanation:
A Initially the electron and proton were at rest so their centre of mass will be rest. When they move toward each other under mutual attraction then centre of mass remains unaffected because external force on the system is zero.
149693
In the figure shown, the blocks have equal masses. Friction, mass of the string and the mass of the pulley are negligible. The magnitude of the acceleration of the centre of mass of the two blocks is (Acceleration due to gravity \(=\underline{g}\) ).
D \(\operatorname{mg} \sin 60^{\circ}\) Force equation for both the blocks \(\mathrm{T}-\mathrm{mg} \sin 30^{\circ}=\mathrm{ma}\) \(\mathrm{T}-\frac{\mathrm{mg}}{2}=\mathrm{ma}\) \(\mathrm{mg} \sin 60^{\circ}-\mathrm{T}=\mathrm{ma}\) \(\frac{\sqrt{3} \mathrm{mg}}{2}-\mathrm{T}=\mathrm{ma}\) Adding eq \({ }^{\mathrm{n}}\) (i) and (ii) we get, \(\mathrm{a}=\left(\frac{\sqrt{3}-1}{4}\right) \mathrm{g}\) \(\therefore \quad\) Magnitude of the acceleration of centre of mass \(\mathrm{a}_{\mathrm{CM}}=\left|\frac{\mathrm{ma}_{1}+m \mathrm{a}_{2}}{\mathrm{~m}+\mathrm{m}}\right|\) \(\mathrm{a}_{\mathrm{CM}}=\left|\frac{\frac{(\sqrt{3}-1) \mathrm{g}}{4} \hat{\mathrm{i}}+\frac{(\sqrt{3}-1) \mathrm{g}}{4} \hat{\mathrm{j}}}{2}\right|\) \(=\frac{\mathrm{g}}{2} \sqrt{\left(\frac{\sqrt{3}-1}{4}\right)^{2}+\left(\frac{\sqrt{3}-1}{4}\right)^{2}}\) \(=\frac{\mathrm{g}}{2} \times \sqrt{2} \times \frac{\sqrt{3}-1}{4}\) \(\mathrm{a}_{\mathrm{CM}}=\left(\frac{\sqrt{3}-1}{4 \sqrt{2}) \mathrm{g}}\right.\)
AP EAMCET (23.04.2019) Shift-I
Rotational Motion
149694
Three identical spheres each of diameter \(2 \sqrt{3} \mathrm{~m}\) are kept on a horizontal surface such that each sphere touches the other two spheres. If one of the sphere is removed, then the shift in the position of the centre of mass of the system is
1 \(12 \mathrm{~m}\)
2 \(1 \mathrm{~m}\)
3 \(2 \mathrm{~m}\)
4 \(\frac{3}{2} \mathrm{~m}\)
Explanation:
B The centre of mass of the spheres is given by \(\mathrm{x}_{\mathrm{CM}}=\frac{\mathrm{m}_{1} \mathrm{x}_{1}+\mathrm{m}_{2} \mathrm{x}_{2}+\mathrm{m}_{3} \mathrm{x}_{3}}{\mathrm{~m}_{1}+\mathrm{m}_{2}+\mathrm{m}_{3}}\) As the spheres are identical \(\mathrm{x}_{\mathrm{CM}} =\frac{\mathrm{m}}{3 \mathrm{~m}}(0+2 \sqrt{3}+\sqrt{3})\) \(=\frac{3 \sqrt{3}}{3}=\sqrt{3}\) Similarly \(\mathrm{y}_{\mathrm{CM}} =\frac{\mathrm{m}}{3 \mathrm{~m}}\left(\mathrm{y}_{1}+\mathrm{y}_{2}+\mathrm{y}_{3}\right)\) \(=\frac{0+0+3}{3}=1\) Center of mass \(\mathrm{c}_{\mathrm{CM}}=(\sqrt{3}, 1)\) Hence, the centre of mass shifted by \(1 \mathrm{~m}\) in the \(y\) direction.
AP EAMCET (21.04.2019) Shift-I
Rotational Motion
149695
A semicircular plate of mass \(m\) has radius \(r\) and centre \(C\). The centre of mass of the plate is at a distance \(x\) from its centre \(C\). Its moment of inertia about an axis passing through its centre of mass and perpendicular to its plane is
1 \(\frac{m r^{2}}{2}\)
2 \(\frac{\mathrm{mr}^{2}}{4}\)
3 \(\frac{\mathrm{mr}^{2}}{2}+\mathrm{mx}^{2}\)
4 \(\frac{m^{2}}{2}-m x^{2}\)
Explanation:
D From the parallel axis theorem, \(\mathrm{I}_{\mathrm{c}}=\mathrm{I}_{\mathrm{cm}}+\mathrm{mx}^{2}\) \(\mathrm{I}_{\mathrm{cm}}=\mathrm{I}_{\mathrm{c}}-\mathrm{mx}^{2}\) We know, moment of inertia of semicircle \(\left(I_{c}\right)=\frac{m^{2}}{2}\) Hence, moment of inertia of semicircular plate about an axis passing through its centre of mass and perpendicular to its plane is \(\mathrm{I}_{\mathrm{cm}}=\frac{\mathrm{mr}^{2}}{2}-\mathrm{mx}^{2}\)
AP EAMCET (20.04.2019) Shift-1
Rotational Motion
149692
Assertion: The centre of mass of a proton and an electron, released from their respective positions remains at rest. Reason: The centre of mass remains at rest, if no external force is applied.
1 If both assertion and reason are true and reason is the correct explanation of assertion.
2 If both assertion and reason are true but reason is not the correct explanation of assertion.
3 If assertion is true but reason is false.
4 If both assertion and reason are false.
Explanation:
A Initially the electron and proton were at rest so their centre of mass will be rest. When they move toward each other under mutual attraction then centre of mass remains unaffected because external force on the system is zero.
149693
In the figure shown, the blocks have equal masses. Friction, mass of the string and the mass of the pulley are negligible. The magnitude of the acceleration of the centre of mass of the two blocks is (Acceleration due to gravity \(=\underline{g}\) ).
D \(\operatorname{mg} \sin 60^{\circ}\) Force equation for both the blocks \(\mathrm{T}-\mathrm{mg} \sin 30^{\circ}=\mathrm{ma}\) \(\mathrm{T}-\frac{\mathrm{mg}}{2}=\mathrm{ma}\) \(\mathrm{mg} \sin 60^{\circ}-\mathrm{T}=\mathrm{ma}\) \(\frac{\sqrt{3} \mathrm{mg}}{2}-\mathrm{T}=\mathrm{ma}\) Adding eq \({ }^{\mathrm{n}}\) (i) and (ii) we get, \(\mathrm{a}=\left(\frac{\sqrt{3}-1}{4}\right) \mathrm{g}\) \(\therefore \quad\) Magnitude of the acceleration of centre of mass \(\mathrm{a}_{\mathrm{CM}}=\left|\frac{\mathrm{ma}_{1}+m \mathrm{a}_{2}}{\mathrm{~m}+\mathrm{m}}\right|\) \(\mathrm{a}_{\mathrm{CM}}=\left|\frac{\frac{(\sqrt{3}-1) \mathrm{g}}{4} \hat{\mathrm{i}}+\frac{(\sqrt{3}-1) \mathrm{g}}{4} \hat{\mathrm{j}}}{2}\right|\) \(=\frac{\mathrm{g}}{2} \sqrt{\left(\frac{\sqrt{3}-1}{4}\right)^{2}+\left(\frac{\sqrt{3}-1}{4}\right)^{2}}\) \(=\frac{\mathrm{g}}{2} \times \sqrt{2} \times \frac{\sqrt{3}-1}{4}\) \(\mathrm{a}_{\mathrm{CM}}=\left(\frac{\sqrt{3}-1}{4 \sqrt{2}) \mathrm{g}}\right.\)
AP EAMCET (23.04.2019) Shift-I
Rotational Motion
149694
Three identical spheres each of diameter \(2 \sqrt{3} \mathrm{~m}\) are kept on a horizontal surface such that each sphere touches the other two spheres. If one of the sphere is removed, then the shift in the position of the centre of mass of the system is
1 \(12 \mathrm{~m}\)
2 \(1 \mathrm{~m}\)
3 \(2 \mathrm{~m}\)
4 \(\frac{3}{2} \mathrm{~m}\)
Explanation:
B The centre of mass of the spheres is given by \(\mathrm{x}_{\mathrm{CM}}=\frac{\mathrm{m}_{1} \mathrm{x}_{1}+\mathrm{m}_{2} \mathrm{x}_{2}+\mathrm{m}_{3} \mathrm{x}_{3}}{\mathrm{~m}_{1}+\mathrm{m}_{2}+\mathrm{m}_{3}}\) As the spheres are identical \(\mathrm{x}_{\mathrm{CM}} =\frac{\mathrm{m}}{3 \mathrm{~m}}(0+2 \sqrt{3}+\sqrt{3})\) \(=\frac{3 \sqrt{3}}{3}=\sqrt{3}\) Similarly \(\mathrm{y}_{\mathrm{CM}} =\frac{\mathrm{m}}{3 \mathrm{~m}}\left(\mathrm{y}_{1}+\mathrm{y}_{2}+\mathrm{y}_{3}\right)\) \(=\frac{0+0+3}{3}=1\) Center of mass \(\mathrm{c}_{\mathrm{CM}}=(\sqrt{3}, 1)\) Hence, the centre of mass shifted by \(1 \mathrm{~m}\) in the \(y\) direction.
AP EAMCET (21.04.2019) Shift-I
Rotational Motion
149695
A semicircular plate of mass \(m\) has radius \(r\) and centre \(C\). The centre of mass of the plate is at a distance \(x\) from its centre \(C\). Its moment of inertia about an axis passing through its centre of mass and perpendicular to its plane is
1 \(\frac{m r^{2}}{2}\)
2 \(\frac{\mathrm{mr}^{2}}{4}\)
3 \(\frac{\mathrm{mr}^{2}}{2}+\mathrm{mx}^{2}\)
4 \(\frac{m^{2}}{2}-m x^{2}\)
Explanation:
D From the parallel axis theorem, \(\mathrm{I}_{\mathrm{c}}=\mathrm{I}_{\mathrm{cm}}+\mathrm{mx}^{2}\) \(\mathrm{I}_{\mathrm{cm}}=\mathrm{I}_{\mathrm{c}}-\mathrm{mx}^{2}\) We know, moment of inertia of semicircle \(\left(I_{c}\right)=\frac{m^{2}}{2}\) Hence, moment of inertia of semicircular plate about an axis passing through its centre of mass and perpendicular to its plane is \(\mathrm{I}_{\mathrm{cm}}=\frac{\mathrm{mr}^{2}}{2}-\mathrm{mx}^{2}\)
AP EAMCET (20.04.2019) Shift-1
Rotational Motion
149692
Assertion: The centre of mass of a proton and an electron, released from their respective positions remains at rest. Reason: The centre of mass remains at rest, if no external force is applied.
1 If both assertion and reason are true and reason is the correct explanation of assertion.
2 If both assertion and reason are true but reason is not the correct explanation of assertion.
3 If assertion is true but reason is false.
4 If both assertion and reason are false.
Explanation:
A Initially the electron and proton were at rest so their centre of mass will be rest. When they move toward each other under mutual attraction then centre of mass remains unaffected because external force on the system is zero.
