365878
The radius of gyration of a rotating circular ring is maximum about following axis of rotation
1 Natural axis
2 Axis passing through diameter of ring
3 Axis passing through tangent of ring in its plane
4 Axis passing through tangent of ring perpendicular to plane of ring.
Explanation:
Conceptual Question
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365879
The moment of inertia of a uniform cylinder of length \(\ell\) and radius \(\mathrm{R}\) about its perpendicular bisector is \(\mathrm{I}\). What is the ratio \(\ell / \mathrm{R}\) such that the moment of inertia is minimum?
1 \(\dfrac{\sqrt{3}}{2}\)
2 1
3 \(\sqrt{\dfrac{3}{2}}\)
4 \(\dfrac{3}{\sqrt{2}}\)
Explanation:
Momentum of inertia about perpendicular bisector is \({\rm{I}} = \frac{{m{\ell ^2}}}{{12}} + \frac{{{\rm{m}}{{\rm{R}}^2}}}{4}\) \( \Rightarrow {\rm{I}} = \frac{m}{4}\left( {\frac{{{\ell ^2}}}{3} + {{\rm{R}}^2}} \right)\,\,\,\,\,\,\,\,\,\,(1)\) Also \(m=\pi \mathrm{R}^{2} \ell \rho\) \(\Rightarrow \mathrm{R}^{2}=\dfrac{m}{\pi \ell \rho}\) Put in equation (1) \(\mathrm{I}=\dfrac{m}{4}\left(\dfrac{\ell^{2}}{3}+\dfrac{m}{\pi \ell \rho}\right)\) For maxima & minima \(\begin{aligned}& \dfrac{\mathrm{dI}}{\mathrm{d} \ell}=\dfrac{m}{4}\left(\dfrac{2 \ell}{3}-\dfrac{m}{\pi \ell^{2} \rho}\right)=0 \\& \Rightarrow \dfrac{2 \ell}{3}=\dfrac{m}{\pi \ell^{2} \rho} \Rightarrow \dfrac{2 \ell}{3}=\dfrac{\pi \mathrm{R}^{2} \ell \rho}{\pi \ell^{2} \rho} \\& \text { or } \dfrac{2 \ell}{3}=\dfrac{\mathrm{R}^{2}}{\ell} \Rightarrow \dfrac{\ell^{2}}{\mathrm{R}^{2}}=\dfrac{3}{2} \\& \Rightarrow \dfrac{\ell}{\mathrm{R}}=\sqrt{\dfrac{3}{2}}\end{aligned}\)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365880
A \(T\) joint is formed by two identical rods \(A\) and \(B\) each of mass \(m\) and length \(L\) in \(X Y\) plane as shown. Its moment of inertia about axis passing along the length of \(\operatorname{rod} A\) is
1 \(\frac{{2\,m{L^3}}}{3}\)
2 \(\frac{{m\,{L^2}}}{{12}}\)
3 \(\dfrac{m L^{2}}{6}\)
4 \(\dfrac{m L^{2}}{3}\)
Explanation:
Total amount of inertia of system \(I=I_{A}+I_{B}=0+\dfrac{m L^{2}}{12}\) \(I=\dfrac{m L^{2}}{12}\)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365881
Two circular loops \(A\) and \(B\) of radii \(R\) and \(N R\) respectively are made from a uniform wire. Moment of inertia of \(B\) about its axis is 3 times that of \(A\) about its axis. The value of \(N\) is
1 \({[2]^{\frac{1}{3}}}\)
2 \({[5]^{\frac{1}{3}}}\)
3 \({[3]^{\frac{1}{3}}}\)
4 \({[4]^{\frac{1}{3}}}\)
Explanation:
Let \(\rho=\) mass per unit length of the wire. Mass of loop A, \(m_{A}=2 \pi R \rho\) Mass of loop B, \(m_{B}=2 \pi N R \rho=N m_{A}\) Given, \(\quad I_{B}=3 I_{A}\) \(\Rightarrow m_{B} r_{B}^{2}=3 \times m_{A} r_{A}^{2}\) \(\Rightarrow\left(N m_{A}\right)(N R)^{2}=3 m_{A}(R)^{2}\) \(\Rightarrow N^{3}=3 \Rightarrow N=(3)^{1 / 3}\)
MHTCET - 2020
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365882
Assertion : The moment of inertia of a rigid body reduces to its minimum value, when the axis of rotation passes through its centre of gravity. Reason : The weight of a rigid body always acts through its centre of gravity.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
365878
The radius of gyration of a rotating circular ring is maximum about following axis of rotation
1 Natural axis
2 Axis passing through diameter of ring
3 Axis passing through tangent of ring in its plane
4 Axis passing through tangent of ring perpendicular to plane of ring.
Explanation:
Conceptual Question
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365879
The moment of inertia of a uniform cylinder of length \(\ell\) and radius \(\mathrm{R}\) about its perpendicular bisector is \(\mathrm{I}\). What is the ratio \(\ell / \mathrm{R}\) such that the moment of inertia is minimum?
1 \(\dfrac{\sqrt{3}}{2}\)
2 1
3 \(\sqrt{\dfrac{3}{2}}\)
4 \(\dfrac{3}{\sqrt{2}}\)
Explanation:
Momentum of inertia about perpendicular bisector is \({\rm{I}} = \frac{{m{\ell ^2}}}{{12}} + \frac{{{\rm{m}}{{\rm{R}}^2}}}{4}\) \( \Rightarrow {\rm{I}} = \frac{m}{4}\left( {\frac{{{\ell ^2}}}{3} + {{\rm{R}}^2}} \right)\,\,\,\,\,\,\,\,\,\,(1)\) Also \(m=\pi \mathrm{R}^{2} \ell \rho\) \(\Rightarrow \mathrm{R}^{2}=\dfrac{m}{\pi \ell \rho}\) Put in equation (1) \(\mathrm{I}=\dfrac{m}{4}\left(\dfrac{\ell^{2}}{3}+\dfrac{m}{\pi \ell \rho}\right)\) For maxima & minima \(\begin{aligned}& \dfrac{\mathrm{dI}}{\mathrm{d} \ell}=\dfrac{m}{4}\left(\dfrac{2 \ell}{3}-\dfrac{m}{\pi \ell^{2} \rho}\right)=0 \\& \Rightarrow \dfrac{2 \ell}{3}=\dfrac{m}{\pi \ell^{2} \rho} \Rightarrow \dfrac{2 \ell}{3}=\dfrac{\pi \mathrm{R}^{2} \ell \rho}{\pi \ell^{2} \rho} \\& \text { or } \dfrac{2 \ell}{3}=\dfrac{\mathrm{R}^{2}}{\ell} \Rightarrow \dfrac{\ell^{2}}{\mathrm{R}^{2}}=\dfrac{3}{2} \\& \Rightarrow \dfrac{\ell}{\mathrm{R}}=\sqrt{\dfrac{3}{2}}\end{aligned}\)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365880
A \(T\) joint is formed by two identical rods \(A\) and \(B\) each of mass \(m\) and length \(L\) in \(X Y\) plane as shown. Its moment of inertia about axis passing along the length of \(\operatorname{rod} A\) is
1 \(\frac{{2\,m{L^3}}}{3}\)
2 \(\frac{{m\,{L^2}}}{{12}}\)
3 \(\dfrac{m L^{2}}{6}\)
4 \(\dfrac{m L^{2}}{3}\)
Explanation:
Total amount of inertia of system \(I=I_{A}+I_{B}=0+\dfrac{m L^{2}}{12}\) \(I=\dfrac{m L^{2}}{12}\)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365881
Two circular loops \(A\) and \(B\) of radii \(R\) and \(N R\) respectively are made from a uniform wire. Moment of inertia of \(B\) about its axis is 3 times that of \(A\) about its axis. The value of \(N\) is
1 \({[2]^{\frac{1}{3}}}\)
2 \({[5]^{\frac{1}{3}}}\)
3 \({[3]^{\frac{1}{3}}}\)
4 \({[4]^{\frac{1}{3}}}\)
Explanation:
Let \(\rho=\) mass per unit length of the wire. Mass of loop A, \(m_{A}=2 \pi R \rho\) Mass of loop B, \(m_{B}=2 \pi N R \rho=N m_{A}\) Given, \(\quad I_{B}=3 I_{A}\) \(\Rightarrow m_{B} r_{B}^{2}=3 \times m_{A} r_{A}^{2}\) \(\Rightarrow\left(N m_{A}\right)(N R)^{2}=3 m_{A}(R)^{2}\) \(\Rightarrow N^{3}=3 \Rightarrow N=(3)^{1 / 3}\)
MHTCET - 2020
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365882
Assertion : The moment of inertia of a rigid body reduces to its minimum value, when the axis of rotation passes through its centre of gravity. Reason : The weight of a rigid body always acts through its centre of gravity.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
365878
The radius of gyration of a rotating circular ring is maximum about following axis of rotation
1 Natural axis
2 Axis passing through diameter of ring
3 Axis passing through tangent of ring in its plane
4 Axis passing through tangent of ring perpendicular to plane of ring.
Explanation:
Conceptual Question
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365879
The moment of inertia of a uniform cylinder of length \(\ell\) and radius \(\mathrm{R}\) about its perpendicular bisector is \(\mathrm{I}\). What is the ratio \(\ell / \mathrm{R}\) such that the moment of inertia is minimum?
1 \(\dfrac{\sqrt{3}}{2}\)
2 1
3 \(\sqrt{\dfrac{3}{2}}\)
4 \(\dfrac{3}{\sqrt{2}}\)
Explanation:
Momentum of inertia about perpendicular bisector is \({\rm{I}} = \frac{{m{\ell ^2}}}{{12}} + \frac{{{\rm{m}}{{\rm{R}}^2}}}{4}\) \( \Rightarrow {\rm{I}} = \frac{m}{4}\left( {\frac{{{\ell ^2}}}{3} + {{\rm{R}}^2}} \right)\,\,\,\,\,\,\,\,\,\,(1)\) Also \(m=\pi \mathrm{R}^{2} \ell \rho\) \(\Rightarrow \mathrm{R}^{2}=\dfrac{m}{\pi \ell \rho}\) Put in equation (1) \(\mathrm{I}=\dfrac{m}{4}\left(\dfrac{\ell^{2}}{3}+\dfrac{m}{\pi \ell \rho}\right)\) For maxima & minima \(\begin{aligned}& \dfrac{\mathrm{dI}}{\mathrm{d} \ell}=\dfrac{m}{4}\left(\dfrac{2 \ell}{3}-\dfrac{m}{\pi \ell^{2} \rho}\right)=0 \\& \Rightarrow \dfrac{2 \ell}{3}=\dfrac{m}{\pi \ell^{2} \rho} \Rightarrow \dfrac{2 \ell}{3}=\dfrac{\pi \mathrm{R}^{2} \ell \rho}{\pi \ell^{2} \rho} \\& \text { or } \dfrac{2 \ell}{3}=\dfrac{\mathrm{R}^{2}}{\ell} \Rightarrow \dfrac{\ell^{2}}{\mathrm{R}^{2}}=\dfrac{3}{2} \\& \Rightarrow \dfrac{\ell}{\mathrm{R}}=\sqrt{\dfrac{3}{2}}\end{aligned}\)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365880
A \(T\) joint is formed by two identical rods \(A\) and \(B\) each of mass \(m\) and length \(L\) in \(X Y\) plane as shown. Its moment of inertia about axis passing along the length of \(\operatorname{rod} A\) is
1 \(\frac{{2\,m{L^3}}}{3}\)
2 \(\frac{{m\,{L^2}}}{{12}}\)
3 \(\dfrac{m L^{2}}{6}\)
4 \(\dfrac{m L^{2}}{3}\)
Explanation:
Total amount of inertia of system \(I=I_{A}+I_{B}=0+\dfrac{m L^{2}}{12}\) \(I=\dfrac{m L^{2}}{12}\)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365881
Two circular loops \(A\) and \(B\) of radii \(R\) and \(N R\) respectively are made from a uniform wire. Moment of inertia of \(B\) about its axis is 3 times that of \(A\) about its axis. The value of \(N\) is
1 \({[2]^{\frac{1}{3}}}\)
2 \({[5]^{\frac{1}{3}}}\)
3 \({[3]^{\frac{1}{3}}}\)
4 \({[4]^{\frac{1}{3}}}\)
Explanation:
Let \(\rho=\) mass per unit length of the wire. Mass of loop A, \(m_{A}=2 \pi R \rho\) Mass of loop B, \(m_{B}=2 \pi N R \rho=N m_{A}\) Given, \(\quad I_{B}=3 I_{A}\) \(\Rightarrow m_{B} r_{B}^{2}=3 \times m_{A} r_{A}^{2}\) \(\Rightarrow\left(N m_{A}\right)(N R)^{2}=3 m_{A}(R)^{2}\) \(\Rightarrow N^{3}=3 \Rightarrow N=(3)^{1 / 3}\)
MHTCET - 2020
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365882
Assertion : The moment of inertia of a rigid body reduces to its minimum value, when the axis of rotation passes through its centre of gravity. Reason : The weight of a rigid body always acts through its centre of gravity.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
365878
The radius of gyration of a rotating circular ring is maximum about following axis of rotation
1 Natural axis
2 Axis passing through diameter of ring
3 Axis passing through tangent of ring in its plane
4 Axis passing through tangent of ring perpendicular to plane of ring.
Explanation:
Conceptual Question
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365879
The moment of inertia of a uniform cylinder of length \(\ell\) and radius \(\mathrm{R}\) about its perpendicular bisector is \(\mathrm{I}\). What is the ratio \(\ell / \mathrm{R}\) such that the moment of inertia is minimum?
1 \(\dfrac{\sqrt{3}}{2}\)
2 1
3 \(\sqrt{\dfrac{3}{2}}\)
4 \(\dfrac{3}{\sqrt{2}}\)
Explanation:
Momentum of inertia about perpendicular bisector is \({\rm{I}} = \frac{{m{\ell ^2}}}{{12}} + \frac{{{\rm{m}}{{\rm{R}}^2}}}{4}\) \( \Rightarrow {\rm{I}} = \frac{m}{4}\left( {\frac{{{\ell ^2}}}{3} + {{\rm{R}}^2}} \right)\,\,\,\,\,\,\,\,\,\,(1)\) Also \(m=\pi \mathrm{R}^{2} \ell \rho\) \(\Rightarrow \mathrm{R}^{2}=\dfrac{m}{\pi \ell \rho}\) Put in equation (1) \(\mathrm{I}=\dfrac{m}{4}\left(\dfrac{\ell^{2}}{3}+\dfrac{m}{\pi \ell \rho}\right)\) For maxima & minima \(\begin{aligned}& \dfrac{\mathrm{dI}}{\mathrm{d} \ell}=\dfrac{m}{4}\left(\dfrac{2 \ell}{3}-\dfrac{m}{\pi \ell^{2} \rho}\right)=0 \\& \Rightarrow \dfrac{2 \ell}{3}=\dfrac{m}{\pi \ell^{2} \rho} \Rightarrow \dfrac{2 \ell}{3}=\dfrac{\pi \mathrm{R}^{2} \ell \rho}{\pi \ell^{2} \rho} \\& \text { or } \dfrac{2 \ell}{3}=\dfrac{\mathrm{R}^{2}}{\ell} \Rightarrow \dfrac{\ell^{2}}{\mathrm{R}^{2}}=\dfrac{3}{2} \\& \Rightarrow \dfrac{\ell}{\mathrm{R}}=\sqrt{\dfrac{3}{2}}\end{aligned}\)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365880
A \(T\) joint is formed by two identical rods \(A\) and \(B\) each of mass \(m\) and length \(L\) in \(X Y\) plane as shown. Its moment of inertia about axis passing along the length of \(\operatorname{rod} A\) is
1 \(\frac{{2\,m{L^3}}}{3}\)
2 \(\frac{{m\,{L^2}}}{{12}}\)
3 \(\dfrac{m L^{2}}{6}\)
4 \(\dfrac{m L^{2}}{3}\)
Explanation:
Total amount of inertia of system \(I=I_{A}+I_{B}=0+\dfrac{m L^{2}}{12}\) \(I=\dfrac{m L^{2}}{12}\)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365881
Two circular loops \(A\) and \(B\) of radii \(R\) and \(N R\) respectively are made from a uniform wire. Moment of inertia of \(B\) about its axis is 3 times that of \(A\) about its axis. The value of \(N\) is
1 \({[2]^{\frac{1}{3}}}\)
2 \({[5]^{\frac{1}{3}}}\)
3 \({[3]^{\frac{1}{3}}}\)
4 \({[4]^{\frac{1}{3}}}\)
Explanation:
Let \(\rho=\) mass per unit length of the wire. Mass of loop A, \(m_{A}=2 \pi R \rho\) Mass of loop B, \(m_{B}=2 \pi N R \rho=N m_{A}\) Given, \(\quad I_{B}=3 I_{A}\) \(\Rightarrow m_{B} r_{B}^{2}=3 \times m_{A} r_{A}^{2}\) \(\Rightarrow\left(N m_{A}\right)(N R)^{2}=3 m_{A}(R)^{2}\) \(\Rightarrow N^{3}=3 \Rightarrow N=(3)^{1 / 3}\)
MHTCET - 2020
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365882
Assertion : The moment of inertia of a rigid body reduces to its minimum value, when the axis of rotation passes through its centre of gravity. Reason : The weight of a rigid body always acts through its centre of gravity.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
365878
The radius of gyration of a rotating circular ring is maximum about following axis of rotation
1 Natural axis
2 Axis passing through diameter of ring
3 Axis passing through tangent of ring in its plane
4 Axis passing through tangent of ring perpendicular to plane of ring.
Explanation:
Conceptual Question
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365879
The moment of inertia of a uniform cylinder of length \(\ell\) and radius \(\mathrm{R}\) about its perpendicular bisector is \(\mathrm{I}\). What is the ratio \(\ell / \mathrm{R}\) such that the moment of inertia is minimum?
1 \(\dfrac{\sqrt{3}}{2}\)
2 1
3 \(\sqrt{\dfrac{3}{2}}\)
4 \(\dfrac{3}{\sqrt{2}}\)
Explanation:
Momentum of inertia about perpendicular bisector is \({\rm{I}} = \frac{{m{\ell ^2}}}{{12}} + \frac{{{\rm{m}}{{\rm{R}}^2}}}{4}\) \( \Rightarrow {\rm{I}} = \frac{m}{4}\left( {\frac{{{\ell ^2}}}{3} + {{\rm{R}}^2}} \right)\,\,\,\,\,\,\,\,\,\,(1)\) Also \(m=\pi \mathrm{R}^{2} \ell \rho\) \(\Rightarrow \mathrm{R}^{2}=\dfrac{m}{\pi \ell \rho}\) Put in equation (1) \(\mathrm{I}=\dfrac{m}{4}\left(\dfrac{\ell^{2}}{3}+\dfrac{m}{\pi \ell \rho}\right)\) For maxima & minima \(\begin{aligned}& \dfrac{\mathrm{dI}}{\mathrm{d} \ell}=\dfrac{m}{4}\left(\dfrac{2 \ell}{3}-\dfrac{m}{\pi \ell^{2} \rho}\right)=0 \\& \Rightarrow \dfrac{2 \ell}{3}=\dfrac{m}{\pi \ell^{2} \rho} \Rightarrow \dfrac{2 \ell}{3}=\dfrac{\pi \mathrm{R}^{2} \ell \rho}{\pi \ell^{2} \rho} \\& \text { or } \dfrac{2 \ell}{3}=\dfrac{\mathrm{R}^{2}}{\ell} \Rightarrow \dfrac{\ell^{2}}{\mathrm{R}^{2}}=\dfrac{3}{2} \\& \Rightarrow \dfrac{\ell}{\mathrm{R}}=\sqrt{\dfrac{3}{2}}\end{aligned}\)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365880
A \(T\) joint is formed by two identical rods \(A\) and \(B\) each of mass \(m\) and length \(L\) in \(X Y\) plane as shown. Its moment of inertia about axis passing along the length of \(\operatorname{rod} A\) is
1 \(\frac{{2\,m{L^3}}}{3}\)
2 \(\frac{{m\,{L^2}}}{{12}}\)
3 \(\dfrac{m L^{2}}{6}\)
4 \(\dfrac{m L^{2}}{3}\)
Explanation:
Total amount of inertia of system \(I=I_{A}+I_{B}=0+\dfrac{m L^{2}}{12}\) \(I=\dfrac{m L^{2}}{12}\)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365881
Two circular loops \(A\) and \(B\) of radii \(R\) and \(N R\) respectively are made from a uniform wire. Moment of inertia of \(B\) about its axis is 3 times that of \(A\) about its axis. The value of \(N\) is
1 \({[2]^{\frac{1}{3}}}\)
2 \({[5]^{\frac{1}{3}}}\)
3 \({[3]^{\frac{1}{3}}}\)
4 \({[4]^{\frac{1}{3}}}\)
Explanation:
Let \(\rho=\) mass per unit length of the wire. Mass of loop A, \(m_{A}=2 \pi R \rho\) Mass of loop B, \(m_{B}=2 \pi N R \rho=N m_{A}\) Given, \(\quad I_{B}=3 I_{A}\) \(\Rightarrow m_{B} r_{B}^{2}=3 \times m_{A} r_{A}^{2}\) \(\Rightarrow\left(N m_{A}\right)(N R)^{2}=3 m_{A}(R)^{2}\) \(\Rightarrow N^{3}=3 \Rightarrow N=(3)^{1 / 3}\)
MHTCET - 2020
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365882
Assertion : The moment of inertia of a rigid body reduces to its minimum value, when the axis of rotation passes through its centre of gravity. Reason : The weight of a rigid body always acts through its centre of gravity.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
