150120
Two discs \(A\) and \(B\) of equal mass and thickness have densities \(6800 \mathrm{~kg} / \mathrm{m}^{3}\) and \(8500 \mathrm{~kg} / \mathrm{m}^{3}\) respectively. The ratio of their moments of inertia ( \(A\) to \(B\)is
1 \(\frac{1}{6.8 \times 8.5}\)
2 \(\frac{4}{5}\)
3 \(\frac{5}{4}\)
4 \(\frac{5}{9}\)
Explanation:
C Given, Density \(\rho_{1}=6800 \mathrm{~kg} / \mathrm{m}^{3}\) and density \(\rho_{2}=8500 \mathrm{~kg} / \mathrm{m}^{3}\) Moment of inertia for disc - \(\mathrm{I}=\frac{1}{2} \mathrm{mr}^{2}\) Volume of \(\operatorname{disc}(\mathrm{V})=\pi \mathrm{r}^{2} \mathrm{t}\) Mass of the \(\operatorname{disc}(\mathrm{m})=\rho \mathrm{V}\) Hence, \(\mathrm{m}=\rho \times \pi \mathrm{r}^{2} \mathrm{t}\) \(\mathrm{r}^{2}=\frac{\mathrm{m}}{\rho \pi \mathrm{t}}\) \(I=\frac{1}{2} m \cdot\left(\frac{m}{\rho \pi t}\right)\) \(I \propto \frac{1}{\rho}\) So, \(\frac{I_{1}}{I_{2}}=\frac{\rho_{2}}{\rho_{1}}=\frac{8500}{6800}=\frac{5}{4}\)
MHT-CET 2020
Rotational Motion
150121
Three-point masses each of mass ' \(M\) ' are placed at the corners of an equilateral triangle of side ' \(a\) '. The moment of inertia of this system about an axis passing through one side of a triangle is
1 \(\frac{\mathrm{Ma}^{2}}{4}\)
2 \(\frac{2 \mathrm{Ma}^{2}}{3}\)
3 \(\frac{\mathrm{Ma}^{2}}{3}\)
4 \(\frac{3 \mathrm{Ma}^{2}}{4}\)
Explanation:
D From Pythagoras theorem \(\mathrm{a}^{2}=\mathrm{h}^{2}+\left(\frac{\mathrm{a}}{2}\right)^{2}\) \(\mathrm{~h}=\frac{\sqrt{3}}{2} \mathrm{a}\) The moment of inertia of this system about an axis passing through one side of triangle is \(I=M h^{2}\) \(I=M\left(\frac{\sqrt{3}}{2} a\right)^{2}\) \(I=\frac{3 M^{2}}{4}\)
MHT-CET 2020
Rotational Motion
150122
A thin wire of length ' \(L\) ' and uniform linear mass density ' \(\rho\) ' is bent into a circular coil with ' \(O\) ' as centre. The moment of inertia of a coil about the axis \(\mathrm{XX}^{\prime}\) is
1 \(3 \rho \mathrm{L}^{3} / 8 \pi^{2}\)
2 \(\rho L^{3} / 4 \pi^{2}\)
3 \(3 \rho \mathrm{L}^{2} / 4 \pi^{2}\)
4 \(\rho \mathrm{L}^{3} / 8 \pi^{2}\)
Explanation:
A The moment of inertia of the loop about X X' axis is Moment of inertia of the thin circular coil, \(I=\frac{\mathrm{MR}^{2}}{2}\) \(\mathrm{I}_{\mathrm{XX}}=\frac{\mathrm{MR}^{2}}{2}+\mathrm{MR}^{2}\) \(\mathrm{I}_{\mathrm{XX}}=\frac{3}{2} \mathrm{MR}^{2}\) Where, \(M=\) mass, \(R=\) radius, \(\rho=\) linear mass density \(\because\) Mass, \((M)=\rho \mathrm{L} \quad\) and \(\quad \mathrm{R}=\frac{\mathrm{L}}{2 \pi}\) \(\therefore \mathrm{I}_{\mathrm{XX}}=\frac{3}{2}(\rho \mathrm{L})\left(\frac{\mathrm{L}}{2 \pi}\right)^{2}\) \(\mathrm{I}_{X X^{\prime}}=\frac{3 \rho L^{3}}{8 \pi^{2}}\)
MHT-CET 2020
Rotational Motion
150124
Two discs \(P\) and \(Q\) having equal masses and thickness but densities ' \(\rho_{1}\) ' and ' \(\rho_{2}\) ' respectively, are such that \(\rho_{1} \lt \rho_{2}\). The moment of inertia of the disc \(P\) and \(Q\) are related as
D We know that, Mass \(=\) volume \(\times\) density \(M=\left(\pi R^{2} t\right) \times \rho \ldots . .(i) \quad(\because t=\) thickness \(\) \(\Rightarrow \mathrm{R}^{2}=\frac{\mathrm{M}}{\pi \rho \mathrm{t}}\) \(\therefore\) M.I of disc, \(\quad \mathrm{I}=\frac{\mathrm{MR}^{2}}{2} \ldots\) From eq \({ }^{\mathrm{n}}\) (i) and (ii), we get \(I=\frac{M}{2} \times \frac{M}{\pi t \rho}\) \(I=\frac{M^{2}}{2 \pi t \rho}\) \(I \propto \frac{1}{\rho}\) For disc \(\mathrm{P}\) density is given \(\rho_{1}\) and disc \(\mathrm{Q}\) density is \(\rho_{2}\) So, \(\mathrm{I}_{\mathrm{P}}>\mathrm{I}_{\mathrm{Q}}\) \(\left[\therefore \rho_{1} \lt \rho_{2}\right]\)
MHT-CET 2020
Rotational Motion
150125
Moment of inertia of the rod about an axis passing through the centre and perpendicular to its length is ' \(I_{1}\) '. The same rod is bent into a ring and its moment of inertia about the diameter is ' \(I_{2}\) ', then \(\frac{I_{2}}{I_{1}}\) is
1 \(\frac{3}{2 \pi^{2}}\)
2 \(\frac{3}{4 \pi^{2}}\)
3 \(\frac{2 \pi^{2}}{3}\)
4 \(\frac{4 \pi^{2}}{3}\)
Explanation:
A Moment of inertia of \(\operatorname{rod}\left(I_{1}\right)=\frac{M L^{2}}{12}\) When the rod is bent into a ring, then Length of rod \(=\) Perimeter of the ring \(\mathrm{L}=2 \pi \mathrm{R}\) \(\mathrm{R}=\frac{\mathrm{L}}{2 \pi}\) Moment of inertia of a ring about its diameter \(\mathrm{I}_{2}=\frac{\mathrm{MR}^{2}}{2}\) \(I_{2}=\frac{M}{2} \cdot\left(\frac{L}{2 \pi}\right)^{2}=\frac{M L^{2}}{8 \pi^{2}}\) Then, \(\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}=\frac{\frac{\mathrm{ML}^{2}}{8 \pi^{2}}}{\frac{\mathrm{ML}^{2}}{12}}=\frac{3}{2 \pi^{2}}\) \(\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}=\frac{3}{2 \pi^{2}}\)
150120
Two discs \(A\) and \(B\) of equal mass and thickness have densities \(6800 \mathrm{~kg} / \mathrm{m}^{3}\) and \(8500 \mathrm{~kg} / \mathrm{m}^{3}\) respectively. The ratio of their moments of inertia ( \(A\) to \(B\)is
1 \(\frac{1}{6.8 \times 8.5}\)
2 \(\frac{4}{5}\)
3 \(\frac{5}{4}\)
4 \(\frac{5}{9}\)
Explanation:
C Given, Density \(\rho_{1}=6800 \mathrm{~kg} / \mathrm{m}^{3}\) and density \(\rho_{2}=8500 \mathrm{~kg} / \mathrm{m}^{3}\) Moment of inertia for disc - \(\mathrm{I}=\frac{1}{2} \mathrm{mr}^{2}\) Volume of \(\operatorname{disc}(\mathrm{V})=\pi \mathrm{r}^{2} \mathrm{t}\) Mass of the \(\operatorname{disc}(\mathrm{m})=\rho \mathrm{V}\) Hence, \(\mathrm{m}=\rho \times \pi \mathrm{r}^{2} \mathrm{t}\) \(\mathrm{r}^{2}=\frac{\mathrm{m}}{\rho \pi \mathrm{t}}\) \(I=\frac{1}{2} m \cdot\left(\frac{m}{\rho \pi t}\right)\) \(I \propto \frac{1}{\rho}\) So, \(\frac{I_{1}}{I_{2}}=\frac{\rho_{2}}{\rho_{1}}=\frac{8500}{6800}=\frac{5}{4}\)
MHT-CET 2020
Rotational Motion
150121
Three-point masses each of mass ' \(M\) ' are placed at the corners of an equilateral triangle of side ' \(a\) '. The moment of inertia of this system about an axis passing through one side of a triangle is
1 \(\frac{\mathrm{Ma}^{2}}{4}\)
2 \(\frac{2 \mathrm{Ma}^{2}}{3}\)
3 \(\frac{\mathrm{Ma}^{2}}{3}\)
4 \(\frac{3 \mathrm{Ma}^{2}}{4}\)
Explanation:
D From Pythagoras theorem \(\mathrm{a}^{2}=\mathrm{h}^{2}+\left(\frac{\mathrm{a}}{2}\right)^{2}\) \(\mathrm{~h}=\frac{\sqrt{3}}{2} \mathrm{a}\) The moment of inertia of this system about an axis passing through one side of triangle is \(I=M h^{2}\) \(I=M\left(\frac{\sqrt{3}}{2} a\right)^{2}\) \(I=\frac{3 M^{2}}{4}\)
MHT-CET 2020
Rotational Motion
150122
A thin wire of length ' \(L\) ' and uniform linear mass density ' \(\rho\) ' is bent into a circular coil with ' \(O\) ' as centre. The moment of inertia of a coil about the axis \(\mathrm{XX}^{\prime}\) is
1 \(3 \rho \mathrm{L}^{3} / 8 \pi^{2}\)
2 \(\rho L^{3} / 4 \pi^{2}\)
3 \(3 \rho \mathrm{L}^{2} / 4 \pi^{2}\)
4 \(\rho \mathrm{L}^{3} / 8 \pi^{2}\)
Explanation:
A The moment of inertia of the loop about X X' axis is Moment of inertia of the thin circular coil, \(I=\frac{\mathrm{MR}^{2}}{2}\) \(\mathrm{I}_{\mathrm{XX}}=\frac{\mathrm{MR}^{2}}{2}+\mathrm{MR}^{2}\) \(\mathrm{I}_{\mathrm{XX}}=\frac{3}{2} \mathrm{MR}^{2}\) Where, \(M=\) mass, \(R=\) radius, \(\rho=\) linear mass density \(\because\) Mass, \((M)=\rho \mathrm{L} \quad\) and \(\quad \mathrm{R}=\frac{\mathrm{L}}{2 \pi}\) \(\therefore \mathrm{I}_{\mathrm{XX}}=\frac{3}{2}(\rho \mathrm{L})\left(\frac{\mathrm{L}}{2 \pi}\right)^{2}\) \(\mathrm{I}_{X X^{\prime}}=\frac{3 \rho L^{3}}{8 \pi^{2}}\)
MHT-CET 2020
Rotational Motion
150124
Two discs \(P\) and \(Q\) having equal masses and thickness but densities ' \(\rho_{1}\) ' and ' \(\rho_{2}\) ' respectively, are such that \(\rho_{1} \lt \rho_{2}\). The moment of inertia of the disc \(P\) and \(Q\) are related as
D We know that, Mass \(=\) volume \(\times\) density \(M=\left(\pi R^{2} t\right) \times \rho \ldots . .(i) \quad(\because t=\) thickness \(\) \(\Rightarrow \mathrm{R}^{2}=\frac{\mathrm{M}}{\pi \rho \mathrm{t}}\) \(\therefore\) M.I of disc, \(\quad \mathrm{I}=\frac{\mathrm{MR}^{2}}{2} \ldots\) From eq \({ }^{\mathrm{n}}\) (i) and (ii), we get \(I=\frac{M}{2} \times \frac{M}{\pi t \rho}\) \(I=\frac{M^{2}}{2 \pi t \rho}\) \(I \propto \frac{1}{\rho}\) For disc \(\mathrm{P}\) density is given \(\rho_{1}\) and disc \(\mathrm{Q}\) density is \(\rho_{2}\) So, \(\mathrm{I}_{\mathrm{P}}>\mathrm{I}_{\mathrm{Q}}\) \(\left[\therefore \rho_{1} \lt \rho_{2}\right]\)
MHT-CET 2020
Rotational Motion
150125
Moment of inertia of the rod about an axis passing through the centre and perpendicular to its length is ' \(I_{1}\) '. The same rod is bent into a ring and its moment of inertia about the diameter is ' \(I_{2}\) ', then \(\frac{I_{2}}{I_{1}}\) is
1 \(\frac{3}{2 \pi^{2}}\)
2 \(\frac{3}{4 \pi^{2}}\)
3 \(\frac{2 \pi^{2}}{3}\)
4 \(\frac{4 \pi^{2}}{3}\)
Explanation:
A Moment of inertia of \(\operatorname{rod}\left(I_{1}\right)=\frac{M L^{2}}{12}\) When the rod is bent into a ring, then Length of rod \(=\) Perimeter of the ring \(\mathrm{L}=2 \pi \mathrm{R}\) \(\mathrm{R}=\frac{\mathrm{L}}{2 \pi}\) Moment of inertia of a ring about its diameter \(\mathrm{I}_{2}=\frac{\mathrm{MR}^{2}}{2}\) \(I_{2}=\frac{M}{2} \cdot\left(\frac{L}{2 \pi}\right)^{2}=\frac{M L^{2}}{8 \pi^{2}}\) Then, \(\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}=\frac{\frac{\mathrm{ML}^{2}}{8 \pi^{2}}}{\frac{\mathrm{ML}^{2}}{12}}=\frac{3}{2 \pi^{2}}\) \(\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}=\frac{3}{2 \pi^{2}}\)
150120
Two discs \(A\) and \(B\) of equal mass and thickness have densities \(6800 \mathrm{~kg} / \mathrm{m}^{3}\) and \(8500 \mathrm{~kg} / \mathrm{m}^{3}\) respectively. The ratio of their moments of inertia ( \(A\) to \(B\)is
1 \(\frac{1}{6.8 \times 8.5}\)
2 \(\frac{4}{5}\)
3 \(\frac{5}{4}\)
4 \(\frac{5}{9}\)
Explanation:
C Given, Density \(\rho_{1}=6800 \mathrm{~kg} / \mathrm{m}^{3}\) and density \(\rho_{2}=8500 \mathrm{~kg} / \mathrm{m}^{3}\) Moment of inertia for disc - \(\mathrm{I}=\frac{1}{2} \mathrm{mr}^{2}\) Volume of \(\operatorname{disc}(\mathrm{V})=\pi \mathrm{r}^{2} \mathrm{t}\) Mass of the \(\operatorname{disc}(\mathrm{m})=\rho \mathrm{V}\) Hence, \(\mathrm{m}=\rho \times \pi \mathrm{r}^{2} \mathrm{t}\) \(\mathrm{r}^{2}=\frac{\mathrm{m}}{\rho \pi \mathrm{t}}\) \(I=\frac{1}{2} m \cdot\left(\frac{m}{\rho \pi t}\right)\) \(I \propto \frac{1}{\rho}\) So, \(\frac{I_{1}}{I_{2}}=\frac{\rho_{2}}{\rho_{1}}=\frac{8500}{6800}=\frac{5}{4}\)
MHT-CET 2020
Rotational Motion
150121
Three-point masses each of mass ' \(M\) ' are placed at the corners of an equilateral triangle of side ' \(a\) '. The moment of inertia of this system about an axis passing through one side of a triangle is
1 \(\frac{\mathrm{Ma}^{2}}{4}\)
2 \(\frac{2 \mathrm{Ma}^{2}}{3}\)
3 \(\frac{\mathrm{Ma}^{2}}{3}\)
4 \(\frac{3 \mathrm{Ma}^{2}}{4}\)
Explanation:
D From Pythagoras theorem \(\mathrm{a}^{2}=\mathrm{h}^{2}+\left(\frac{\mathrm{a}}{2}\right)^{2}\) \(\mathrm{~h}=\frac{\sqrt{3}}{2} \mathrm{a}\) The moment of inertia of this system about an axis passing through one side of triangle is \(I=M h^{2}\) \(I=M\left(\frac{\sqrt{3}}{2} a\right)^{2}\) \(I=\frac{3 M^{2}}{4}\)
MHT-CET 2020
Rotational Motion
150122
A thin wire of length ' \(L\) ' and uniform linear mass density ' \(\rho\) ' is bent into a circular coil with ' \(O\) ' as centre. The moment of inertia of a coil about the axis \(\mathrm{XX}^{\prime}\) is
1 \(3 \rho \mathrm{L}^{3} / 8 \pi^{2}\)
2 \(\rho L^{3} / 4 \pi^{2}\)
3 \(3 \rho \mathrm{L}^{2} / 4 \pi^{2}\)
4 \(\rho \mathrm{L}^{3} / 8 \pi^{2}\)
Explanation:
A The moment of inertia of the loop about X X' axis is Moment of inertia of the thin circular coil, \(I=\frac{\mathrm{MR}^{2}}{2}\) \(\mathrm{I}_{\mathrm{XX}}=\frac{\mathrm{MR}^{2}}{2}+\mathrm{MR}^{2}\) \(\mathrm{I}_{\mathrm{XX}}=\frac{3}{2} \mathrm{MR}^{2}\) Where, \(M=\) mass, \(R=\) radius, \(\rho=\) linear mass density \(\because\) Mass, \((M)=\rho \mathrm{L} \quad\) and \(\quad \mathrm{R}=\frac{\mathrm{L}}{2 \pi}\) \(\therefore \mathrm{I}_{\mathrm{XX}}=\frac{3}{2}(\rho \mathrm{L})\left(\frac{\mathrm{L}}{2 \pi}\right)^{2}\) \(\mathrm{I}_{X X^{\prime}}=\frac{3 \rho L^{3}}{8 \pi^{2}}\)
MHT-CET 2020
Rotational Motion
150124
Two discs \(P\) and \(Q\) having equal masses and thickness but densities ' \(\rho_{1}\) ' and ' \(\rho_{2}\) ' respectively, are such that \(\rho_{1} \lt \rho_{2}\). The moment of inertia of the disc \(P\) and \(Q\) are related as
D We know that, Mass \(=\) volume \(\times\) density \(M=\left(\pi R^{2} t\right) \times \rho \ldots . .(i) \quad(\because t=\) thickness \(\) \(\Rightarrow \mathrm{R}^{2}=\frac{\mathrm{M}}{\pi \rho \mathrm{t}}\) \(\therefore\) M.I of disc, \(\quad \mathrm{I}=\frac{\mathrm{MR}^{2}}{2} \ldots\) From eq \({ }^{\mathrm{n}}\) (i) and (ii), we get \(I=\frac{M}{2} \times \frac{M}{\pi t \rho}\) \(I=\frac{M^{2}}{2 \pi t \rho}\) \(I \propto \frac{1}{\rho}\) For disc \(\mathrm{P}\) density is given \(\rho_{1}\) and disc \(\mathrm{Q}\) density is \(\rho_{2}\) So, \(\mathrm{I}_{\mathrm{P}}>\mathrm{I}_{\mathrm{Q}}\) \(\left[\therefore \rho_{1} \lt \rho_{2}\right]\)
MHT-CET 2020
Rotational Motion
150125
Moment of inertia of the rod about an axis passing through the centre and perpendicular to its length is ' \(I_{1}\) '. The same rod is bent into a ring and its moment of inertia about the diameter is ' \(I_{2}\) ', then \(\frac{I_{2}}{I_{1}}\) is
1 \(\frac{3}{2 \pi^{2}}\)
2 \(\frac{3}{4 \pi^{2}}\)
3 \(\frac{2 \pi^{2}}{3}\)
4 \(\frac{4 \pi^{2}}{3}\)
Explanation:
A Moment of inertia of \(\operatorname{rod}\left(I_{1}\right)=\frac{M L^{2}}{12}\) When the rod is bent into a ring, then Length of rod \(=\) Perimeter of the ring \(\mathrm{L}=2 \pi \mathrm{R}\) \(\mathrm{R}=\frac{\mathrm{L}}{2 \pi}\) Moment of inertia of a ring about its diameter \(\mathrm{I}_{2}=\frac{\mathrm{MR}^{2}}{2}\) \(I_{2}=\frac{M}{2} \cdot\left(\frac{L}{2 \pi}\right)^{2}=\frac{M L^{2}}{8 \pi^{2}}\) Then, \(\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}=\frac{\frac{\mathrm{ML}^{2}}{8 \pi^{2}}}{\frac{\mathrm{ML}^{2}}{12}}=\frac{3}{2 \pi^{2}}\) \(\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}=\frac{3}{2 \pi^{2}}\)
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Rotational Motion
150120
Two discs \(A\) and \(B\) of equal mass and thickness have densities \(6800 \mathrm{~kg} / \mathrm{m}^{3}\) and \(8500 \mathrm{~kg} / \mathrm{m}^{3}\) respectively. The ratio of their moments of inertia ( \(A\) to \(B\)is
1 \(\frac{1}{6.8 \times 8.5}\)
2 \(\frac{4}{5}\)
3 \(\frac{5}{4}\)
4 \(\frac{5}{9}\)
Explanation:
C Given, Density \(\rho_{1}=6800 \mathrm{~kg} / \mathrm{m}^{3}\) and density \(\rho_{2}=8500 \mathrm{~kg} / \mathrm{m}^{3}\) Moment of inertia for disc - \(\mathrm{I}=\frac{1}{2} \mathrm{mr}^{2}\) Volume of \(\operatorname{disc}(\mathrm{V})=\pi \mathrm{r}^{2} \mathrm{t}\) Mass of the \(\operatorname{disc}(\mathrm{m})=\rho \mathrm{V}\) Hence, \(\mathrm{m}=\rho \times \pi \mathrm{r}^{2} \mathrm{t}\) \(\mathrm{r}^{2}=\frac{\mathrm{m}}{\rho \pi \mathrm{t}}\) \(I=\frac{1}{2} m \cdot\left(\frac{m}{\rho \pi t}\right)\) \(I \propto \frac{1}{\rho}\) So, \(\frac{I_{1}}{I_{2}}=\frac{\rho_{2}}{\rho_{1}}=\frac{8500}{6800}=\frac{5}{4}\)
MHT-CET 2020
Rotational Motion
150121
Three-point masses each of mass ' \(M\) ' are placed at the corners of an equilateral triangle of side ' \(a\) '. The moment of inertia of this system about an axis passing through one side of a triangle is
1 \(\frac{\mathrm{Ma}^{2}}{4}\)
2 \(\frac{2 \mathrm{Ma}^{2}}{3}\)
3 \(\frac{\mathrm{Ma}^{2}}{3}\)
4 \(\frac{3 \mathrm{Ma}^{2}}{4}\)
Explanation:
D From Pythagoras theorem \(\mathrm{a}^{2}=\mathrm{h}^{2}+\left(\frac{\mathrm{a}}{2}\right)^{2}\) \(\mathrm{~h}=\frac{\sqrt{3}}{2} \mathrm{a}\) The moment of inertia of this system about an axis passing through one side of triangle is \(I=M h^{2}\) \(I=M\left(\frac{\sqrt{3}}{2} a\right)^{2}\) \(I=\frac{3 M^{2}}{4}\)
MHT-CET 2020
Rotational Motion
150122
A thin wire of length ' \(L\) ' and uniform linear mass density ' \(\rho\) ' is bent into a circular coil with ' \(O\) ' as centre. The moment of inertia of a coil about the axis \(\mathrm{XX}^{\prime}\) is
1 \(3 \rho \mathrm{L}^{3} / 8 \pi^{2}\)
2 \(\rho L^{3} / 4 \pi^{2}\)
3 \(3 \rho \mathrm{L}^{2} / 4 \pi^{2}\)
4 \(\rho \mathrm{L}^{3} / 8 \pi^{2}\)
Explanation:
A The moment of inertia of the loop about X X' axis is Moment of inertia of the thin circular coil, \(I=\frac{\mathrm{MR}^{2}}{2}\) \(\mathrm{I}_{\mathrm{XX}}=\frac{\mathrm{MR}^{2}}{2}+\mathrm{MR}^{2}\) \(\mathrm{I}_{\mathrm{XX}}=\frac{3}{2} \mathrm{MR}^{2}\) Where, \(M=\) mass, \(R=\) radius, \(\rho=\) linear mass density \(\because\) Mass, \((M)=\rho \mathrm{L} \quad\) and \(\quad \mathrm{R}=\frac{\mathrm{L}}{2 \pi}\) \(\therefore \mathrm{I}_{\mathrm{XX}}=\frac{3}{2}(\rho \mathrm{L})\left(\frac{\mathrm{L}}{2 \pi}\right)^{2}\) \(\mathrm{I}_{X X^{\prime}}=\frac{3 \rho L^{3}}{8 \pi^{2}}\)
MHT-CET 2020
Rotational Motion
150124
Two discs \(P\) and \(Q\) having equal masses and thickness but densities ' \(\rho_{1}\) ' and ' \(\rho_{2}\) ' respectively, are such that \(\rho_{1} \lt \rho_{2}\). The moment of inertia of the disc \(P\) and \(Q\) are related as
D We know that, Mass \(=\) volume \(\times\) density \(M=\left(\pi R^{2} t\right) \times \rho \ldots . .(i) \quad(\because t=\) thickness \(\) \(\Rightarrow \mathrm{R}^{2}=\frac{\mathrm{M}}{\pi \rho \mathrm{t}}\) \(\therefore\) M.I of disc, \(\quad \mathrm{I}=\frac{\mathrm{MR}^{2}}{2} \ldots\) From eq \({ }^{\mathrm{n}}\) (i) and (ii), we get \(I=\frac{M}{2} \times \frac{M}{\pi t \rho}\) \(I=\frac{M^{2}}{2 \pi t \rho}\) \(I \propto \frac{1}{\rho}\) For disc \(\mathrm{P}\) density is given \(\rho_{1}\) and disc \(\mathrm{Q}\) density is \(\rho_{2}\) So, \(\mathrm{I}_{\mathrm{P}}>\mathrm{I}_{\mathrm{Q}}\) \(\left[\therefore \rho_{1} \lt \rho_{2}\right]\)
MHT-CET 2020
Rotational Motion
150125
Moment of inertia of the rod about an axis passing through the centre and perpendicular to its length is ' \(I_{1}\) '. The same rod is bent into a ring and its moment of inertia about the diameter is ' \(I_{2}\) ', then \(\frac{I_{2}}{I_{1}}\) is
1 \(\frac{3}{2 \pi^{2}}\)
2 \(\frac{3}{4 \pi^{2}}\)
3 \(\frac{2 \pi^{2}}{3}\)
4 \(\frac{4 \pi^{2}}{3}\)
Explanation:
A Moment of inertia of \(\operatorname{rod}\left(I_{1}\right)=\frac{M L^{2}}{12}\) When the rod is bent into a ring, then Length of rod \(=\) Perimeter of the ring \(\mathrm{L}=2 \pi \mathrm{R}\) \(\mathrm{R}=\frac{\mathrm{L}}{2 \pi}\) Moment of inertia of a ring about its diameter \(\mathrm{I}_{2}=\frac{\mathrm{MR}^{2}}{2}\) \(I_{2}=\frac{M}{2} \cdot\left(\frac{L}{2 \pi}\right)^{2}=\frac{M L^{2}}{8 \pi^{2}}\) Then, \(\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}=\frac{\frac{\mathrm{ML}^{2}}{8 \pi^{2}}}{\frac{\mathrm{ML}^{2}}{12}}=\frac{3}{2 \pi^{2}}\) \(\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}=\frac{3}{2 \pi^{2}}\)
150120
Two discs \(A\) and \(B\) of equal mass and thickness have densities \(6800 \mathrm{~kg} / \mathrm{m}^{3}\) and \(8500 \mathrm{~kg} / \mathrm{m}^{3}\) respectively. The ratio of their moments of inertia ( \(A\) to \(B\)is
1 \(\frac{1}{6.8 \times 8.5}\)
2 \(\frac{4}{5}\)
3 \(\frac{5}{4}\)
4 \(\frac{5}{9}\)
Explanation:
C Given, Density \(\rho_{1}=6800 \mathrm{~kg} / \mathrm{m}^{3}\) and density \(\rho_{2}=8500 \mathrm{~kg} / \mathrm{m}^{3}\) Moment of inertia for disc - \(\mathrm{I}=\frac{1}{2} \mathrm{mr}^{2}\) Volume of \(\operatorname{disc}(\mathrm{V})=\pi \mathrm{r}^{2} \mathrm{t}\) Mass of the \(\operatorname{disc}(\mathrm{m})=\rho \mathrm{V}\) Hence, \(\mathrm{m}=\rho \times \pi \mathrm{r}^{2} \mathrm{t}\) \(\mathrm{r}^{2}=\frac{\mathrm{m}}{\rho \pi \mathrm{t}}\) \(I=\frac{1}{2} m \cdot\left(\frac{m}{\rho \pi t}\right)\) \(I \propto \frac{1}{\rho}\) So, \(\frac{I_{1}}{I_{2}}=\frac{\rho_{2}}{\rho_{1}}=\frac{8500}{6800}=\frac{5}{4}\)
MHT-CET 2020
Rotational Motion
150121
Three-point masses each of mass ' \(M\) ' are placed at the corners of an equilateral triangle of side ' \(a\) '. The moment of inertia of this system about an axis passing through one side of a triangle is
1 \(\frac{\mathrm{Ma}^{2}}{4}\)
2 \(\frac{2 \mathrm{Ma}^{2}}{3}\)
3 \(\frac{\mathrm{Ma}^{2}}{3}\)
4 \(\frac{3 \mathrm{Ma}^{2}}{4}\)
Explanation:
D From Pythagoras theorem \(\mathrm{a}^{2}=\mathrm{h}^{2}+\left(\frac{\mathrm{a}}{2}\right)^{2}\) \(\mathrm{~h}=\frac{\sqrt{3}}{2} \mathrm{a}\) The moment of inertia of this system about an axis passing through one side of triangle is \(I=M h^{2}\) \(I=M\left(\frac{\sqrt{3}}{2} a\right)^{2}\) \(I=\frac{3 M^{2}}{4}\)
MHT-CET 2020
Rotational Motion
150122
A thin wire of length ' \(L\) ' and uniform linear mass density ' \(\rho\) ' is bent into a circular coil with ' \(O\) ' as centre. The moment of inertia of a coil about the axis \(\mathrm{XX}^{\prime}\) is
1 \(3 \rho \mathrm{L}^{3} / 8 \pi^{2}\)
2 \(\rho L^{3} / 4 \pi^{2}\)
3 \(3 \rho \mathrm{L}^{2} / 4 \pi^{2}\)
4 \(\rho \mathrm{L}^{3} / 8 \pi^{2}\)
Explanation:
A The moment of inertia of the loop about X X' axis is Moment of inertia of the thin circular coil, \(I=\frac{\mathrm{MR}^{2}}{2}\) \(\mathrm{I}_{\mathrm{XX}}=\frac{\mathrm{MR}^{2}}{2}+\mathrm{MR}^{2}\) \(\mathrm{I}_{\mathrm{XX}}=\frac{3}{2} \mathrm{MR}^{2}\) Where, \(M=\) mass, \(R=\) radius, \(\rho=\) linear mass density \(\because\) Mass, \((M)=\rho \mathrm{L} \quad\) and \(\quad \mathrm{R}=\frac{\mathrm{L}}{2 \pi}\) \(\therefore \mathrm{I}_{\mathrm{XX}}=\frac{3}{2}(\rho \mathrm{L})\left(\frac{\mathrm{L}}{2 \pi}\right)^{2}\) \(\mathrm{I}_{X X^{\prime}}=\frac{3 \rho L^{3}}{8 \pi^{2}}\)
MHT-CET 2020
Rotational Motion
150124
Two discs \(P\) and \(Q\) having equal masses and thickness but densities ' \(\rho_{1}\) ' and ' \(\rho_{2}\) ' respectively, are such that \(\rho_{1} \lt \rho_{2}\). The moment of inertia of the disc \(P\) and \(Q\) are related as
D We know that, Mass \(=\) volume \(\times\) density \(M=\left(\pi R^{2} t\right) \times \rho \ldots . .(i) \quad(\because t=\) thickness \(\) \(\Rightarrow \mathrm{R}^{2}=\frac{\mathrm{M}}{\pi \rho \mathrm{t}}\) \(\therefore\) M.I of disc, \(\quad \mathrm{I}=\frac{\mathrm{MR}^{2}}{2} \ldots\) From eq \({ }^{\mathrm{n}}\) (i) and (ii), we get \(I=\frac{M}{2} \times \frac{M}{\pi t \rho}\) \(I=\frac{M^{2}}{2 \pi t \rho}\) \(I \propto \frac{1}{\rho}\) For disc \(\mathrm{P}\) density is given \(\rho_{1}\) and disc \(\mathrm{Q}\) density is \(\rho_{2}\) So, \(\mathrm{I}_{\mathrm{P}}>\mathrm{I}_{\mathrm{Q}}\) \(\left[\therefore \rho_{1} \lt \rho_{2}\right]\)
MHT-CET 2020
Rotational Motion
150125
Moment of inertia of the rod about an axis passing through the centre and perpendicular to its length is ' \(I_{1}\) '. The same rod is bent into a ring and its moment of inertia about the diameter is ' \(I_{2}\) ', then \(\frac{I_{2}}{I_{1}}\) is
1 \(\frac{3}{2 \pi^{2}}\)
2 \(\frac{3}{4 \pi^{2}}\)
3 \(\frac{2 \pi^{2}}{3}\)
4 \(\frac{4 \pi^{2}}{3}\)
Explanation:
A Moment of inertia of \(\operatorname{rod}\left(I_{1}\right)=\frac{M L^{2}}{12}\) When the rod is bent into a ring, then Length of rod \(=\) Perimeter of the ring \(\mathrm{L}=2 \pi \mathrm{R}\) \(\mathrm{R}=\frac{\mathrm{L}}{2 \pi}\) Moment of inertia of a ring about its diameter \(\mathrm{I}_{2}=\frac{\mathrm{MR}^{2}}{2}\) \(I_{2}=\frac{M}{2} \cdot\left(\frac{L}{2 \pi}\right)^{2}=\frac{M L^{2}}{8 \pi^{2}}\) Then, \(\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}=\frac{\frac{\mathrm{ML}^{2}}{8 \pi^{2}}}{\frac{\mathrm{ML}^{2}}{12}}=\frac{3}{2 \pi^{2}}\) \(\frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}=\frac{3}{2 \pi^{2}}\)
