NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365857
Four holes of radius \(\mathrm{R}\) are cut from a thin square plate of side \(4 \mathrm{R}\) and mass \(\mathrm{M}\). The moment of inertia of the remaining portion about \(z\) - axis is:-
Mass of each disc \(m=\dfrac{M}{16 R^{2}} \times \pi R^{2}=\dfrac{\pi}{16} \mathrm{M}\) Moment of inertia of the remaining part is \(\mathrm{I}=\mathrm{I}_{\text {square }}-4 I_{\text {disc }}\) \( = \frac{{M{{(4R)}^2}}}{6} - 4\left[ {\frac{{m{R^2}}}{2} + m{{(\sqrt 2 R)}^2}} \right]\) \( = \frac{8}{3}M{R^2} - 10m{R^2}\left[ {{\rm{ here }}m = \frac{{M\pi }}{{16}}} \right]\) \( = \left( {\frac{8}{3} - \frac{{5\pi }}{8}} \right){\rm{M}}{{\rm{R}}^2}\)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365858
A uniform square plate has a small piece \(Q\) of an irregular shape removed and glued to the centre of the plate leaving a hole behind. The moment of inertia about the \(z\)-aixs is then
1 Decreased
2 Increased
3 Changed in unpredicted manner
4 The same
Explanation:
If we keep the small piece at the centre then its moment of inertia about the central axis is zero. The mass of the remaining portion decreases and hence moment of inertia also decreases.
NCERT Exemplar
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365859
From a disc of radius \(R\) and mass \(M\), a circular hole of diameter \(R\), whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through centre?
1 \(11 M R^{2} / 32\)
2 \(9 M R^{2} / 32\)
3 \(15 M R^{2} / 32\)
4 \(13 M R^{2} / 32\)
Explanation:
Mass per unit area of disc \(=\dfrac{M}{\pi R^{2}}\) Mass of removed portion of disc, \(M^{\prime}=\dfrac{M}{\pi R^{2}} \times \pi\left(\dfrac{R}{2}\right)^{2}=\dfrac{M}{4}\) Moment of inertia of removed portion about an axis passing through centre of disc \(O\) and perpendicular to the plane of disc. \(\begin{gathered}I_{O}^{\prime}=I_{o^{\prime}}+M^{\prime} d^{2} \\=\dfrac{1}{2} \times \dfrac{M}{4} \times\left(\dfrac{R}{2}\right)^{2}+\dfrac{M}{4} \times\left(\dfrac{R}{2}\right)^{2} \\=\dfrac{M R^{2}}{32}+\dfrac{M R^{2}}{16}=\dfrac{3 M R^{2}}{32}\end{gathered}\) The moment of inertia of complete disc about centre \(O\) is \(I_{O}=\dfrac{1}{2} M R^{2}\) So, moment of inertia of the disc with removed portion is \(I=I_{O}=I_{O}^{\prime}=\dfrac{1}{2} M R^{2}-\dfrac{3 M R^{2}}{32}=\dfrac{13 M R^{2}}{32}\)
NEET - 2016
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365860
From a solid sphere of mass \(M\) and radius \(\mathrm{R}\) a cube of maximum possible volume is cut. Moment of inertia of cube about an axis passing through its centre and perpendicular to one of its faces is :
1 \(\dfrac{M R^{2}}{16 \sqrt{2} \pi}\)
2 \(\dfrac{4 M R^{2}}{9 \sqrt{3} \pi}\)
3 \(\dfrac{4 M R^{2}}{3 \sqrt{3} \pi}\)
4 \(\dfrac{M R^{2}}{32 \sqrt{2} \pi}\)
Explanation:
For maximum possible volume of cube \(2 R=\sqrt{3} a, a\) is side of the cube. Moment of inertia about the required axis \(I=\dfrac{m a^{2}}{6}=\rho a^{3} \dfrac{a^{2}}{6}\), where \(\rho=\dfrac{M}{\dfrac{4}{3} \pi R^{3}}\) \(I = \frac{{3M}}{{4\pi {R^3}}}\frac{1}{6}{\left( {\frac{{2R}}{{\sqrt 3 }}} \right)^5}\) \( = \frac{{3M}}{{4\pi {R^3}}}\frac{1}{6}\frac{{32{R^5}}}{{9\sqrt 3 }} = \frac{{4M{R^2}}}{{9\sqrt 3 \pi }}\)
365857
Four holes of radius \(\mathrm{R}\) are cut from a thin square plate of side \(4 \mathrm{R}\) and mass \(\mathrm{M}\). The moment of inertia of the remaining portion about \(z\) - axis is:-
Mass of each disc \(m=\dfrac{M}{16 R^{2}} \times \pi R^{2}=\dfrac{\pi}{16} \mathrm{M}\) Moment of inertia of the remaining part is \(\mathrm{I}=\mathrm{I}_{\text {square }}-4 I_{\text {disc }}\) \( = \frac{{M{{(4R)}^2}}}{6} - 4\left[ {\frac{{m{R^2}}}{2} + m{{(\sqrt 2 R)}^2}} \right]\) \( = \frac{8}{3}M{R^2} - 10m{R^2}\left[ {{\rm{ here }}m = \frac{{M\pi }}{{16}}} \right]\) \( = \left( {\frac{8}{3} - \frac{{5\pi }}{8}} \right){\rm{M}}{{\rm{R}}^2}\)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365858
A uniform square plate has a small piece \(Q\) of an irregular shape removed and glued to the centre of the plate leaving a hole behind. The moment of inertia about the \(z\)-aixs is then
1 Decreased
2 Increased
3 Changed in unpredicted manner
4 The same
Explanation:
If we keep the small piece at the centre then its moment of inertia about the central axis is zero. The mass of the remaining portion decreases and hence moment of inertia also decreases.
NCERT Exemplar
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365859
From a disc of radius \(R\) and mass \(M\), a circular hole of diameter \(R\), whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through centre?
1 \(11 M R^{2} / 32\)
2 \(9 M R^{2} / 32\)
3 \(15 M R^{2} / 32\)
4 \(13 M R^{2} / 32\)
Explanation:
Mass per unit area of disc \(=\dfrac{M}{\pi R^{2}}\) Mass of removed portion of disc, \(M^{\prime}=\dfrac{M}{\pi R^{2}} \times \pi\left(\dfrac{R}{2}\right)^{2}=\dfrac{M}{4}\) Moment of inertia of removed portion about an axis passing through centre of disc \(O\) and perpendicular to the plane of disc. \(\begin{gathered}I_{O}^{\prime}=I_{o^{\prime}}+M^{\prime} d^{2} \\=\dfrac{1}{2} \times \dfrac{M}{4} \times\left(\dfrac{R}{2}\right)^{2}+\dfrac{M}{4} \times\left(\dfrac{R}{2}\right)^{2} \\=\dfrac{M R^{2}}{32}+\dfrac{M R^{2}}{16}=\dfrac{3 M R^{2}}{32}\end{gathered}\) The moment of inertia of complete disc about centre \(O\) is \(I_{O}=\dfrac{1}{2} M R^{2}\) So, moment of inertia of the disc with removed portion is \(I=I_{O}=I_{O}^{\prime}=\dfrac{1}{2} M R^{2}-\dfrac{3 M R^{2}}{32}=\dfrac{13 M R^{2}}{32}\)
NEET - 2016
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365860
From a solid sphere of mass \(M\) and radius \(\mathrm{R}\) a cube of maximum possible volume is cut. Moment of inertia of cube about an axis passing through its centre and perpendicular to one of its faces is :
1 \(\dfrac{M R^{2}}{16 \sqrt{2} \pi}\)
2 \(\dfrac{4 M R^{2}}{9 \sqrt{3} \pi}\)
3 \(\dfrac{4 M R^{2}}{3 \sqrt{3} \pi}\)
4 \(\dfrac{M R^{2}}{32 \sqrt{2} \pi}\)
Explanation:
For maximum possible volume of cube \(2 R=\sqrt{3} a, a\) is side of the cube. Moment of inertia about the required axis \(I=\dfrac{m a^{2}}{6}=\rho a^{3} \dfrac{a^{2}}{6}\), where \(\rho=\dfrac{M}{\dfrac{4}{3} \pi R^{3}}\) \(I = \frac{{3M}}{{4\pi {R^3}}}\frac{1}{6}{\left( {\frac{{2R}}{{\sqrt 3 }}} \right)^5}\) \( = \frac{{3M}}{{4\pi {R^3}}}\frac{1}{6}\frac{{32{R^5}}}{{9\sqrt 3 }} = \frac{{4M{R^2}}}{{9\sqrt 3 \pi }}\)
365857
Four holes of radius \(\mathrm{R}\) are cut from a thin square plate of side \(4 \mathrm{R}\) and mass \(\mathrm{M}\). The moment of inertia of the remaining portion about \(z\) - axis is:-
Mass of each disc \(m=\dfrac{M}{16 R^{2}} \times \pi R^{2}=\dfrac{\pi}{16} \mathrm{M}\) Moment of inertia of the remaining part is \(\mathrm{I}=\mathrm{I}_{\text {square }}-4 I_{\text {disc }}\) \( = \frac{{M{{(4R)}^2}}}{6} - 4\left[ {\frac{{m{R^2}}}{2} + m{{(\sqrt 2 R)}^2}} \right]\) \( = \frac{8}{3}M{R^2} - 10m{R^2}\left[ {{\rm{ here }}m = \frac{{M\pi }}{{16}}} \right]\) \( = \left( {\frac{8}{3} - \frac{{5\pi }}{8}} \right){\rm{M}}{{\rm{R}}^2}\)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365858
A uniform square plate has a small piece \(Q\) of an irregular shape removed and glued to the centre of the plate leaving a hole behind. The moment of inertia about the \(z\)-aixs is then
1 Decreased
2 Increased
3 Changed in unpredicted manner
4 The same
Explanation:
If we keep the small piece at the centre then its moment of inertia about the central axis is zero. The mass of the remaining portion decreases and hence moment of inertia also decreases.
NCERT Exemplar
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365859
From a disc of radius \(R\) and mass \(M\), a circular hole of diameter \(R\), whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through centre?
1 \(11 M R^{2} / 32\)
2 \(9 M R^{2} / 32\)
3 \(15 M R^{2} / 32\)
4 \(13 M R^{2} / 32\)
Explanation:
Mass per unit area of disc \(=\dfrac{M}{\pi R^{2}}\) Mass of removed portion of disc, \(M^{\prime}=\dfrac{M}{\pi R^{2}} \times \pi\left(\dfrac{R}{2}\right)^{2}=\dfrac{M}{4}\) Moment of inertia of removed portion about an axis passing through centre of disc \(O\) and perpendicular to the plane of disc. \(\begin{gathered}I_{O}^{\prime}=I_{o^{\prime}}+M^{\prime} d^{2} \\=\dfrac{1}{2} \times \dfrac{M}{4} \times\left(\dfrac{R}{2}\right)^{2}+\dfrac{M}{4} \times\left(\dfrac{R}{2}\right)^{2} \\=\dfrac{M R^{2}}{32}+\dfrac{M R^{2}}{16}=\dfrac{3 M R^{2}}{32}\end{gathered}\) The moment of inertia of complete disc about centre \(O\) is \(I_{O}=\dfrac{1}{2} M R^{2}\) So, moment of inertia of the disc with removed portion is \(I=I_{O}=I_{O}^{\prime}=\dfrac{1}{2} M R^{2}-\dfrac{3 M R^{2}}{32}=\dfrac{13 M R^{2}}{32}\)
NEET - 2016
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365860
From a solid sphere of mass \(M\) and radius \(\mathrm{R}\) a cube of maximum possible volume is cut. Moment of inertia of cube about an axis passing through its centre and perpendicular to one of its faces is :
1 \(\dfrac{M R^{2}}{16 \sqrt{2} \pi}\)
2 \(\dfrac{4 M R^{2}}{9 \sqrt{3} \pi}\)
3 \(\dfrac{4 M R^{2}}{3 \sqrt{3} \pi}\)
4 \(\dfrac{M R^{2}}{32 \sqrt{2} \pi}\)
Explanation:
For maximum possible volume of cube \(2 R=\sqrt{3} a, a\) is side of the cube. Moment of inertia about the required axis \(I=\dfrac{m a^{2}}{6}=\rho a^{3} \dfrac{a^{2}}{6}\), where \(\rho=\dfrac{M}{\dfrac{4}{3} \pi R^{3}}\) \(I = \frac{{3M}}{{4\pi {R^3}}}\frac{1}{6}{\left( {\frac{{2R}}{{\sqrt 3 }}} \right)^5}\) \( = \frac{{3M}}{{4\pi {R^3}}}\frac{1}{6}\frac{{32{R^5}}}{{9\sqrt 3 }} = \frac{{4M{R^2}}}{{9\sqrt 3 \pi }}\)
365857
Four holes of radius \(\mathrm{R}\) are cut from a thin square plate of side \(4 \mathrm{R}\) and mass \(\mathrm{M}\). The moment of inertia of the remaining portion about \(z\) - axis is:-
Mass of each disc \(m=\dfrac{M}{16 R^{2}} \times \pi R^{2}=\dfrac{\pi}{16} \mathrm{M}\) Moment of inertia of the remaining part is \(\mathrm{I}=\mathrm{I}_{\text {square }}-4 I_{\text {disc }}\) \( = \frac{{M{{(4R)}^2}}}{6} - 4\left[ {\frac{{m{R^2}}}{2} + m{{(\sqrt 2 R)}^2}} \right]\) \( = \frac{8}{3}M{R^2} - 10m{R^2}\left[ {{\rm{ here }}m = \frac{{M\pi }}{{16}}} \right]\) \( = \left( {\frac{8}{3} - \frac{{5\pi }}{8}} \right){\rm{M}}{{\rm{R}}^2}\)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365858
A uniform square plate has a small piece \(Q\) of an irregular shape removed and glued to the centre of the plate leaving a hole behind. The moment of inertia about the \(z\)-aixs is then
1 Decreased
2 Increased
3 Changed in unpredicted manner
4 The same
Explanation:
If we keep the small piece at the centre then its moment of inertia about the central axis is zero. The mass of the remaining portion decreases and hence moment of inertia also decreases.
NCERT Exemplar
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365859
From a disc of radius \(R\) and mass \(M\), a circular hole of diameter \(R\), whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through centre?
1 \(11 M R^{2} / 32\)
2 \(9 M R^{2} / 32\)
3 \(15 M R^{2} / 32\)
4 \(13 M R^{2} / 32\)
Explanation:
Mass per unit area of disc \(=\dfrac{M}{\pi R^{2}}\) Mass of removed portion of disc, \(M^{\prime}=\dfrac{M}{\pi R^{2}} \times \pi\left(\dfrac{R}{2}\right)^{2}=\dfrac{M}{4}\) Moment of inertia of removed portion about an axis passing through centre of disc \(O\) and perpendicular to the plane of disc. \(\begin{gathered}I_{O}^{\prime}=I_{o^{\prime}}+M^{\prime} d^{2} \\=\dfrac{1}{2} \times \dfrac{M}{4} \times\left(\dfrac{R}{2}\right)^{2}+\dfrac{M}{4} \times\left(\dfrac{R}{2}\right)^{2} \\=\dfrac{M R^{2}}{32}+\dfrac{M R^{2}}{16}=\dfrac{3 M R^{2}}{32}\end{gathered}\) The moment of inertia of complete disc about centre \(O\) is \(I_{O}=\dfrac{1}{2} M R^{2}\) So, moment of inertia of the disc with removed portion is \(I=I_{O}=I_{O}^{\prime}=\dfrac{1}{2} M R^{2}-\dfrac{3 M R^{2}}{32}=\dfrac{13 M R^{2}}{32}\)
NEET - 2016
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365860
From a solid sphere of mass \(M\) and radius \(\mathrm{R}\) a cube of maximum possible volume is cut. Moment of inertia of cube about an axis passing through its centre and perpendicular to one of its faces is :
1 \(\dfrac{M R^{2}}{16 \sqrt{2} \pi}\)
2 \(\dfrac{4 M R^{2}}{9 \sqrt{3} \pi}\)
3 \(\dfrac{4 M R^{2}}{3 \sqrt{3} \pi}\)
4 \(\dfrac{M R^{2}}{32 \sqrt{2} \pi}\)
Explanation:
For maximum possible volume of cube \(2 R=\sqrt{3} a, a\) is side of the cube. Moment of inertia about the required axis \(I=\dfrac{m a^{2}}{6}=\rho a^{3} \dfrac{a^{2}}{6}\), where \(\rho=\dfrac{M}{\dfrac{4}{3} \pi R^{3}}\) \(I = \frac{{3M}}{{4\pi {R^3}}}\frac{1}{6}{\left( {\frac{{2R}}{{\sqrt 3 }}} \right)^5}\) \( = \frac{{3M}}{{4\pi {R^3}}}\frac{1}{6}\frac{{32{R^5}}}{{9\sqrt 3 }} = \frac{{4M{R^2}}}{{9\sqrt 3 \pi }}\)
