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CBSE Questions for Class 11 Engineering Physics Systems Of Particles And Rotational Motion Quiz 10 - MCQExams.com
CBSE
Class 11 Engineering Physics
Systems Of Particles And Rotational Motion
Quiz 10
The velocity of centre of mass of the system as shown in the figure.
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$${(\dfrac{2-2\sqrt{3}}{3})}\hat{i}-\dfrac{1}{3} \hat{j}$$
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$${(\dfrac{2+2\sqrt{3}}{3})}\hat{i}-\dfrac{2}{3} \hat{j}$$
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$$4\hat{i}$$
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None of these
Explanation
Let $$V_x$$ be the $$x-$$ component of velocity of CM, $$V_x$$
$$=\dfrac{m_1\times v_{1x}+m_2\times v_{2x}}{m_1+m_2}=\dfrac{1\times 2+2\times 2\times \dfrac{\sqrt 3}{2}}{1+2}=\dfrac{2+2\sqrt 3}{3}\,m/s$$
Let $$V_y$$ be the $$y-$$ component of velocity of CM, $$V_y$$
$$=\dfrac{m_1\times v_{1y}+m_2\times v_{2y}}{m_1+m_2}=\dfrac{1\times 0+2\times 2\times \dfrac 12}{1+2}=\dfrac 23\,m/s$$
Velocity of CM
$$\sqrt{V_x^2-V_y^2}=(\dfrac{2+2\sqrt 3}{2})\hat i-\dfrac 23 \hat j$$
The M.I of a thin rod of length $$l$$ about the perpendicular axis through its centre is $$I$$. The M.I of the square structure made by four such rods about a perpendicular axis to the plane and through the centre will be-
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$$\dfrac{4}{3}I$$
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$$\dfrac{8}{3}I$$
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$$\dfrac{1}{2}I$$
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$$16I$$
Explanation
$$\begin{array}{l}\text { Given- The moment of Inertia of a Rod } \\\text { about perpendicular Bisector }=I=\frac{M L^{2}}{12}-0 \\\Rightarrow\text { I about the given axis will be- } \\ I_{A}=\frac{M L^{2}}{12}+\frac{M L^{2}}{4}=\frac{M L^{2}}{3} \\\text { (By parallel axis theorem) }\end{array} $$
$$\begin{array}{l}\Rightarrow \text { For four such Rods } \\\qquad \quad I_{\text {Total }}=4 \times I_{A}=\frac{4}{3} M L^{2} \\\Rightarrow I_{\text {Total }}=\frac{4}{3} \times 12 I=16 \text { I From eq } 0 \\\text { Hence, option (D)is correct. }\end{array}$$
A uniform rod of mass $$m$$ and length $$l$$ hinged at its end is released from rest when it is in the horizontal position. The normal reaction at the hinged when the rod becomes vertical is
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$$\cfrac {mg}4$$
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$$\cfrac 5 2 mg$$
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$$\cfrac {mg} 6$$
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$$\cfrac 72 mg$$
Explanation
$$\begin{array}{l} mg\frac { l }{ 2 } =\dfrac { l }{ 2 } \left( { \dfrac { { m{ l^{ 2 } } } }{ 3 } } \right) { w^{ 2 } } \\ \Rightarrow { w^{ 2 } }=\dfrac { { 3g } }{ l } ....\left( 1 \right) \\ N-mg=m{ w^{ 2 } }\left( { \dfrac { l }{ 2 } } \right) \\ \Rightarrow N=mg+\dfrac { { m.3g } }{ 2 } \\ =\dfrac { 5 }{ 2 } mg \\ Hence, \\ option\, \, B\, \, is\, \, correct\, answer. \end{array}$$
If the kinetic energy of a body is increased by 300% then determine the percentage increase in the momentum
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100 %
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150 %
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$$ \sqrt {300%} $$
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175 %
Explanation
When the kinetic energy increases by $$300$$%,new kinetic energy will be
$$K'=K+300$$%K
$$K'=K+3K$$
$$K'=4K$$.............(i)
We know that,
K.E, $$K = \frac{{{P^2}}}{{2m}}$$ 9$$m$$ is the mass of object0
Therefore, eq.(i) become
$$\begin{array}{l} \frac { { P{ '^{ 2 } } } }{ { 2m } } =4\frac { { { p^{ 2 } } } }{ { 2m } } \\ p{ '^{ 2 } }=4{ p^{ 2 } } \\ p'=\sqrt { 4 } { p^{ 2 } } \\ p'=2p...........\left( { ii } \right) \end{array}$$
% change in momentum $$ = \frac{{p' - p}}{p} \times 100$$
$$ = \frac{{2p - p}}{p} \times 100$$%$$=1 \times 100=100$$%
Therefore, increases in momentum is $$100$$%.
The moment of inertia of a uniform thin rod of length $$L$$ and mass $$M$$ about an axis passing through a point at a distance of $$\cfrac{L}{3}$$ from one of its ends and perpendicular to the rod is
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$$\cfrac{7M{L}^{2}}{48}$$
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$$\cfrac{M{L}^{2}}{9}$$
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$$\cfrac{M{L}^{2}}{12}$$
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$$\cfrac{M{L}^{2}}{3}$$
Explanation
$$\begin{array}{l}\text { know - } \\\text { moment of Inertia of a rod about an axis } \\\text { passing through its centre- } \\\qquad I=\frac{M L^{2}}{12}\end{array}$$
$$\begin{array}{l}\text { Therefore moment of Inertia about the given } a x \text { is } \\\text { by parallel axis theorem- } \\\qquad I^{\prime}=I+M d^{2}=\frac{M L^{2}}{12}+M\left(\frac{L}{6}\right)^{2} \\ I^{\prime}=\frac{M L^{2}}{9}\end{array} $$
A rod of moment of inertia I and length L is suspended from a fixed end and given small oscillations about the point of suspension , the restoring torque is found to be -(mgL/2) $$sin\theta $$ What will be the angular equation of motion of the SHM
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$$\dfrac { { d }^{ 2 }\theta }{ { dt }^{ 2 } } +\left( mgL/I \right) \theta =0$$
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$$\dfrac { { d }^{ 2 }\theta }{ { dt }^{ 2 } } +\left( mgL/2I \right) \theta =0$$
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$$\dfrac { { d }^{ 2 }\theta }{ { dt }^{ 2 } } +\left( 2mgL/I \right) \theta =0$$
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$$\dfrac { { d }^{ 2 }\theta }{ { dt }^{ 2 } } +\left( mgL/4I \right) \theta =0$$
Explanation
$$\begin{aligned}\text { Given }-& * \text { Restoring tooque } \tau=-\frac{m g L}{2} \sin\theta \\ & * \text { Moment of Inertia }=I\end{aligned} $$
$$\begin{array}{l}\text { Restoring torque- } \\\qquad \tau=-\frac{m g L}{2} \sin \theta \\\text { for very small deflection }\sin\theta\approx\theta \\\Rightarrow \tau=\frac{m g L}{2} \theta \\\text { and }\alpha=\frac{d^{2}\theta}{d t^{2}} \\\Rightarrow\quad I \cdot \frac{d^{2} \theta}{d t^{2}}=\frac{-m g L}{2} \theta \\\Rightarrow\quad\frac{d^{2} \theta}{d t^{2}}+\frac{m g L}{2 I} \theta=0\end{array}$$
If a square of side $$R/2$$ is removed from a uniform circular disc of radius $$R$$ as shown in the figure, the shift in centre of mass is
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$$\dfrac{R}{4 \pi - 1}$$
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$$\dfrac{R}{2(4 \pi - 1)}$$
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$$\dfrac{R}{3(4 \pi - 1)}$$
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$$\dfrac{R}{4(4 \pi - 1)}$$
The moment of inertia of a metre scale of mass 0.6 kg about an axis perpendicular to the scale and located at the 20 cm position on the scale in kg $$m^2$$ is (Breadth of the scale is negligible)
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0.074
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0.104
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0.148
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0.208
Explanation
Hence, the option $$B$$ is the correct answer.
Three identical rods, each of length I, are joined to form a rigid equilateral triangle. Its radius of gyration about an axis passing through a corner and perpendicular to the plane of the triangle is
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$$\dfrac { \ell }{ \sqrt { 3 } } $$
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$$\dfrac { \ell }{ \sqrt { 2 } } $$
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$$\dfrac { \ell }{ \sqrt { 5 } } $$
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$$\dfrac { \ell }{ \sqrt { 7 } } $$
Explanation
$$\begin{array}{l}\text { we know } \\\text { Moment of Inertia of a Rod about its axis } \\\text { and perpendicular is - } \\\qquad I=\frac{M L^{2}}{3} \\\Rightarrow I_{0 A}=I_{0 B}=\frac{M L^{2}}{3} \\\text { * Moment of Inertia of a Rod about its centre } \\\qquad I_{c}=\frac{M L^{2}}{12}\end{array}$$
$$\begin{array}{l}\Rightarrow \text { By parallel axis theorem- } \\\qquad \begin{array}{rl} I_{A B}=I_{c}+M d^{2} & d=\text { distanu from the Axis } \\ I_{A B}=\frac{M L^{2}}{12}+M\left(\frac{\sqrt{3} L}{2} L\right)^{2}=\frac{5 M L^{2}}{6}\end{array} \\\Rightarrow \text { Total moment of Inertia I }=I_{O A}+I_{D B}+I_{A B}=\frac{3 M L^{2}}{2} \\\Rightarrow \text { Radius of gyration (R)- }\end{array}$$
$$ 3 M R^{2}=\frac{3}{2} M L^{2}\quad\Rightarrow\left[R=\frac{L}{\sqrt{2}}\right] \text { option(B) } $$
Force F=300 N acting vertically upward at x=2 m, y=2 m. The magnitude of moment of force about origin is
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600 Nm
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660 Nm
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300 Nm
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330 Nm
Explanation
$$\begin{array}{l}\text { We have force vector }\vec{F}=300 \hat{j} \\\text { and radial vector of point } \vec{\mu}=2 \hat{i}+2 \hat{j} \\\Rightarrow\text { Force moment }=\vec{r}\times\vec{F} \\ =(2 \hat{i}+2 \hat{j}) \times(300 \hat{j}) \\ =600 \hat{k} \\\text { Force moment }\bar{\tau}=600\mathrm{Nm} \\\text { Hence, option }(A) \text { is correct. }\end{array}$$
A solid cylinder of mass $$500 \mathrm{g} $$ and radius $$10 \mathrm{cm} $$ .has radius of gyration about the axis $$ y y' $$ as given
below is
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$$10 \sqrt{2} \mathrm{cm} $$
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$$ \dfrac{10}{\sqrt{2}} \mathrm{cm} $$
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$$5 \sqrt{2} \mathrm{cm} $$
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$$ \frac{5}{\sqrt{2}} \mathrm{cm} $$
Explanation
$$\begin{aligned}\text { Given - } & * \text { mass of the cylinder } M=500 \mathrm{~g} \\ & * \text { Radius of the cylinder } R=10\mathrm{~cm} .\end{aligned} $$
$$\begin{array}{l}\text { we know that moment of Inertia of Cylindur - } \\\qquad I=\frac{M R^{2}}{2} \\\text { Let the Radius of gyration of the cylindur be (K) } \\\Rightarrow M k^{2}=\frac{M R^{2}}{2} \\\Rightarrow k=\frac{R}{\sqrt{2}}=\frac{10}{\sqrt{2}} \mathrm{~cm} .\end{array}$$
Two identical uniform rod each of mass $$m$$ and length $$l$$ joined perpendicular to each other. An axis passes through junction and in the plane of rods. Then $$M.I.$$ of system about the axis is
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$$\dfrac{1}{3} ml^2$$
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$$\dfrac{1}{3\sqrt{2}} ml^2$$
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$$ ml^2$$
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$$\dfrac{ml^2}{2}$$
Explanation
$$\begin{array}{l}\text { MOI of rod at an end making }\theta \text { degree angle with } \\\text { axis is given by } \frac{m L^{2}}{3} \operatorname{sin}^{2}\theta\end{array}$$
$$\begin{aligned} M .0 . I \text { af system } &=\frac{m L^{2}}{3} \sin ^{2}\theta \times 2 \\ &=\frac{m L^{2}}{3} \times \frac{1}{2} \times 2 \\ &=\frac{m L^{2}}{3}\end{aligned}$$
The temperature of a thin uniform rod increase by $$\Delta t$$ If moment of intertia 1 about an axis perpendicular to its length then its moment of increase by
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0
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$$\alpha I\Delta t$$
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$$2\alpha I\Delta t$$
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$${ \alpha }^{ 2 }I\Delta t$$
Explanation
$$\begin{array}{l}\text { Given - } * \text { Temperatwe of Rod increased by } \Delta t \\\text { Let the Length of the Rod be (L) and its } \\\text { coefficient of linear expansion be ( }\alpha\text { ) } \\\text { Therefore its change in Length will be } \\\qquad \Delta L=L_{0} \cdot \alpha \Delta t\end{array}$$
$$\begin{array}{l}\text { we have moment of Inertia of Rod about its } \\\text { centre perpendicular to its Length }=I \\ I=\frac{M L^{2}}{12}\end{array} $$
$$\begin{array}{l}\Rightarrow \text { By Error analysis we get- } \\\qquad \frac{\Delta I}{I}=2\frac{\Delta L}{L} \quad \Delta I=\text { change in moment of } \\\text { Inertia }\end{array}$$
$$\begin{aligned}\Rightarrow \Delta I &=\frac{2\Delta L}{L_{0}} \cdot I \\\Delta I &=2\left(\frac{\left.L_{0} \alpha \Delta t\right)}{L_{0}} \cdot I=2 \alpha I \Delta t\right.\\\Delta I &=2 \alpha I \alpha t \\\text { Hence, option }(C)\text { is correct. }\end{aligned}$$
Ratio of radii of gyration of a hollow & solid sphere of the same radii about the axis which is tangent to the sphere is
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$$\sqrt { \dfrac73 } $$
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$$\dfrac {5}{\sqrt {21}} $$
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$$\sqrt { \dfrac { 4 }{ 5 } } $$
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$$\dfrac{25}9$$
Explanation
$$\begin{array}{l}\text { We know that moment of Inertia of a Solid } \\\text { sphere about its axis is - } \\\qquad I_{\text {solid sphere }}=\frac{2}{5}\mathrm{MR}^{2} \\\text { Theretore moment of Inertia about an axis } \\\text { Tangent to the sphere- } \\\qquad I^{\prime}=\frac{2}{5} M R^{2}+M R^{2}=\frac{7}{5} M R^{2} \\\text { I }_{\text {Hollow sphere }}=\frac{2}{3} M R^{2}\end{array}$$
$$\begin{array}{l}\text { Therefore Moment of Inertia about an axis tangent } \\\text { to the sphere - } \\\qquad I^{\prime}=\frac{2}{3}\mathrm{MR}^{2}+\mathrm{MR}^{2}=\frac{5}{3} \mathrm{MR}^{2} \\\text { Radius of gyration of solid sphere be }\mathrm{K}_{5} \\\Rightarrow \quad M \cdot K_{3}^{2}=\frac{7}{5} M R^{2} \\\Rightarrow k_{5}=\sqrt{\frac{7}{5}} R\end{array}$$
$$\begin{array}{l}\text { Radius of gyration of Hollow sphere be } k_{H} \\\Rightarrow M K_{H}^{2}=\frac{5}{3} M R^{2} \\\Rightarrow K_{H}=\sqrt{\frac{5}{3}} R \\\Rightarrow \frac{k_{H}}{K_{S}}=\frac{5}{\sqrt{21}} \\\Rightarrow\text { Hence, option (B) is correct. }\end{array}$$
A shown in figure two uniform rods of mass M and length L each are joined . what is the moment of inertia of system about an axis perpendicular to plane through end A?
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$$ \frac { 5ML^2}{3} $$
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$$ \frac { 7ML^2}{3} $$
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$$ \frac { 5ML^2}{12} $$
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$$ \frac { 7ML^2}{12} $$
Explanation
$$\begin{array}{l}\text { Mass of each rod }=m \\\text { lengthof each rod }=L \\\text { Let o be the centre of rod } C B \text { . }\end{array}$$
$$\begin{aligned} I_{1} &=I_{\text {of } A C \text { about } A} \\ &=\frac{m l^{2}}{3} \\ I_{2} &=I \text { of } C B \text { a bout } A \\ &\left.=I_{C M}+M(O A)^{2} \quad \quad \quad (: A C^{2}+O C^{2}=A D^{2}\right)\end{aligned}$$
$$\begin{array}{l} =\frac{m l^{2}}{12}+m\left(l^{2}+\left(\frac{l}{2}\right)^{2}\right) \\ =\frac{16}{12} m l^{2}=\frac{4}{3} m l^{2} \\ I=\frac{m l^{2}}{3}+\frac{4}{3} m l^{2} \\ I \text { total }=\frac{5}{3} m l^{2}\end{array}$$
If moment of inertia of a point particle at a distance r from an axis would have defined as $$L = mr$$ instead of L=$${ mr }^{ 2 }$$, then moment of inertia of a uniform rod of mass M and length L about an axis passing through centre of mass and perpendicular to the rod will be (moment of inertia is still a scalar )
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$$\dfrac { ML }{ 12 } $$
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$$\dfrac { { ML }^{ 2 } }{ 12 } $$
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$$\dfrac { ML }{ 4 } $$
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Zero
Explanation
$$\begin{array}{l}\text { Let a small element on the rod of Length }(d x) \text { at } \\\text { a distance } x \text { from the centre- } \\\text { mass of the element will be }=\frac{m \cdot d x=d m}{L} d x=d m \\\text { Moment of Inertia of mall element will be } \\ d I=d m \cdot r=\left(\frac{m}{L} \cdot d x\right)\cdot(x)\end{array} $$
$$\begin{array}{l}\Rightarrow d I=\frac{m}{L} \cdot x \cdot d x \\\text { Integrating both Sides with limits - } \\\qquad \begin{array}{l}\text { If } \\\int_{0}^{L} d I=\int_{-L / 2}^{L / 2} \frac{m}{L} \cdot x \cdot d x \quad=2 \int_{0}^{L / 2}\frac{m}{L} \cdot x \cdot d x \text { (Scalar) } \\ I=\left.2 \frac{m}{L} \frac{x^{2}}{2}\right|_{0} ^{L / 2}=\frac{m}{L} \cdot\left(\frac{L^{2}}{4}\right)=\frac{m L}{4}\end{array}\end{array} $$
Three identical uniform rods each of length 1 m and mass 2 kg are arranged to form an equilateral triangle. What is the moment of inertia of the system about an axis passing through one corner ans perpendicular to the plane of the triangle:-
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$$4 kg-{ m }^{ 2 }$$
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$$3 kg-{ m }^{ 2 }$$
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$$2 kg-{ m }^{ 2 }$$
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$$5 kg-{ m }^{ 2 }$$
Explanation
$$\begin{array}{l}\text { On calculating d we get } d=\frac{\sqrt{3}}{2} \text { e } \\\text { Moment of Inertia of whole system is equal to } \\\text { moment of Inertia of all the three Rods about c. } \\\text { I }_{A C}=I_{B C}=\frac{m l^{2}}{3}\end{array}$$
$$\begin{array}{l}\text { Calculating I }_{A B} \text { - } \\\text { By Parallel axis theorem we can soly- } \\\qquad I_{A B}=\frac{m e^{2}}{12}+m d^{2}=\frac{m\ell^{2}}{12}+\frac{3}{4} m l^{2} \\ I_{A B}=\frac{5}{6} m l^{2}\end{array} $$
$$\begin{array}{l}\Rightarrow \text { Moment of Inertia of System } I=I_{A B}+I_{B C}+I_{A C} \\ I=\frac{3}{2} m l^{2}=3\mathrm{~kg}\mathrm{~m}^{2}\end{array} $$
The moment of inertia of uniform thin rod of mass m and length l about two axis PQ and RS passing through centre of rod C and in the plane of the rod are $$ I_{PQ} and I_{RS} $$ respectively. then $$ I_{PQ}+I_{RS}$$ is equal to
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$$ \frac {ml^2}{3} $$
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$$ \frac {ml^2}{2} $$
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$$ \frac {ml^2}{4} $$
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$$ \frac {ml^2}{12} $$
Explanation
$$\begin{array}{l}\text { we know that in } F i g-2 \\\text { momentot Inertia of Rod about an axis } \\\text { inclined at an angle }(\theta) \text { is given as - } \\\qquad I=I_{0} \cdot \sin ^{2} \theta\quad \quad I_{0}=\text { Moment of Invetia about centre } \\ I=\frac{M L^{2}}{12} \sin ^{2} \theta \\\text { and } I_{R S}=\frac{M L^{2}}{12}\sin ^{2}\left(72^{\circ}\right)=\frac{M L^{2}}{12} \cos ^{2}\left(18^{\circ}\right)\end{array} $$
$$\begin{aligned}\Rightarrow I_{P Q}+I_{R S} &=\frac{M L^{2}}{12}\left[\sin ^{2}\left(18^{\circ}\right)+\cos ^{2}\left(18^{\circ}\right)\right] \\ &=\frac{M L^{2}}{12}\end{aligned} $$
A point at which a whole weight of body act vertically downward is ________.
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centre of gravity
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centre of mass
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centre of force
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centre of acceleration
A uniform rod of mass $$m$$ is bent into the form of a semicircle of radius $$R$$. The moment of inertia of the rod about an axis passing through $$A$$ and perpendicular to the plane of the paper is
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$$mR^2$$
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$$mR$$
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$$3mR$$
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$$2mR$$
Explanation
$$\begin{array}{l}\text { Given a rod of mass } m \text { is bent into a semicircle } \\\text { of radius } R \text { . }\end{array}$$
$$\begin{array}{l}\text { Let us consider another same piece and Connect } \\\text { them to forma circlular rang of mass } 2\mathrm{~m} \text { and radrus } R \text { . }\end{array}$$
$$\begin{array}{l}\text { Now let MOI of each semicircle be I' about centre } \\\text { MOI of ring (system of two semicircle's) }\end{array}$$
$$\begin{array}{l} I_{\text {System }}=I^{\prime}+I^{\prime} \\ (2 m) R^{2}=2 I^{\prime} \\\qquad\underline{I^{\prime}=m R^{2}} \\\text { So moment of Inertia of Semicircle about centre is } \\ I^{\prime}=m R^{2}\end{array}$$
Two particles which are initially at rest, move towards each other under the action of their intrnal attraction. If their speeds are V and 2V at any instant, then the speed of centre of mass of the system will be
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$$V$$
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$$2V$$
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$$Zero$$
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$$1.5V$$
Explanation
Two particle of masses $$ m_1 and m_2 $$ initially at rest start moving towards each other under their mutual force of attraction, The speed of the center of mass at any time t, when they are at distance r apart , is
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zero
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$$ \left( G\frac { m_{ 1 }m_{ 2 } }{ r^{ 2 } } ,\frac { 1 }{ \quad m_{ 1 } } \right) t $$
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$$ \left( G\frac { m_{ 1 }m_{ 2 } }{ r^{ 2 } } ,\frac { 1 }{ \quad m_{ 2 } } \right) t $$
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$$ \left( G\frac { m_ 1m_ 2 }{ r^ 2 } ,\frac { 1 }{ m_ 1\quad +\quad m_ 2 } \right) t $$
Explanation
$$F_{ext} = 0 $$
$$P_{i} = P_{f} $$
$$ 0 = m_{1} V_{1} + m_{2} V_{2} $$
$$ = m_{1} V_{1} - m_{2} V_{2} $$
Now, $$V_{CM} = \dfrac{m_{1} V_{1} + m_{2} V_{2}}{m_{1} + m_{2}} = \dfrac{m_{1} V_{1} - m_{2} V_{2}}{m_{1} + m{2}} = 0$$
The following bodies are made to roll up (without slipping) the same inclined plane from a horizontal plane: (i) a ring of radius $$R$$, (ii) a solid cylinder of radius $$\cfrac{R}{2}$$ and (iii) a solid sphere of radius $$\cfrac{R}{4}$$. If in each case, the speed of the centre of mass at the bottom of the incline is same, the ratio of the maximum heights they climb is:
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$$4:3:2$$
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$$14:15:20$$
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$$10:15:7$$
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$$2:3:4$$
Explanation
$$\cfrac { 1 }{ 2 } \left( m+\cfrac { 1 }{ { R }^{ 2 } } \right) { v }^{ 2 }=mgh\quad $$
if radius of gyration is $$k$$, then
$$h=\cfrac { \left( 1+\cfrac { { k }^{ 2 } }{ { R }^{ 2 } } \right) { v }^{ 2 } }{ 2g } ,\cfrac { { k }_{ ring } }{ { R }_{ ring } } =1,\cfrac { { k }_{ solid\quad cylinder } }{ { R }_{ solid\quad cylinder } } =\cfrac { 1 }{ \sqrt { 2 } } $$
$$\cfrac { { k }_{ solid\quad sphere } }{ { R }_{ solid\quad sphere } } =\sqrt { \cfrac { 2 }{ 5 } } $$
$${ h }_{ 1 }:{ h }_{ 2 }:{ h }_{ 3 }::(1+2)\left( 1+\cfrac { 1 }{ 2 } \right) :\left( 1+\cfrac { 2 }{ 5 } \right) ::20:15:14\quad $$
A solid sphere of radius $$R$$ has total charge $$2Q$$ and volume charge density $$\rho = kr$$ where $$r$$ is distance from centre. Now charges $$Q$$ and $$-Q$$ are placed diametrically opposite at distance $$2a$$ where $$a$$ is distance form centre of sphere such that net force on charge $$Q$$ is zero then relation between $$a$$ and $$R$$ is
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$$a = R/2$$
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$$a = R$$
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$$a = 2R$$
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$$a = 3R/4$$
Explanation
$$\int_{0}^{R} kr 4\pi r^{2} dr = 2Q$$
$$k\pi R^{4} =2Q ..... (1)$$
$$\dfrac {KQ^{2}}{4a^{2}} = Q \int_{0}^{a} \dfrac {Kkr4\pi r^{2}dr}{r^{2}}$$
$$\dfrac {KQ^{2}}{4a^{2}} = QKk4\pi \dfrac {a^{2}}{2} \Rightarrow \dfrac {KQ^{2}}{4a^{2}} = Q.K. \dfrac {2Q}{\pi R^{4}} 4\pi \dfrac {a^{2}}{2}$$
$$\dfrac {1}{4a^{2}} = \dfrac {4a^{2}}{R^{2}} \Rightarrow R^{4} = 16a^{4}$$
$$R = 2a \Rightarrow a = \dfrac {R}{2}$$.
Two forces, each of magnitude $$F$$, act at points $$V$$ and $$W$$ on an object.
The two forces form a couple. The shape of the object is a right-angled triangle with sides of
lengths $$x$$ and $$y$$, as shown.
Which expression gives the torque exerted by the couple?
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$$Fx$$
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$$Fy$$
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$$2Fx$$
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$$2Fy$$
Explanation
Couple refers to two parallel forces that are equal in magnitude, opposite in sense and are at a perpendicular distance from each other.
Since the perpendicular distance is $$y$$.
So, the torque exerted by the couple is $$F \times y$$
Two particles $$A$$ and $$B$$ of equal masses have velocities $$v_A = 2i + j$$ and $$v_B = - i + 2j$$. The particles move with accelerations $$a_A = - 4i- j$$ and $$a_B = - 2i + 3j$$ respectively. The centre of mass of the two particles move along
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A straight line
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a parabola
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a circle
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an ellipse LINE
Explanation
So they will form parabola
A wooden plank rests in equilibrium on two rocks on opposite sides of a narrow stream. Three forces P, Q and R act on the plank. How are the sizes of the forces related?
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P$$+$$Q$$=$$R
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P$$+$$R$$=$$Q
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P$$=$$Q$$=$$R
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P$$=$$Q$$+$$R
Explanation
Given that ,
A wooden plank rests in equilibrium on two rocks
We know that
In equilibrium the total force acting on the body will be zero
which means that ,
Total force acting upwards= Total force acting upwards
$$P+R= Q$$
Distance of the centre of mass of a solid uniform cone from its vertex is $$Z_{0}$$ If the radius and its height is h then $$Z_{0}$$ is equal to
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$$\dfrac{h^{2}}{4R}$$
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$$\dfrac{3h}{4}$$
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$$\dfrac{5h}{8}$$
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$$\dfrac{3h^{r}}{8R}$$
Explanation
Water is drawn from a well in a $$5\ kg$$ drum of capacity $$55\ L$$ by two ropes connected to the top of the drum. The linear mass density of each rope is $$0.5\ kgm^{-1}$$. The work done in lifting water to the ground from the surface of water in the well $$20\ m$$ below is $$[g=10\ ms^{-2}]$$
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$$1.4\times 10^{4}\ J$$
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$$1.5\times 10^{4}\ J$$
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$$9.8\times 10\times 6\ J$$
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$$18\ J$$
A uniform rod of length of $$1$$m and mass of $$2$$kg is attached to a side support at $$0$$ as shown in the figure. The rod is at equilibrium due to upward force T acting at P. Assume the acceleration due to gravity as $$10m/s^2$$. The value of T is?
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$$0$$
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$$2$$N
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$$5$$N
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$$10$$N
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$$20$$N
Explanation
The one end of the uniform rod is fixed and force $$T$$ is acting in upward direction.
Torque about O,
$$ mg \times \dfrac{l}{2} = T \times l $$
$$T=\dfrac{mg}{2}$$
$$=\dfrac{2\times 10}{2}$$
$$=10N$$.
Find the velocity of centre of mass of the system shown in the figure
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$$\left(\dfrac{2+2\sqrt{3}}{3}\right)\hat{i}-\dfrac{2}{3}\hat{j}$$
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$$4\hat{i}$$
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$$\left(\dfrac{2-2\sqrt{3}}{3}\right)\hat{i}-\dfrac{1}{3}\hat{j}$$
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$$None\ of\ these$$
A block of mass $$m$$ having coefficient of friction $$\mu$$ with the floor $$F$$ is placed at one end of the spring. The spring is attached to this block and a vertical shafts. The floor with the shaft is given an angular acceleration $$\alpha$$ Then:
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The spring cannot elongate before $$t=\dfrac {\sqrt {\mu g}}{l \alpha^2}$$
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The spring elongates as soon as the rotation starts
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The stored energy in the spring goes on increasing right from $$t=0$$ onwards
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The maximum spring force acts at $$t=\dfrac {\sqrt {\mu g}}{l \alpha^2}$$
Mark correct option or options:
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Nagpur can be said to the geographical centre of India
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The population centre of India may be Uttar Pradesh
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The population centre may be coincided with geographical centre
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All the above
Explanation
Correct option is $$D$$ i.e. All the above.
For option $$A$$,
Nagpur is exactly at India 's geographic middle point and here is the zero mile marker.
For option $$B$$
Kanpur is a city in the Indian state of Uttar Pradesh.The greater metropolis is split into two districts: the Kanpur Nagar urban district, and the Kanpur Dehat rural district.
For option $$C$$,
I
n demographics, a population center is a geographic point which describes a center point of the population of the region.
In which of the following cases the centre of mass of a rod is certainly not at its geometrical centre?
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The density continuously decreases from left to right
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The density continuously increases from left to right
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The density decreases from left to right upto the centre and then increases
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Both $$(a)$$ and $$(b)$$ are correct
A cracker is thrown into air with a velocity of $$10$$ m/s at an angle of $$45^o$$ with the vertical. When it is at a height of $$0.5$$m from the ground, it explodes into a number of pieces which follow different parabolic paths. What is the velocity of centre of mass, when it is at a height of $$1$$m from the ground? ($$g=10m/s^2$$)
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$$4\sqrt{5}$$ m/s
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$$2\sqrt{5}$$ m/s
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$$5\sqrt{4}$$ m/s
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$$10$$ m/s
Explanation
Motion of centre of mass is exactly similar to the motion of a body had it not exploded.
$$u_x=u\cos\theta =\dfrac{10}{\sqrt{2}}m/s, u_y=u\sin\theta =\dfrac{10}{\sqrt{2}}m/s$$
$$v_x=u_x=\dfrac{10}{\sqrt{2}}m/s$$
(since there is no change in the horizontal velocity)
$$v^2_y-u^2_y=2(-g)(h)$$
$$\Rightarrow v^2_y=\dfrac{100}{2}-2\times 10\times 1=30$$
Therefore, net velocity of $$CM=\sqrt{v^2_x+v^2_y}$$
$$=\sqrt{\dfrac{100}{2}+30}=\sqrt{80}=4\sqrt{5}m/s$$.
A rigid body is in pure rotation , that is undergoing fixed axis rotation. then which of the following statements are true ?
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You can find two points in the body in a plane perpendicular to the axis of rotation having the same velocity
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You can find two points in the body in a plane perpendicular to the axis of rotation having the same acceleration.
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Speed of all the particles lying on the curved surface of a cylinder whose axis coincides with the axis of rotation is the same
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Angular speed of the body is the same seen from any point in the body
Explanation
All points in the body , in plane perpendicular to the axis or rotation , revolve in concentric circles all points lying on the circle of same radius have same speed ( and also same magnitude of acceleration ) but different directions of velocity
Hence there can not be two points in the given plane with same velocity or with same acceleration. as mention above,points lying on circle of same radius have same speed.
Angular speed of body at any instant w.r.t.any point on the body is same by definition
Option C and D are correct.
The mathematical statement $$\vec{v}=\vec{v_c}+\vec{v'}$$, where $$\vec{v_c}$$ is the velocity of centre of mass, $$\vec{v'}$$ is the velocity of the point with respect to the centre of mass and $$\vec{v}$$ is the total velocity of the point with respect to the ground.
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Is true for a rolling sphere
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Is true for a block moving on a frictionless horizontal surface
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Is true for a rolling cylinder
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None of these
Explanation
with respect to the ground, the toal velocity of the point is
$$ \vec{v} = \vec{v_c} + \vec{{v}'} $$
this equation is for rolling condition (no slipping) and there is no friction between surface and rolling block.
If there will be friction, this will affect the velocity equation.
So, option 'B' is true
The rolling object may be any type so sphere and rolling cylinder both will satisfy this equation so option 'A' and option 'C' is true.
A ball falls vertically onto a floor with momentum p, and then bounces repeatedly. If the coefficient of restitution is e, then the total momentum impareted by the ball on the floor till the ball comes to rest is?
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$$p(1+e)$$
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$$\dfrac{p}{1-e}$$
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$$p\left(1+\dfrac{1}{e}\right)$$
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$$p\left(\dfrac{1+e}{1-e}\right)$$
Which of the following statements are correct for instantaneous axis of rotation?
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Acceleration of every point lying on the axis must be equal to zero.
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velocity of a point at a distance r from the axis is equal to $$ r \omega $$
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if moment of inertia of a body about the axis is I and angular velocity is $$ \omega $$ then kinetic energy of the body is equal $$ I \omega^2 / 2$$
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Moment of inertia of a body is least about instantaneous axis of rotation among all the parallel axes.
Three-point masses $$ m_1, m_2\ and\ m_3 $$ are located at the vertices of an equilateral triangle of side $$ á'$$ what is the moment of inertia of the system about an axis along the altitude of the triangle passing through $$ m_1 $$?
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$$ (m_1 +m_2) \dfrac {a^2}{4} $$
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$$ ( m_2 +m_2) \dfrac {a^2}{4} $$
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$$ ( m_1+ m_3) \dfrac {a^2}{4} $$
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$$ (m_1+ m_2 + m_3) \dfrac {a^2}{4} $$
Explanation
Since the axis passes through $$m_1$$ and it's a point mass, it 's contribution to moment of Inertia is zero .
For $$m_1$$ and $$m_2$$ distance from the axis is $$\dfrac{a}{2}$$ as altitude divides the base of equilateral triangle into half.
So, $$I=(m_2+m_3)\dfrac{a^2}{4}$$
Hence Option B is correct
A ball kept in a closed box moves in the box making collisions with the walls. The box is kept on a smooth surface. The velocity of the centre of mass.
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Of the box remains constant
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Of the (box $$+$$ ball) system remains constant
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Of the ball remains constant
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Of the ball relative to the box remains constant
Explanation
As no external force acts on the ball $$+$$ box system, hence velocity of the system remains constant.
A train of mass M is moving on a circular track of radius 'R' with constant speed V. The length of the train is half of the perimeter of the track. The linear momentum of the train will be?
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$$0$$
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$$\dfrac{2MV}{\pi}$$
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$$MVR$$
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$$MV$$
Explanation
If we treat the train as a ring of mass 'M', then its CM will be at a distance $$2R/\pi$$ from the centre of the circle. Velocity of the centre of mass is
$$V_{CM}=R_{CM}\omega =\dfrac{2R}{\pi}\dfrac{V}{R}$$
$$\left(\because \omega =\dfrac{V}{R}\right)$$
$$=\dfrac{2V}{\pi}\Rightarrow MV_{CM}=\dfrac{2MV}{\pi}$$
As the linear momentum of any system is $$MV_{CM}$$, the linear momentum of the train is $$2MV/\pi$$.
The graph between kinetic energy and momentum of a particle is plotted as shown in Fig. The mass of the moving particle is?
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$$1$$ kg
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$$2$$ kg
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$$3$$ kg
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$$4$$ kg
Explanation
Kinetic Energy is given as:
$$K=\dfrac{P^2}{2m}$$
From the graph, $$4=\dfrac{4^2}{2m}\Rightarrow m=2 kg$$.
Two identical rods are joined to form an $$'X'$$. The smaller angle between the rods is $$\theta$$. The moment of inertia of the system about an axis passing through point of intersection of the rods and perpendicular to their plane is proportional to :
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$$\theta$$
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$$\sin^2 \theta$$
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$$\cos^2 \theta$$
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independent of $$\theta$$
Explanation
The moment of inertia of the two rods will always be the summation of their individual moment of inertia irrespective of the angle between them.
$$I_{total}=I_A+I_B$$
$$I_{total}=\dfrac{ml^2}{12}+\dfrac{ml^2}{12}$$
Hence correct option is D, independent of $$\theta$$
Distance moved by cylinder during time taken by it to complete one rotation is:
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$$2\pi R$$
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$$\pi R$$
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$$3\pi R$$
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$$\dfrac{4\pi R}{3}$$
Explanation
Total distance moved by the string:
$$s=ut+\dfrac{1}{2}at^2$$
$$s=\dfrac{1}{2}\times \dfrac{3g}{4}\times \dfrac{8\pi R}{g}=3\pi R$$
Distance moved in unwinding:
$$s'=\dfrac{1}{2}\times \dfrac{g}{4}\times \dfrac{8\pi R}{g}=\pi R$$
Distance moves by the cylinder in one complete rotation= $$3\pi R-\pi R=2\pi R$$
Two particles A and B, initially at rest, move towards each other under a mutual force of attraction. At the instant when the speed of A is V and the speed of B is 2V, the speed of the centre of mass of the system is
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3V
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V
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1.5 V
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Zero
Explanation
We know that $$ F_{ext} = Ma_{cm} $$
We consider the two particles in a system . mutual force of attraction in an internal force . there are no external forces acting on the system from Eq(I)
we get
$$ A_{CM} = 0$$
Since intial $$ v_{CM} = 0 $$
therefore final $$v_{CM} = 0 $$
Distance moved by hanging mass during the above time interval is :
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$$2\pi R$$
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$$\pi R$$
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$$3\pi R$$
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$$\dfrac{4\pi R}{3}$$
Explanation
$$s=ut+\dfrac{1}{2}at^2$$
$$s=\dfrac{1}{2}\times \dfrac{3g}{4}\times \dfrac{8\pi R}{g}=3\pi R$$
Velocity of the center of mass of the rod after collision is
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12 m/s
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9 m/s
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6 m/s
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3 m/s
Acceleration of cylinder is :
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$$g$$
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$$\dfrac{g}{2}$$
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$$\dfrac{g}{4}$$
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$$\dfrac{3g}{4}$$
A wire of length $$l$$ and mass $$m$$ is first bent into a circle, then in a square and then in an equilateral triangle. The moment of inertia in these three cases about an axis perpendicular to their planes and passing through their centers of masses are $$I_1$$ $$I_{2}$$ & $$I_{3}$$ respectively. Then maximum of them is:
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$$I_{1}$$
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$$I_{2}$$
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$$I_{3}$$
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data insufficient
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Practice Class 11 Engineering Physics Quiz Questions and Answers
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