System of Particles and Rotational Motion Test

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System of Particles and Rotational Motion Test - 58

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Question 1
1 / -0

The velocity of centre of mass of the system as shown in the figure is

A
$$\left( \frac { 2 - 2 \sqrt { 3 } } { 3 } \right) \hat { i } - \frac { 1 } { 3 } \hat { j }$$

B
$$\left( \frac { 2 + 2 \sqrt { 3 } } { 3 } \right) \hat { i } - \frac { 3 } { 3 }\hat j$$

C
both

D
None of these

Solution

Let $$x$$ component of velocity for mass $$1kg$$ is $${ v }_{ 1x }$$
Similarly  $$x$$ component of velocity for mass $$2kg$$ is $${ v }_{ 2x }$$
$$y$$ component of velocity for mass $$1kg$$ is $${ v }_{ 1y }$$
Similarly  $$y$$ component of velocity for mass $$2kg$$ is $${ v }_{ 2y }$$
Now
$$x$$ component of velocity of mass $$2kg$$ is,
$$2cos{ 30 }^{ 0 }=2\frac { \sqrt { 3 } }{ 2 } =\sqrt { 3 } $$
    $$y$$ component of velocity is
$$2sin{ 30 }^{ 0 }=2\frac { 1 } { 2 } =1 $$

$$M=\frac { { m }_{ 1 }{ v }_{ 1 }+{ m }_{ 2 }{ v }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \\ M=\frac { 1\left( 2\hat { i } \right) +2\left( \sqrt { 3 } \hat { i } -\hat { j } \right) }{ 1+2 } \\ M=\frac { 2\hat { i } +2\sqrt { 3 } \hat { i } -2\hat { j } }{ 3 } \\ M=\left( \frac { 2+2\sqrt { 3 } }{ 3 } \right) \hat { i } -\frac { 2 }{ 3 } \hat { j }$$
Correct option is $$(D)$$
Question 2
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What is the moment of force about the apex of triangle, if 3 forces of 40 N each acting along the sides of equilateral triangle of side 2 m taken in order

A
51.96 Nm

B
69.3 Nm

C
30.6 Nm

D
6.67 Nm

Solution

$$\begin{array}{l}\text { Force moment due to single Foru is } \\\qquad\tau=40\mathrm{~h}\mathrm{Nm} \\\text { on calculating the value of } h \text { - } \\\qquad\tan 30^{\circ}=\frac{h}{1} \quad n=\frac{1}{\sqrt{3}} \mathrm{~m} \\\Rightarrow\quad t=\frac{40}{\sqrt{3}} \mathrm{~N} m\end{array}$$
$$\begin{array}{l}\text { Force moment of three such forces will be } \\\qquad \begin{array}{c} t_{\text {Tota }}=3 x \frac{40}{\sqrt{3}}\mathrm{Nm}=40 \sqrt{3} \mathrm{Nm} \\ T_{\text {Total }}=69.3\mathrm{Nm}\end{array}\end{array}$$
Question 3
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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-

A
$$\dfrac{4}{3}I$$

B
$$\dfrac{8}{3}I$$

C
$$\dfrac{1}{2}I$$

D
$$16I$$

Solution

$$\begin{array}{l}\text { Given- The moment of Inertia of a Rod } \\\text { about perpendicular Bisector }=I=\frac{M L^{2}}{12}-0 \\\Rightarrow\text { I about the given axis will be- } \\ I_{A}=\frac{M L^{2}}{12}+\frac{M L^{2}}{4}=\frac{M L^{2}}{3} \\\text { (By parallel axis theorem) }\end{array} $$
$$\begin{array}{l}\Rightarrow \text { For four such Rods } \\\qquad \quad I_{\text {Total }}=4 \times I_{A}=\frac{4}{3} M L^{2} \\\Rightarrow I_{\text {Total }}=\frac{4}{3} \times 12 I=16 \text { I From eq } 0 \\\text { Hence, option (D)is correct. }\end{array}$$
Question 4
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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

A
$$\cfrac {mg}4$$

B
$$\cfrac 5 2 mg$$

C
$$\cfrac {mg} 6$$

D
$$\cfrac 72 mg$$

Solution

$$\begin{array}{l} mg\frac { l }{ 2 } =\dfrac { l }{ 2 } \left( { \dfrac { { m{ l^{ 2 } } } }{ 3 } } \right) { w^{ 2 } } \\ \Rightarrow { w^{ 2 } }=\dfrac { { 3g } }{ l } ....\left( 1 \right) \\ N-mg=m{ w^{ 2 } }\left( { \dfrac { l }{ 2 } } \right) \\ \Rightarrow N=mg+\dfrac { { m.3g } }{ 2 } \\ =\dfrac { 5 }{ 2 } mg \\ Hence, \\ option\, \, B\, \, is\, \, correct\, answer. \end{array}$$
Question 5
1 / -0

If the kinetic energy of a body is increased by 300% then determine the percentage increase in the momentum

A
100 %

B
150 %

C
$$ \sqrt {300%} $$

D
175 %

Solution

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 $$100$$%.
Question 6
1 / -0

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

A
0

B
$$\alpha I\Delta t$$

C
$$2\alpha I\Delta t$$

D
$${ \alpha }^{ 2 }I\Delta t$$

Solution

$$\begin{array}{l}\text { Given - } * \text { Temperatwe of Rod increased by } \Delta t \\\text { Let the Length of the Rod be (L) and its } \\\text { coefficient of linear expansion be ( }\alpha\text { ) } \\\text { Therefore its change in Length will be } \\\qquad \Delta L=L_{0} \cdot \alpha \Delta t\end{array}$$
$$\begin{array}{l}\text { we have moment of Inertia of Rod about its } \\\text { centre perpendicular to its Length }=I \\ I=\frac{M L^{2}}{12}\end{array} $$
$$\begin{array}{l}\Rightarrow \text { By Error analysis we get- } \\\qquad \frac{\Delta I}{I}=2\frac{\Delta L}{L} \quad \Delta I=\text { change in moment of } \\\text { Inertia }\end{array}$$
$$\begin{aligned}\Rightarrow \Delta I &=\frac{2\Delta L}{L_{0}} \cdot I \\\Delta I &=2\left(\frac{\left.L_{0} \alpha \Delta t\right)}{L_{0}} \cdot I=2 \alpha I \Delta t\right.\\\Delta I &=2 \alpha I \alpha t \\\text { Hence, option }(C)\text { is correct. }\end{aligned}$$
Question 7
1 / -0

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

A
$$\cfrac{7M{L}^{2}}{48}$$

B
$$\cfrac{M{L}^{2}}{9}$$

C
$$\cfrac{M{L}^{2}}{12}$$

D
$$\cfrac{M{L}^{2}}{3}$$

Solution

$$\begin{array}{l}\text { know - } \\\text { moment of Inertia of a rod about an axis } \\\text { passing through its centre- } \\\qquad I=\frac{M L^{2}}{12}\end{array}$$
$$\begin{array}{l}\text { Therefore moment of Inertia about the given } a x \text { is } \\\text { by parallel axis theorem- } \\\qquad I^{\prime}=I+M d^{2}=\frac{M L^{2}}{12}+M\left(\frac{L}{6}\right)^{2} \\ I^{\prime}=\frac{M L^{2}}{9}\end{array} $$
Question 8
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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

A
$$10 \sqrt{2} \mathrm{cm} $$

B
$$ \dfrac{10}{\sqrt{2}} \mathrm{cm} $$

C
$$5 \sqrt{2} \mathrm{cm} $$

D
$$ \frac{5}{\sqrt{2}} \mathrm{cm} $$

Solution

$$\begin{aligned}\text { Given - } & * \text { mass of the cylinder } M=500 \mathrm{~g} \\ & * \text { Radius of the cylinder } R=10\mathrm{~cm} .\end{aligned} $$
$$\begin{array}{l}\text { we know that moment of Inertia of Cylindur - } \\\qquad I=\frac{M R^{2}}{2} \\\text { Let the Radius of gyration of the cylindur be (K) } \\\Rightarrow M k^{2}=\frac{M R^{2}}{2} \\\Rightarrow k=\frac{R}{\sqrt{2}}=\frac{10}{\sqrt{2}} \mathrm{~cm} .\end{array}$$
Question 9
1 / -0

Ratio of radii of gyration of a hollow & solid sphere of the same radii about the axis which is tangent to the sphere is

A
$$\sqrt { \dfrac73 } $$

B
$$\dfrac {5}{\sqrt {21}} $$

C
$$\sqrt { \dfrac { 4 }{ 5 } } $$

D
$$\dfrac{25}9$$

Solution

$$\begin{array}{l}\text { We know that moment of Inertia of a Solid } \\\text { sphere about its axis is - } \\\qquad I_{\text {solid sphere }}=\frac{2}{5}\mathrm{MR}^{2} \\\text { Theretore moment of Inertia about an axis } \\\text { Tangent to the sphere- } \\\qquad I^{\prime}=\frac{2}{5} M R^{2}+M R^{2}=\frac{7}{5} M R^{2} \\\text { I }_{\text {Hollow sphere }}=\frac{2}{3} M R^{2}\end{array}$$
$$\begin{array}{l}\text { Therefore Moment of Inertia about an axis tangent } \\\text { to the sphere - } \\\qquad I^{\prime}=\frac{2}{3}\mathrm{MR}^{2}+\mathrm{MR}^{2}=\frac{5}{3} \mathrm{MR}^{2} \\\text { Radius of gyration of solid sphere be }\mathrm{K}_{5} \\\Rightarrow \quad M \cdot K_{3}^{2}=\frac{7}{5} M R^{2} \\\Rightarrow k_{5}=\sqrt{\frac{7}{5}} R\end{array}$$$$\begin{array}{l}\text { Radius of gyration of Hollow sphere be } k_{H} \\\Rightarrow M K_{H}^{2}=\frac{5}{3} M R^{2} \\\Rightarrow K_{H}=\sqrt{\frac{5}{3}} R \\\Rightarrow \frac{k_{H}}{K_{S}}=\frac{5}{\sqrt{21}} \\\Rightarrow\text { Hence, option (B) is correct. }\end{array}$$
Question 10
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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

A
$$\dfrac{R}{4 \pi - 1}$$

B
$$\dfrac{R}{2(4 \pi - 1)}$$

C
$$\dfrac{R}{3(4 \pi - 1)}$$

D
$$\dfrac{R}{4(4 \pi - 1)}$$