System of Particles and Rotational Motion Test

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System of Particles and Rotational Motion Test - 43

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Question 1
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The ABC is a triangular plate of uniform thickness. The sides are in the ratio shown in the figure. $${I}_{AB},{I}_{BC}$$ and $${I}_{CA}$$ are the moments if inertia of the plate about AB,BC and CA respectively?

A
$${I}_{AB}+{I}_{BC}={I}_{CA}$$

B
$${I}_{CA}$$ is maximum

C
$${I}_{AB}> {I}_{BC}$$

D
$${I}_{BC}> {I}_{AB}$$

Solution

Depends upon farthest point as $$bc$$ is the farthest point from axis and $$AB$$ is nearest.
$${I}_{BC}> {I}_{AB}$$
Question 2
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Different minerals are being mined from within the earth and multi-storeyed are being constructed. Due to this activity theoretically

A
Angular speed of earth would increase

B
Angular momentum would increase

C
Time period of earth would decrease

D
Length of day would increase

Solution

Consider the problem,
In the case of mineral,
coverage area $$(r)$$ increases
Therefore, Inertia $$I=Mr^2$$
If $$r$$ increases, then $$I$$ increases
and, also $$L=I\omega $$
$$L$$ is conserved, then if $$I$$ increases, $$\omega $$ decreases.
And also, $$\omega=\frac{2 \pi }{T}$$
If $$\omega $$ decreases, then Time period $$(T)$$ increases.
Therefore, length of the day increases.
Question 3
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A disc rolls down a plane of length $$L$$ and inclined at angle q, without slipping. its velocity on reaching the bottom will be :-

A
$$\sqrt{\dfrac{4gL \sin \theta}{3}}$$

B
$$\sqrt{\dfrac{2gL \sin \theta}{3}}$$

C
$$\sqrt{\dfrac{10gL \sin \theta}{7}}$$

D
$$\sqrt{4gL \sin \theta}$$

Solution

Here,
h=Lsinθ
Question 4
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A spherical shell and a solid cylinder of same radius rolls down an inclined plane. The ratio of their accelerations will be :-

A
$$15 : 14$$

B
$$9 : 10$$

C
$$2 : 3$$

D
$$3 : 5$$

Solution

The formula for $$ac{c^n}$$ of an object $$i.e,$$ sphere cylender etc$$.)$$ of mass $$,$$ radius and $$MOI$$ given respectively is given by
$$a = \dfrac{{g\sin \theta }}{{1 + \dfrac{I}{{M{R^2}}}}}$$
$$I$$ of spherical shell$$,$$ $${I_{shell}} = \dfrac{{2M{R^2}}}{3}$$
$$I$$ of solid cylinder$$,$$ $${I_{cyl}} = \dfrac{{M{R^2}}}{2}$$
Therefore$$,$$ $${a_{shell}} = \dfrac{{g\,\sin \theta }}{{1 + \dfrac{{2M{R^2}}}{{3M{R^2}}}}} = \dfrac{{g\,\sin \theta }}{{\dfrac{5}{3}}}$$
$${a_{cylinder}} = \dfrac{{g\,\sin \theta }}{{1 + \dfrac{{M{R^2}}}{{2M{R^2}}}}}$$
Therefore$$, =$$ $$\dfrac{3}{5} \times \dfrac{3}{2} = \dfrac{9}{{10}} = 9:10$$
hence$$,$$
option (B) is correct
Question 5
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Initially two stable particles $$x$$ and $$y$$ start moving towards each other under mutual attraction. If at one time the velocities of $$x$$ and $$y$$ are $$V$$ and $$2V$$ respectively, what will be the velocity of centre of mass of the system?

A
$$V$$

B
Zero

C
$$\cfrac{V}{3}$$

D
$$\cfrac{V}{5}$$

Solution

Since they are moving under mutual force at attraction their internal forces cancel each other and $$M\overrightarrow { { a }_{ cm } } ={ F }_{ external }\\ \overrightarrow { { F }_{ external } } =0\quad \\ \overrightarrow { { a }_{ cm } } =0\\ Since\quad initial\quad velocity\quad =0\\ Final\quad velocity\quad =0\quad \\ Since\quad \overrightarrow { a } =0$$
Question 6
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A force $$\vec F=(2\hat i+3\hat j-5\hat k)\ N$$ acts at a point $$\vec {r}_{1}=(2\hat i+4\hat j+7\hat k)\ m$$. The torque of the force about the point $$\vec {r}_{2}=(\hat i+2\hat j+3\hat k)\ m$$ is :

A
$$17\hat j+5\hat k -3\hat i)Nm$$

B
$$2\hat i+4\hat j -6\hat k)Nm$$

C
$$12\hat i-5\hat k +7\hat k)Nm$$

D
$$13\hat j-22\hat i -\hat k)Nm$$

Solution

Consider the problem

Force $$\vec F = (2\hat i + 3\hat j - 5\hat k){\rm{}}$$ is acted at a point $${{\vec r}_1} = (2\hat i + 4\hat j + 7\hat k){\rm{}}$$

Torque about the point $$0,{{\vec r}_2} = (\hat i + 2\hat j + 3\hat k){\rm{}}$$ is

$$\vec r \times \vec F=(\vec r_1-\vec r_2) \times \vec F$$

And

$$(\vec r_1-\vec r_2)=\hat i+2\hat j+4 \hat k$$

$$\vec { r } \times \vec { F } =\left| { \begin{array} { *{ 20 }{ c } }{ \hat { i } } & { \hat { j } } & { \hat { k } } \\ 1 & 2 & 4 \\ 2 & 3 & { -5 } \end{array} } \right| $$

$$ = \left( { - 10 - 12} \right)\hat i + \left( {8 - \left( { - 5} \right)} \right)\hat j + \left( {3 - 4} \right)\hat k$$

$$=-22 \hat i+13 \hat j-\hat k$$

Therefore, the torque of the force about the point

$$=13\hat j-22 \hat i-\hat k)Nm$$

Hence, Option $$D$$ is the correct answer.
Question 7
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Consider the following two statements:
(A) The linear momentum of a particle is independent of the frame of reference
(B) The kinetic energy of a particle is independent of the frame of reference

A
Both A and B are true

B
A is true but B is false

C
A is false but B is true

D
both A and B are false

Solution

Explanation:
With the application of pseudo force in a noninertial frame of reference velocities change, so do linear momentum and kinetic energy because both depend on velocity. Hence both statements are false.
Question 8
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A thin rod of mass $$6m$$ and length $$6L$$ is bent into regular hexagon. The $$M.I$$ of the hexagon about a normal axis to its plane and through centre of system is

A
$$m{ L }^{ 2 }$$

B
$$3m{ L }^{ 2 }$$

C
$$5m{ L }^{ 2 }$$

D
$$11m{ L }^{ 2 }$$

Solution

Each rod has mass $$m$$ and length $$L$$ and distance of their C.O.M from center of hexagon $$(d) = L\sin 60^{o}=\dfrac{L\sqrt{3}}{2}$$ (Internal angle of regular hexagon $$= 120^{o})$$
Moment of inertia of one rod about axis through its C.O.M $$(I) =\dfrac{mL^{2}}{12} $$
So its moment of inertia about axis through center of hexagon and perpendicular to the plan is:
$$ I_{centre} = I+md^{2} $$ (Parallel axis theorem)
$$ =\displaystyle \frac{mL^{2}}{12} + \frac{3mL^{2}}{4}=\frac{5mL^{2}}{6} $$
So for $$6$$ rods the net moment of inertia this axis will be:
$$6 \times \dfrac{5mL^2}{6} = 5mL^2$$
Question 9
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The moment of inertia of a thin square plate ABCD of uniform thickness about an axis passing through the centre O and perpendicular to the plane of the plate is l. Which of the following is true?

A
(1) l = $$l_{3}+l_{4}$$

B
(2) l = $$l_{1}+l_{3}$$

C
(3) l = $$l_{4}+l_{2}$$

D
(4) l = $$l_{1}+l_{2}+l_{3}+l_{4}$$

Solution

R.E.F image
According to parallel axis theorem:-
"The moment of inertia of a body about
an axis perpendicular to the plane of body
is equal to the sum of moment of inertia of
body about mutually perpendicular axis
buying in the plane of body and con-
sent to the axis perpendicular to the
plane of body."
$$ \Rightarrow I = I_{1}+I_{2} $$
$$ I = I_{3}+I_{4} $$
Question 10
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A spherically symmetric gravitational system of particles has a mass density
$$\rho =\left\{\begin{matrix} \rho_0,\ \ r\leq R \\ 0,\ \ r > R\end{matrix}\right.$$ where $$\rho_0$$ is a constant. A test mass can undergo circular motion under the influence of the gravitational field of particles. Its speed V as a function of distance $$r(0 < r < \infty)$$ from the centre of the system is represented by.

A

B

C

D

Solution

Given mass density $$P=\left\{ { P }_{ 0 }\quad for\quad r\le R\\ O\quad for\quad r>R \right\} $$
Since gravitational depends on mass, which is zero for $$r>R$$, hence it only acts for $$r\le R$$.
$${ \vec { F } }_{ graw }={ \vec { F } }_{ circular }$$
$$\Rightarrow \dfrac { { GM }_{ sphere }\times M }{ { R }^{ 2 } } =\dfrac { M{ \left( e \right) }^{ 2 } }{ R } $$
$$\Rightarrow { V }^{ 2 }\alpha \dfrac { { M }_{ sphere } }{ R } $$ since $$r\le R$$
$$\Rightarrow { V }^{ 2 }\alpha \dfrac { { R }^{ 3 } }{ R } \alpha { R }^{ 2 }\Rightarrow V\alpha R$$ straight line $$+ve$$ slope
Beyond $$r=R$$, mass density $$=0\Rightarrow $$ acceleration due to gravity $$=0\Rightarrow $$ velocity stays constant.