System of Particles and Rotational Motion Test

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System of Particles and Rotational Motion Test - 59

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Question 1
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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:-

A
$$4 kg-{ m }^{ 2 }$$

B
$$3 kg-{ m }^{ 2 }$$

C
$$2 kg-{ m }^{ 2 }$$

D
$$5 kg-{ m }^{ 2 }$$

Solution

$$\begin{array}{l}\text { On calculating d we get } d=\frac{\sqrt{3}}{2} \text { e } \\\text { Moment of Inertia of whole system is equal to } \\\text { moment of Inertia of all the three Rods about c. } \\\text { I }_{A C}=I_{B C}=\frac{m l^{2}}{3}\end{array}$$
$$\begin{array}{l}\text { Calculating I }_{A B} \text { - } \\\text { By Parallel axis theorem we can soly- } \\\qquad I_{A B}=\frac{m e^{2}}{12}+m d^{2}=\frac{m\ell^{2}}{12}+\frac{3}{4} m l^{2} \\ I_{A B}=\frac{5}{6} m l^{2}\end{array} $$
$$\begin{array}{l}\Rightarrow \text { Moment of Inertia of System } I=I_{A B}+I_{B C}+I_{A C} \\ I=\frac{3}{2} m l^{2}=3\mathrm{~kg}\mathrm{~m}^{2}\end{array} $$
Question 2
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A point at which a whole weight of body act vertically downward is ________.

A
centre of gravity

B
centre of mass

C
centre of force

D
centre of acceleration

Solution
Question 3
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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

A
$$\dfrac{1}{3} ml^2$$

B
$$\dfrac{1}{3\sqrt{2}} ml^2$$

C
$$ ml^2$$

D
$$\dfrac{ml^2}{2}$$

Solution

$$\begin{array}{l}\text { MOI of rod at an end making }\theta \text { degree angle with } \\\text { axis is given by } \frac{m L^{2}}{3} \operatorname{sin}^{2}\theta\end{array}$$
$$\begin{aligned} M .0 . I \text { af system } &=\frac{m L^{2}}{3} \sin ^{2}\theta \times 2 \\ &=\frac{m L^{2}}{3} \times \frac{1}{2} \times 2 \\ &=\frac{m L^{2}}{3}\end{aligned}$$
Question 4
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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?

A
$$0$$

B
$$2$$N

C
$$5$$N

D
$$10$$N

E
$$20$$N

Solution

The one end of the uniform rod is fixed and force $$T$$ is acting in upward direction.
Torque about O,
$$ mg \times \dfrac{l}{2} = T \times l $$

$$T=\dfrac{mg}{2}$$

$$=\dfrac{2\times 10}{2}$$

$$=10N$$.
Question 5
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Find the velocity of centre of mass of the system shown in the figure

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

B
$$4\hat{i}$$

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

D
$$None\ of\ these$$

Solution
Question 6
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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:

A
The spring cannot elongate before $$t=\dfrac {\sqrt {\mu g}}{l \alpha^2}$$

B
The spring elongates as soon as the rotation starts

C
The stored energy in the spring goes on increasing right from $$t=0$$ onwards

D
The maximum spring force acts at $$t=\dfrac {\sqrt {\mu g}}{l \alpha^2}$$

Solution
Question 7
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Mark correct option or options:

A
Nagpur can be said to the geographical centre of India

B
The population centre of India may be Uttar Pradesh

C
The population centre may be coincided with geographical centre

D
All the above

Solution

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 $$C$$,
In demographics, a population center is a geographic point which describes a center point of the population of the region.
Question 8
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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?

A
$$ \frac { 5ML^2}{3} $$

B
$$ \frac { 7ML^2}{3} $$

C
$$ \frac { 5ML^2}{12} $$

D
$$ \frac { 7ML^2}{12} $$

Solution

$$\begin{array}{l}\text { Mass of each rod }=m \\\text { lengthof each rod }=L \\\text { Let o be the centre of rod } C B \text { . }\end{array}$$
$$\begin{aligned} I_{1} &=I_{\text {of } A C \text { about } A} \\ &=\frac{m l^{2}}{3} \\ I_{2} &=I \text { of } C B \text { a bout } A \\ &\left.=I_{C M}+M(O A)^{2} \quad \quad \quad (: A C^{2}+O C^{2}=A D^{2}\right)\end{aligned}$$
$$\begin{array}{l} =\frac{m l^{2}}{12}+m\left(l^{2}+\left(\frac{l}{2}\right)^{2}\right) \\ =\frac{16}{12} m l^{2}=\frac{4}{3} m l^{2} \\ I=\frac{m l^{2}}{3}+\frac{4}{3} m l^{2} \\ I \text { total }=\frac{5}{3} m l^{2}\end{array}$$
Question 9
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$$L=I\omega$$ formula is:

A
always correct

B
sometimes correct

C
always wrong

D
physically correct but dimensionally wrong

Solution
Question 10
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Ram says, $$'A$$ body may be in pure rotation in the presence of a single external force, 'Shyam says, 'This is possible only for a rigid body', then:

A
Ram's statement is correct

B
both statements are correct in different situations

C
both statements are wrong

D
both statements are stated by physicists

Solution