System of Particles and Rotational Motion Test

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System of Particles and Rotational Motion Test - 83

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Question 1
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From a circular card board of uniform thickness and mass $$M$$, a square disc of maximum possible area is cut. If the moment of inertia of the square with the axis of rotation at the centre and perpendicular to the plane of the disc is $$\dfrac{Ma^2}{6}$$, the radius of the circular card board is

A
$$\sqrt{2}a$$

B
$$\dfrac{a}{\sqrt{2}}$$

C
$$2a$$

D
$$\dfrac{a}{2}$$

E
None of the above

Solution

Let, side of the square$$=b$$, and radius of disc$$=$$r
Mass of circular card board$$=$$M
Mass density, $$\sigma =\dfrac{M}{\pi r^2}$$
Mass of square, $$m'=\sigma b^2$$
from diagram
$$\cos 45^o=\dfrac{\left(\dfrac{b}{2}\right)}{r}=\dfrac{b}{2r}$$
$$b=\sqrt{2}r$$ ......$$(1)$$
We know moment of inertia of square$$=\dfrac{m'b^2}{6}$$ and in question moment of inertia of square is given $$=\dfrac{Ma^2}{6}$$
so,
$$\dfrac{m'b^2}{6}=\dfrac{Ma^2}{6}$$
$$\dfrac{\sigma b^2\times b^2}{6}=\dfrac{Ma^2}{6}$$
$$\dfrac{Mb^4}{\pi r^2\times 6}=\dfrac{Ma^2}{6}$$, put $$b=\sqrt{2}r$$
$$\dfrac{M\times (\sqrt{2}r)^4}{6\pi r^2}=\dfrac{Ma^2}{6}$$
$$r^2=\dfrac{\pi a^2}{4}$$
$$r=\sqrt{\pi}\dfrac{a}{2}$$.
Question 2
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$$F = a\bar i + 3\bar j + 6\bar k\,\,and\;\,\bar r = 2\bar i - 6\bar j - 12\bar k.$$ The of $$'a'$$ for which the angular momentum same,duration of day will be

A
$$-1$$

B
$$0$$

C
$$1$$

D
$$6h$$

Solution
Question 3
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Two bodies $$A$$ 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 :

A
$$zero$$

B
$$3v$$

C
$$1.5v$$

D
$$v$$

Solution
Question 4
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A canon shell moving along a straight line bursts into two parts. Just after the burst one part moves with momentum $$20$$ N s making an angle $$30^o$$ with the original line of motion. The minimum momentum of the other part of shell just after the burst is?

A
$$0$$ N s

B
$$5$$ N s

C
$$10$$ N s

D
$$17.32$$ N s

Solution
Question 5
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An inverted T-shaped object is placed on a smooth horizontal floor as shown in Fig.
A force F is applied on the system as shown in Fig. The value of x so that the system performs pure translational motion is?

A
$$\dfrac{L}{4}$$

B
$$\dfrac{3L}{4}$$

C
$$\dfrac{L}{2}$$

D
$$\dfrac{3L}{2}$$

Solution
Question 6
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Two blocks of masses $$5$$kg and $$2$$kg are placed on a frictionless surface and connected by a spring. An external kick gives a velocity of $$14$$ m/s to the heavier block in the direction of lighter one. The magnitudes of velocities of two blocks in the centre of mass frame after the kick are, respectively.

A
$$4$$ m/s, $$4$$ m/s

B
$$10$$ m/s, $$4$$ m/s

C
$$4$$ m/s, $$10$$ m/s

D
$$10$$ m/s, $$10$$ m/s

Solution
Question 7
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Four identical rods are joined end to end from a square. The mass of each rod is $$M$$ and length of the rod is $$l$$. the moment of inertia of the square about the median line is

A
$$ \dfrac {Ml^2}{3} $$

B
$$ \dfrac {Ml^2}{4} $$

C
$$ \dfrac {Ml^2}{6} $$

D
None of these

Solution
Question 8
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A thin rod of length $$4\ l$$ and mass 4 m is bent at the points as shown in fig. what is moment of inertia of the rod about the axis passing through point O and perpendicular to the plane of the paper.

A
$$ \frac { Ml^2}{3} $$

B
$$ \frac {10 Ml^2}{3} $$

C
$$ \frac { Ml^2}{12} $$

D
$$ \frac { Ml^2}{24} $$

Solution

The moment of Inertia of whole system
$$I=I_{AD}+I_{AO}+I_{OB}+I_{BC}$$
Also from the figure $$I_{OA}=I_{OB}$$ and $$I_{AD}=I_{BC}$$
We know
$$I_{OA}=I_{OB}=\dfrac{1}{3}ml^2$$

And $$I_{AD}=I_{BC}=I_{CM}+ mr^2$$

here $$r=\sqrt{l^2+(\dfrac{l}{2})^2}$$

$$I_{AD}=\dfrac{1}{12}ml^2 +m(l^2 + (\dfrac{l}{2})^2)$$

$$I_{AD}=\dfrac{16}{12}ml^2$$

$$\implies I=2(I_{AD}+I_{OA})$$

$$\implies I=\dfrac{10}{3}ml^2$$
Hence Option B is correct
Question 9
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Four identical rod are joined end to end from a square. The mass of each rod is M, the moment of inertia of the system about one of the diagonal is

A
$$ \dfrac {2MI^2}{3} $$

B
$$ \dfrac { 13 MI^2}{3} $$

C
$$ \dfrac {MI^2}{6} $$

D
$$ \dfrac {13 MI^2}{6} $$

Solution
Question 10
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Directions For Questions

A rod AB of mass M and length 8 $$l$$ lies on a smooth horizontal surface. A particle of mass m and velocity $$ v_0 $$ strikes the rod perpendicular to its length, as shown in fig. As a result of a collision, the center of mass of rod attains a speed of $$ V_0/ 8 $$ and the particle rebounds back with a speed of $$ v_0 /4 $$ find the following

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The ratio$$ \frac {M}{m} $$,

A
$$ \frac {M}{m} = 10 $$

B
$$ \frac {M}{m} = 4 $$

C
$$ \frac {M}{m} = 8 $$

D
$$ \frac {M}{m} = 5 $$

Solution