System of Particles and Rotational Motion Test

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System of Particles and Rotational Motion Test - 76

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Question 1
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Particle $$A $$ of mass $$m$$ is moving with a velocity $$- 6 i + 8 j \mathrm { m } / \mathrm { s } $$ experiencing a uniform acceleration of $$ 4 i - 2 \hat { j } \mathrm { m } / \mathrm { s } ^ { 2 } .$$ At that moment particle $$ B $$ of mass $$ m$$ is moving with a velocity of $$ 10\hat { i } - 4 \hat { j } \mathrm { m } / \mathrm { s } $$ and experiencing uniform acceleration of $$ - 2 i + 4 j \mathrm { m } / \mathrm { s } ^ { 2 }$$. The path traced by the center of mass of the $$A, B$$ system is

A
Parabola

B
Circle

C
A straight line

D
Rectangular hyperbola

Solution
Question 2
1 / -0

A square of side 4 cm and uniform thickness is divided into four equal squares. If one of them is cut off find the position of the center of mass of the remaining portion from its geometric center.

A
$$3\sqrt2 cm$$

B
$$\cfrac{3}{\sqrt2}cm$$

C
$$\cfrac{\sqrt2}{3}cm$$

D
$$\cfrac{1}{\sqrt2}cm$$

Solution
Question 3
1 / -0

A small particle of mass m is projected at an angle $$\theta $$ with the x-axis, with initial velocity $${{v_o}}$$ in the x-y plane. Just before time $$\frac{{2{v_o}\sin \theta }}{g}$$, the angular momentum of the particle about the point of projection is

A
$$\dfrac{{2mv_0^3\sin \theta \cdot {{\cos }^2}\theta }}{g}$$

B
$$\dfrac{{2mv_0^3{{\sin }^2}\theta \cdot \cos \theta }}{g}$$

C
$$\dfrac{{mv_0^3\sin \theta \cdot \cos \theta }}{g}$$

D
$$\dfrac{{2mv_0^3{{\sin }^2}\theta \cdot {{\cos }^2}\theta }}{g}$$
Question 4
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A uniform circular disc of radius is taken. A circular portion of radius b has been removed from its as shown in the figure. If the centre of hole is at a distance c from the centre of the disc, the distance $$x_2$$ of the centre of mass of the remaining part from the initial centre of mass O is given by :

A
$$\frac{\pi b^2}{a^2-c^2}$$

B
$$\frac{{{b^2}c}}{{\left( {{a^2} - {b^2}} \right)}}$$

C
$$\frac{\pi c^2}{a^2-b^2}$$

D
$$\frac{ca^2}{c^2-b^2}$$

Solution

$$x_{cm} = \dfrac{A_1x_1 - A_2 x_2}{A_1 - A_2}$$
$$A_1 = \pi \times a^2 $$ {area of whole disc}
$$A_2 = \pi b^2$$ {area of remove d disc}
$$x_1 = 0$$ (centre of mass of whole disc)
$$x_2 = c$$ (centre of mass of removed disc)
$$x_{cm} = \dfrac{(xa^2) (0) - (\pi b^2) (c)}{(\pi a^2 - \pi b^2)} = \dfrac{- \pi b^2 c}{\pi (a^2 - b^2)}$$
$$x_{cm} = \dfrac{-b^2 c}{(a^2 - b^2)}$$
From point 0, centre of mass l is at left at a distance $$= \dfrac{b^2 c}{(a^2 - b^2)}$$
Hence, option $$(B)$$ is correct answer.
Question 5
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A thick hollow sphere rolls down a rough inclined plane without slipping and reaches the bottom with speed $$v _ { 0 } ,$$ when it is again released on a similar but smooth inclined plane, it reaches the bottom with $$\dfrac { 5 v _ { 0 } } { 4 }$$ the radius of gyration of sphere about an axis through its center is ( $$R$$ is the radius of outer surface of the sphere)

A
$$\dfrac { 4 R } { 5 }$$

B
$$\dfrac { 3 R } { 5 }$$

C
$$\dfrac { 3 R } { 4 }$$

D
$$\dfrac { R } { 2 }$$

Solution
Question 6
1 / -0

$$\begin{array} { l } { \text { Three identical uniform rods each of length } 1 \mathrm { m } \text { and } } \\ { \text { mass } 2 \mathrm { kg } \text { are arranged to form an equilateral triangle. } } \\ { \text { What is the moment of inertia of the system about an to } } \\ { \text { axis passing through one corner and perpendicular to } } \\ { \text { the plane of the triangle: } } \end{array}$$

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

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

C
$$2 k g - m ^ { 2 }$$

D
$$3 / 2 \mathrm { kg } - m ^ { 2 }$$

Solution

$$\begin{array}{l} I=\dfrac { { m{ l^{ 2 } } } }{ 3 } +\dfrac { { m{ l^{ 2 } } } }{ 3 } +\left[ { \dfrac { { m{ l^{ 2 } } } }{ { 12 } } +m{ { \left( { \dfrac { { \sqrt { 3 } l } }{ 2 } } \right) }^{ 2 } } } \right] \\ =\dfrac { { 2m{ c^{ 2 } } } }{ 3 } +\left[ { \dfrac { { m{ c^{ 2 } } } }{ { 12 } } +\dfrac { { 3m{ l^{ 2 } } } }{ 4 } } \right] \\ =\dfrac { { 2m{ l^{ 2 } } } }{ 3 } +\dfrac { { 10m{ l^{ 2 } } } }{ { 12 } } \\ =\dfrac { { 18m{ l^{ 2 } } } }{ { 12 } } \\ =\dfrac { { 3m{ l^{ 2 } } } }{ 2 } \\ =3\, kg\, -{ m^{ 2 } } \\ Hence,\, the\, option\, B\, is\, the\; correct\, answer. \end{array}$$
Question 7
1 / -0

A cotton reel rolls without sliding such that the pint P of the string has velocity v = 6 m/s. If r =10 cm and R = 0 cm then the velocity of its centre C is:

A
$$2.5 m/s$$

B
$$5 m/s$$

C
$$4 m/s$$

D
$$2 m/s$$
Question 8
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The radius of gyration of a body about an axis passing through its centre of mass is $$24 cm$$ . Calculate the radius of gyration of the body about a parallel axis passing through a point of a distance of $$7 \mathrm { cm }$$ from its centre of mass

A
$$25 \mathrm { cm }$$

B
$$20 \mathrm { cm }$$

C
$$0.25 \mathrm { cm }$$

D
$$2.5 \mathrm { cm }$$

Solution
Question 9
1 / -0

Four particles each of mass $$m$$ are placed at the corners of a square of side length. The radius of gyration of the system about an axis perpendicular at the square and passing through centre is

A
$$\frac{l}{{\sqrt 2 }}$$

B
$$\frac{l}{2}$$

C
$$l$$

D
$$l\sqrt 2 $$

Solution
Question 10
1 / -0

A disc rolls on ground without slipping. Velocity of centre of mass is v. The speed of particle P at circumference $$(v_p)$$ is

A
v

B
$$\sqrt 2 v$$

C
2v

D
$$\sqrt 3 v$$