System of Particles and Rotational Motion Test

Result

System of Particles and Rotational Motion Test - 45

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Two rods equal mass $$m$$ and length $$\ell$$ lie along the $$x$$ axis and $$y$$ axis with their centres origin. What is the moment of inertia of both about the line $$x = y$$ :

A
$$\dfrac{m\ell^2}{3}$$

B
$$\dfrac{m\ell^2}{4}$$

C
$$\dfrac{m\ell^2}{12}$$

D
$$\dfrac{m\ell^2}{6}$$

Solution

Moment of inertial of a rod of about an axis passing through centre and perpendicular to length :
$$I=\dfrac{M{l}^{2}}{12}$$
moment of inertia of the rod about an axis inclined an angle $$\theta$$ to the original axis:
$${I}^{\prime}=I{\sin}^{2}{\theta}$$
The line $$x=y$$ makes an angle of $${45}^{\circ}$$ about $$x$$ and $$y$$ axes.
Hence $${I}^{\prime}=I{\sin}^{2}{{45}^{\circ}}=\dfrac{I}{2}$$
moment of inertia of both rods about the line passing through $$x=y$$
$${I}_{two/, rods}=2{I}^{\prime}$$
$$=2\times\dfrac{I}{2}=I$$
$$=\dfrac{M{l}^{2}}{12}$$
Question 2
1 / -0

The block of mass $$m_{2}=10\ kg$$ is given a sharp impulse so that it acquires a velocity $$v_{0}=30\ m/s$$ towards right. Find the velocity of the centre of mass.($$m_{2}=5\ kg$$ and $$k=30\ N/m$$)

A
$$5\ m/s$$

B
$$10\ m/s$$

C
$$15\ m/s$$

D
$$20\ m/s$$

Solution

Mass of block 1 & 2, $${{m}_{1}}=10\,kg\,\,and\,{{m}_{2}}=5\,kg$$
Initial velocity 1 & 2, $${{v}_{1}}=30\,m{{s}^{-1}}\,\,and\,\,{{v}_{2}}=0$$
From conservation of momentum
Initial momentum = momentum of center of mass
$$ {{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}=({{m}_{1}}+{{m}_{2}}){{V}_{cm}} $$
$$ \Rightarrow {{V}_{cm}}=\dfrac{{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}}{{{m}_{1}}+{{m}_{2}}}=\dfrac{10\times 30+5\times 0}{10+0}=20\,m{{s}^{-1}} $$
Hence, center of mass velocity is $$20\,m{{s}^{-1}}$$
Question 3
1 / -0

Moment of inertia of a uniform rod of length L and mass M, about an axis passing through L/4 from one end and perpendicular to its length is

A
$$\dfrac{7}{36} ML^2$$

B
$$\dfrac{7}{48} ML^2$$

C
$$\dfrac{11}{48} ML^2$$

D
$$\dfrac{ML^2}{12}$$

Solution
Question 4
1 / -0

Two like parallel forces $$20\ N$$ and $$30\ N$$ act at the ends $$A$$ and $$B$$ of a rod $$1.5\ m$$ long. The resultant of the forces will act at the point:

A
$$90\ cm$$ from A

B
$$75\ cm$$ from B

C
$$20\ cm$$ from B

D
$$85\ cm$$ from A

Solution

Consider the point at which resultant of the forces will act, is 'x' m from the point A.

Thus the point is (1.5 - x) m far from the point B.

Thus, using the formula Torque = Distance x Force acting

$$20x = 30(1.5 - x)$$

or,$$ 2x = 4.5 - 3x$$

or,$$ 5x = 4.5$$

or, $$x = 0.9m or 90cm$$
Thus the resultant point will be 0.9 m or 90 cm far from point A
Question 5
1 / -0

The moment of inertia of a uniform rod of length $$2l$$ and mass $$m$$ about an axis $$xx'$$ passing through its centre and inclined at an angle $$\alpha $$ is

A
$$\cfrac{ml^{2}}{3}\sin^{2}\alpha $$

B
$$\cfrac{ml^{2}}{12}\sin^{2}\alpha $$

C
$$\cfrac{ml^{2}}{6}\cos^{2}\alpha $$

D
$$\cfrac{ml^{2}}{2}\cos^{2}\alpha $$

Solution
Question 6
1 / -0

two identical particles move towards each other with velocity $$2v$$ and $$v$$ respectively. The velocity of centre of mass is-

A
$$v$$

B
$$v/3$$

C
$$v/2$$

D
$$zero$$

Solution
Question 7
1 / -0

Directions For Questions

Two bodies $$A$$ and $$B$$ of masses $$m$$ and $$2m$$ respectively are connected by a spring of spring constant $$k$$. The masses are moving to the right with the right with the uniform velocity $$v_{0}$$ each, the heavier mass leading the fghiter one. The spring has its natural length during this motion. Block $$B$$ collides head on with a third block $$C$$ of mass $$2m$$ at rest, the collision being completely inelastic.

...view full instructions

The velocity of centre of mass of system of block $$A,B$$ & $$C$$ is-

A
$$\dfrac {v_{0}}{2}$$

B
$$\dfrac {3v_{0}}{5}$$

C
$$\dfrac {2v_{0}}{5}$$

D
$$\dfrac {v_{0}}{5}$$

Solution

We know that even after collision net momentum of system in horizontal direction will be conserved ,
By definition $$v_{cm}= \dfrac{\int vdm}{\int dm}=\dfrac{mv_0+2mv_0}{2m+2m+m}=\dfrac{3v_0}{5}$$
Question 8
1 / -0

$$Wb/m^{-2}$$ making an angle $$30^\circ$$ with the field. Find the couple acting on it.

A
$$2.5 \ Nm$$

B
$$5.5 \ Nm$$

C
$$7.5 \ Nm$$

D
$$9.0 \ Nm$$
Question 9
1 / -0

In moment of inertia of a uniform rod of length $$'l'$$ and mass $$"m"$$ above an axis passing through one end of rod and inclined at angle $$\theta$$ to rod is:

A
$$\dfrac {ml^{2}}{3}\cos^{2}\theta$$

B
$$\dfrac {ml^{2}}{3}\sin^{2}\theta$$

C
$$\dfrac {ml^{2}}{12}\cos^{2}\theta$$

D
$$\dfrac {ml^{2}}{12}\sin^{2}\theta$$

Solution
Question 10
1 / -0

A wheel having moment of inertia $$2\ kgm^{-2}$$ about its axis, rotates at $$50\ rpm$$ abut this axis. The angular retardation that can stop the wheel in one minute is

A
$$\dfrac {\pi}{36}rad\ s^{-2}$$

B
$$\dfrac {\pi}{18}rad\ s^{-2}$$

C
$$\dfrac {\pi}{72}rad\ s^{-2}$$

D
$$\dfrac {\pi}{9}rad\ s^{-2}$$

Solution