System of Particles and Rotational Motion Test

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System of Particles and Rotational Motion Test - 31

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Question 1
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A projectile of mass $$m$$ is thrown with velocity $$v$$ making an angle $$ 60^{\circ} $$ with the horizontal. Neglecting air resistance, the change in momentum from the departure A to its arrival at B, along the vertical direction is :

A
$$2 mv $$

B
$$\sqrt{3} mv $$

C
$$ mv $$

D
$$\dfrac{mv}{\sqrt{3}} $$

Solution

As the figure drawn above shows that at points $$A$$ and $$B$$ the vertical component of velocity is $$ v \sin 60 ^{\circ} $$ but their direction are opposite.
Hence , change in momentum is given by:
$$\Delta p= mv \sin 60 ^{\circ}-(-mv \sin 60^{\circ})= 2mv \sin 60^{\circ} $$
$$=2mv\dfrac{\sqrt{3}}{2} =\sqrt{3}mv $$
Question 2
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Four identical rods each of mass $$M$$ are joined to form as square frame, the moment of inertia of the system about one of the diagonals is

A
$$\displaystyle\ \frac{13Ml^{2}}{3}$$

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

C
$$\displaystyle\ \frac{Ml^{2}}{6}$$

D
$$\displaystyle\ \frac{13Ml^{2}}{6}$$

Solution

Consider the frame made using four rods as shown.
We have the moment of inertia of rod AB about the center of mass as $$\dfrac{Ml^2}{12}+M(\dfrac{l}{2})^2=\dfrac{Ml^2}{3}$$
Thus we get the moment of inertia of all the four rods as $$\dfrac{4}{3}Ml^2$$
Now using perpendicular axis theorem we see the the moment of inertia about the two diagonals (which are perpendicular to each other) is given as $$2I=\dfrac{4}{3}Ml^2$$
Thus we get $$I=\dfrac{2}{3}Ml^2$$
Question 3
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Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An Impulse gives a velocity of 14 m/s to the heavier block in the direction of the lighter block. The velocity of the centre of mass is :

A
30 m/s

B
20 m/s

C
10 m/s

D
5 m/s

Solution

The velocity of the heavier block (10kg) = 14m/s.

The velocity of COM is given by: $$\dfrac { { m }_{ 1 }{ v }_{ 1 }+{ m }_{ 2 }{ v }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } }$$

$$ =\dfrac { 10\times 14+4\times 0 }{ 10+4 } =10m/s$$ since spring force is an internal force.
Question 4
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Directions For Questions

Fig. shows a uniform metre rule placed on a fulcrum at its mid-point $$O$$ and having a weight $$40gf$$ at the $$10cm$$ mark and a weight of $$20gf$$ at the $$90cm$$ mark.

...view full instructions

How can be rule be brought in equilibrium by using an additional weight of $$40gf$$?

A
By placing the additional weight of $$80gf$$ at the $$30cm$$ mark.

B
By placing the additional weight of $$80gf$$ at the $$70cm$$ mark.

C
By placing the additional weight of $$40gf$$ at the $$30cm$$ mark.

D
By placing the additional weight of $$40gf$$ at the $$70cm$$ mark.

Solution

Let us find individual moments of force.Considering 40 gf and 10 cm mark then 40 gf is at a distance of 40 cm from point O. we get moment of force is equal to $$ 40*40 = 1600 gf cm.$$
Now let us consider 20 gf and 90 cm mark then 20 gf is at a distance of 40 cm to the right to O. We get moment of force as $$20*40 = 800 gf cm $$. Both are not equal hence not in equilibrium.As we have to balance right side as it has less moment of force we must add 800 gf cm more.GIven that we have to add 40 gf weight.$$800 = 40*x$$
$$x = 800/40 = 20 cm.$$Hence mark must be at a distance of 20 cm and it must be right to O.Therefore that mark would be 70 cm mark.
Question 5
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Two particles $$A$$ and $$B$$, initially at rest, moves towards each other under a mutual force of attraction. At the instant when the speed of $$A$$ is $$v$$ and the speed $$B$$ is $$2v$$, the speed of centre of mass is :

A
$$zero$$

B
$$v$$

C
$$1.5v$$

D
$$3v$$

Solution

Force $$F_A$$ on particle $$A$$ is given by
$$\quad F_A = m_Aa_A = \displaystyle\dfrac{m_Av}{t} \quad ...(1)$$
Similarly $$F_B = m_Ba_B = \displaystyle\dfrac{m_B\times2v}{t} \quad ...(2)$$
Now $$\displaystyle\dfrac{m_Av}{t} = \displaystyle\dfrac{m_B\times2v}{t} (\because F_A = F_B)$$
So $$m_A = 2m_B$$
For the centre of mass of the system
$$\quad v = \displaystyle\dfrac{m_Av_A + m_Bv_B}{m_A + m_B}$$
Or $$\quad v = \displaystyle\dfrac{2m_Bv - m_B\times2v}{2m_B + m_B} = 0$$
Negative sign is used bacause the particles are travelling in opposite directions.
Alternatively, if
we consider the two masses in a system then no external force is acting
on the system. Mutual forces are internal forces. Since the centre of
mass is initially at rest, it well remain at rest.
Question 6
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The radius of gyration of a disc of radius 25 cm about an axis passing through the center of the disc and perpendicular to the disc is approximately equal to:

A
$$17.7 cm$$

B
$$12.5cm$$

C
$$36cm$$

D
$$50cm$$

Solution

$$\displaystyle K=\sqrt{\frac{I}{M}}=\sqrt{\frac {\frac{1}{2} mR^{2}}{m}}=\frac{R}{\sqrt{2}}=\frac{25}{\sqrt{2}} \approx 17.68 cm$$
Question 7
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A couple produces

A
Linear motion

B
Rotational motion

C
Both $$A$$ and $$B$$

D
Nether $$A$$ nor $$B$$

Solution
Question 8
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A rigid spherical body is spinning around an axis without any external torque. Due to change in temperature, the volume increases by $$\mbox{1 %}$$. Its angular speed will

A
Increase approximately by $$\mbox{1 %}$$

B
Decrease approximately by $$\mbox{1 %}$$

C
Decrease approximately by $$\mbox{0.67 %}$$

D
Decrease approximately by $$\mbox{0.33 %}$$

Solution

$$\dfrac{4}{3}\pi R^3 = V$$
$$\implies 3 \dfrac{\Delta R}{R} \times 100= \dfrac{\Delta V}{V} \times 100 $$
$$ 3 \dfrac{\Delta R}{R} \times 100= 1$$% i.e $$R$$ will increase by $$0.33$$%
As $$I= K mR^2$$
$$\implies 2 \dfrac{\Delta R}{R} \times 100= \dfrac{\Delta I}{I} \times 100 $$
$$ \dfrac{\Delta I}{I} \times 100= 0.67$$% i.e $$I$$ increases by $$0.67$$%
As there is no external torque, thus $$L= Iw= constant$$
Hence, $$w$$ must decrease by $$0.67$$%.
Question 9
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A projectile of mass $$m$$ is fired with velocity $$v$$ from a point as shown in figure, neglecting air resistance, what is the change in momentum when leaving $$P$$ and arriving at $$Q$$ ?

A
$$0$$

B
$$mv$$

C
$$4 mv$$

D
$$\displaystyle \frac {2mv}{\sqrt {3}}$$

Solution

Horizontal component will be same but vertical component will be completely -ve of initial so change in momentum $$= mvsin\theta - (-mvsin \theta)= 2mvsin\theta =2mv*sin30^o =2mv\frac 12=mv $$
Question 10
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The figure represents two cases. In first case a block of mass $$M$$ is attached to a string which is lightly wound on a disc of mass $$M$$ and radius $$R$$. In second case $$F = Mg$$ Initially, the disc is stationary in each case. If the same length of string is unwound from the disc, then

A
same amount ofwork is done on both discs

B
angular velocities of both the discs are equal

C
both the discs have unequal angular accelerations

D
All of the above

Solution

In first case $$T < Mg$$ In second case $$T = Mg$$
Torques are different, so angular accelerations are different.