System of Particles and Rotational Motion Test

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System of Particles and Rotational Motion Test - 24

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Weekly Quiz Competition

Question 1
1 / -0

$$K$$ is the radius of gyration of a thin rod when it is rotating about an axis perpendicular to the length of the rod and is passing through it's centre. The length of that rod is :

A
$$12K$$

B
$$3K$$

C
$$2\sqrt{3}K$$

D
$$\sqrt{3}K$$

Solution

$$I= \dfrac{ML^2}{12}= MK^2$$
above equation is relevent because radius of gyration is equivalent distance from axis of rotation,finally;
$$K= \dfrac{L}{\sqrt{12}}$$
$$L= 2\sqrt{3}K$$
Question 2
1 / -0

$$ABC$$ is a right angled triangular plate of uniform thickness. The sides are such that $$AB > BC$$ as shown in the figure above. $$I_{1},I_{2},I_{3}$$ are moments of inertia about $$AB, BC$$ and $$AC$$ respectively. Then which of the following relations is correct ?

A
$$I_{1}=I_{2}=I_{3}$$

B
$$I_{2}>I_{1}>I_{3}$$

C
$$I_{3}>I_{2}>I_{1}$$

D
$$I_{3}>I_{1}>I_{2}$$

Solution

If distance of mass distribution from axis is large then, MOI will also become larger.
$$\therefore $$ by observing the differential masses and their distance from the axis,hence we can say that
$$I_2 > I_1 > I_3$$
Question 3
1 / -0

The moment of inertia of a meter scale of mass $$0.6 \ kg$$ about an axis perpendicular to the scale and located at the $$20 \ cm$$ position on the scale in $$kg\ m^{2}$$ is :

A
$$0.074$$

B
$$0.104$$

C
$$0.148$$

D
$$0.208$$

Solution

Moment of Inertia of the scale around the center of mass is

$$I_{cm} = \dfrac{ML^2}{12}$$

Moment of inertia of the scale around an axis at a distance of 'x' from the CM axis is

$$I = I_{cm} + Mx^2 $$

$$x = (50 - 20)$$ $$cm$$

$$I = \dfrac{ML^2}{12} +Mx^2 $$

$$ I = 0.6 (\dfrac{100^2}{12} +30^2)\ kg\ cm^2$$

$$= 1040\ kg\ cm^2 $$ $$ = 0.104 \ kg\ m^2$$
Question 4
1 / -0

The radius of gyration of a rotating circular ring is maximum about which of the following axis of rotation:

A
neutral axis

B
axis passing through diameter of ring

C
axis passing through tangent of ring in its plane

D
axis passing through tangent of ring perpendicular to plane of ring.

Solution

MI about a diameter : $$I= \frac{1}{2}MR^2$$ by perpendicular axis theorem
MI about an axis passing through tangent to the ring: $$I = \frac{1}{2}MR^2 + MR^2= \frac{3}{2}MR^2$$ by parallel axis theorem
For maximum radius of gyration ring should have maximum moment of inertia about a axis here, option [D] has maximum moment of inertial which is $$2MR^2$$
Question 5
1 / -0

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

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

B
$$\dfrac{l}{2}$$

C
$$l$$

D
$$(\sqrt{2}) l$$

Solution

The distance of mass from the center of square is $$r=\dfrac{l}{\sqrt{2}}$$

So the total moment of inertia of system $$I=4mr^2=2ml^2$$

Let the radius of gyration be k

$$I=2ml^2=4mk^2\Rightarrow k=\dfrac{l}{\sqrt{2}}$$
Question 6
1 / -0

A thin rod of mass $$M$$ and length $$L$$ is rotating about an axis perpendicular to the length of the rod and passing through its centre. If half of the total length rod that lies at one side of axis is cut and removed then the moment of inertia of remaining rod is:

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

B
$$\dfrac{ML^{2}}{18}$$

C
$$\dfrac{ML^{2}}{24}$$

D
$$\dfrac{ML^{2}}{36}$$

Solution

$$I= \dfrac{ML^2}{12}$$
If a half rod is removed. mass becomes $$\dfrac{M}{2}$$ and the length of the rod becomes $$\dfrac{L}{2}$$
$$I'= \dfrac{1}{3}\left ( \dfrac{M}{2} \right )\left ( \dfrac{L}{2} \right )^2$$
$$= \dfrac{1}{24}ML^2$$
Question 7
1 / -0

A ceiling fan is rotating about its own axis with uniform angular velocity $$\omega $$. The electric current is switched off then due to constant opposing torque its angular velocity is reduced to $$\dfrac{2\omega }{3}$$ as it completes $$30$$ rotations. The number of rotations further it makes before coming to rest is :

A
$$18$$

B
$$12$$

C
$$9$$

D
$$24$$

Solution

$$\omega_1^2-\omega_2^2=2\alpha \theta $$ (governing equation as per question)
$$\therefore \omega ^2-\left ( \dfrac{2\omega }{3} \right )^2=2\times \alpha \times 30$$
          $$\dfrac{5\omega ^2}{9}=60\alpha $$
similarly let the no of rotations further made be $$n$$
$$\left( \dfrac{2\omega }{3} \right )^2=2\alpha n$$
   $$\dfrac{4\omega^2 }{9}=2n\alpha$$

$$\Rightarrow \dfrac{5}{4}=\dfrac{30}{n}$$

      $$n=24$$
Question 8
1 / -0

$$K$$ is radius of gyration of a thin square plate about an axis perpendicular to the plane of plate and passing through its centre. The radius of gyration of plate about a parallel axis to the first and passing through a corner of plate is :

A
$$\sqrt{2}K$$

B
$$\sqrt{3}K$$

C
$$2 K$$

D
$$3 K$$

Solution

MI about an axis perpendicular to the plane of plate and passing through its center is $$I_{cm}=\dfrac{Ml^2}{6}$$ ,
Also $$MK^2=Ml^2/6\Rightarrow K^2=l^2/6$$...(1)
Distance of corner from the center of plate $$d=\dfrac{l}{\sqrt{2}}$$
Using parallel axis theorem $$I_{1}=I_{cm}+Md^2=\dfrac{2Ml^2}{3}$$
Also $$MK_1^2=\dfrac{2Ml^2}{3}\Rightarrow K_1^2=2l^2/3$$...(2)
From equation 1 and 2 we get $$K_1=2K$$
Question 9
1 / -0

The moment of inertia of a flywheel is $$0.2 kgm^{2}$$ which is initially stationary. A constant external torque $$5 Nm$$ acts on the wheel. The work done by this torque during $$10 sec$$ is:

A
$$1250 J$$

B
$$2500 J$$

C
$$5000 J$$

D
$$6250 J$$

Solution

Change in momentum $$L=\int\tau.dt$$ (basic property of torque)
                                 $$=5\times10=50$$
$$Work\: done=change\: in \:kinetic\: energy$$
              $$=\dfrac{L^2}{2I}=\dfrac{(50)^2}{2\times 0.2}=\dfrac{25000}{4}$$
              $$=6250 J$$
Question 10
1 / -0

The radius of a flywheel is $$10 cm$$. It is rotated at the rate of $$25 rads^{-2}$$. When a constant force $$40 N$$ is applied on the rim of wheel along its tangent. The moment of inertia of wheel is:

A
$$0.01 kg m^{2}$$

B
$$0.1 kg m^{2}$$

C
$$0.16 kg m^{2}$$

D
$$1.6 kgm^{2}$$

Solution

Torque on wheel $$={F}\times{R}$$ as it is applied at tangent
                           $$=40\times0.1$$
                           $$=4Nm$$
Torque $$\tau =I\alpha $$
                $$4=I\times25$$
                $$\Rightarrow I=\dfrac{4}{25}=0.16kgm^2$$

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System of Particles and Rotational Motion Test - 24

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System of Particles and Rotational Motion Test - 24

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SHARING IS CARING

Question 1

$$12K$$

$$3K$$

$$2\sqrt{3}K$$

$$\sqrt{3}K$$

Solution

Question 2

$$I_{1}=I_{2}=I_{3}$$

$$I_{2}>I_{1}>I_{3}$$

$$I_{3}>I_{2}>I_{1}$$

$$I_{3}>I_{1}>I_{2}$$

Solution

Question 3

$$0.074$$

$$0.104$$

$$0.148$$

$$0.208$$

Solution

Question 4

neutral axis

axis passing through diameter of ring

axis passing through tangent of ring in its plane

axis passing through tangent of ring perpendicular to plane of ring.

Solution

Question 5

$$\dfrac{l}{\sqrt{2}}$$

$$\dfrac{l}{2}$$

$$l$$

$$(\sqrt{2}) l$$

Solution

Question 6

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

$$\dfrac{ML^{2}}{18}$$

$$\dfrac{ML^{2}}{24}$$

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

Solution

Question 7

$$18$$

$$12$$

$$9$$

$$24$$

Solution

Question 8

$$\sqrt{2}K$$

$$\sqrt{3}K$$

$$2 K$$

$$3 K$$

Solution

Question 9

$$1250 J$$

$$2500 J$$

$$5000 J$$

$$6250 J$$

Solution

Question 10

$$0.01 kg m^{2}$$

$$0.1 kg m^{2}$$

$$0.16 kg m^{2}$$

$$1.6 kgm^{2}$$

Solution

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