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Test: System of Particles and Rotational Motion (September 10) - NEET MCQ


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15 Questions MCQ Test Daily Test for NEET Preparation - Test: System of Particles and Rotational Motion (September 10)

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Test: System of Particles and Rotational Motion (September 10) - Question 1

A constant torque of 1000 N-m turns a wheel of the moment of inertia 200 kg-m2 about an axis through its centre. Its angular velocity after 3 seconds is:
[2001]

Detailed Solution for Test: System of Particles and Rotational Motion (September 10) - Question 1

Torque, T = 1000N - m & I = 200 Kg-m2
► T = Iα = 1000
► α = 1000 / 200 = 5 rad/s2
We know that, ω = ωo + αt
= 0 + 3 x 5 = 15 rad/s

Test: System of Particles and Rotational Motion (September 10) - Question 2

If the linear density (mass per unit length) of a rod of length 3m is proportional to x, where x is the distance from one end of the rod, the distance of the centre of gravity of the rod from this end is:
[2002]

Detailed Solution for Test: System of Particles and Rotational Motion (September 10) - Question 2

Consider an element of length dx at a distance x from end A.
Here mass per unit length λ of rod is:

λ ∝ x ⇒ λ = kx
∴ dm = λdx = kx dx

Position of centre of gravity of rod from end A is:

 

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Test: System of Particles and Rotational Motion (September 10) - Question 3

A composite disc is to be made using equal masses of aluminium and iron so that it has as high a moment of inertia as possible. This is possible when:
[2002]

Detailed Solution for Test: System of Particles and Rotational Motion (September 10) - Question 3

Density of iron > Density of aluminium
Moment of inertia = ∫ r2 dm


Since, ρiron > ρaluminium
So, the whole of aluminium is kept in the core and the iron at the outer rim of the disc.

Test: System of Particles and Rotational Motion (September 10) - Question 4

Two persons of masses 55 kg and 65 kg respectively, are at the opposite ends of a boat. The length of the boat is 3.0 m and weighs 100 kg. The 55 kg man walks up to the 65 kg man and sits with him. If the boat is in still water the centre of mass of the system shifts by:
[2012]

Detailed Solution for Test: System of Particles and Rotational Motion (September 10) - Question 4

There is no external force so centre of mass of the system will not shift.

Test: System of Particles and Rotational Motion (September 10) - Question 5

A circular platform is mounted on a frictionless vertical axle. Its radius R = 2 m and its moment of inertia about the axle is 200 kgm2. It is initially at rest. A 50 kg man stands on the edge of the platform and begins to walk along the edge at the speed of 1 ms–1 relative to the ground. Time taken by the man to complete one revolution is:

Detailed Solution for Test: System of Particles and Rotational Motion (September 10) - Question 5

By applying the law of conservation of angular momentum,

Test: System of Particles and Rotational Motion (September 10) - Question 6

The moment of inertia of a uniform circular disc is maximum about an axis perpendicular to the disc and passing through:
[2012M]

Detailed Solution for Test: System of Particles and Rotational Motion (September 10) - Question 6

According to the parallel axis theorem of the moment of Inertia,   

I = Icm + md2

So, d is maximum for point B so Imax about B.

Test: System of Particles and Rotational Motion (September 10) - Question 7

A small object of uniform density rolls up a curved surface with an initial velocity  ‘ν’. It reaches up to a maximum height of 3v2 / 4g with respect to the initial position. The object is a: 
[2013]

Detailed Solution for Test: System of Particles and Rotational Motion (September 10) - Question 7

From law of conservation of mechanical energy:
► 1 / 2 Iω2 + 0 + 1 / 2 mv2 = mg x 3v2 / 4g
► 1 / 2 Iω2 = 3 / 4 mv2 - 1 / 2 mv2
► 1 / 2 Iω= mv2 / 4
► (1 / 2 )* I * (V2 / R2) = mv2 / 4
► I = 1 / 2 mR2
Hence, object is a disc.

Test: System of Particles and Rotational Motion (September 10) - Question 8

A rod PQ of mass M and length L is hinged at end P. The rod is kept horizontal by a massless string tied to point Q as shown in the figure. When the string is cut, the initial angular acceleration of the rod is:
[2013]

Detailed Solution for Test: System of Particles and Rotational Motion (September 10) - Question 8

Weight of the rod will produce the torque,
τ = mg x L / 2 = I α = mL2 / 3 α (∵ Irod = ML2 / 3)
Hence, angular acceleration, α = 3g / 2L

Test: System of Particles and Rotational Motion (September 10) - Question 9

Two discs are rotating about their axes, normal to the discs and passing through the centres of the discs. Disc D1 has 2 kg mass and 0.2 m radius and initial angular velocity of  50 rad s–1. Disc D2 has 4kg mass, 0.1 m radius and initial angular velocity of 200 rad s–1. The two discs are brought in contact face to face, with their axes of rotation coincident. The final angular velocity (in rad s–1) of the system is:
[2013]

Detailed Solution for Test: System of Particles and Rotational Motion (September 10) - Question 9

Given: m1 = 2 kg, m2 = 4 kg, r1 = 0.2 m, r2 = 0.1 m, w1 = 50 rad s–1, w2 = 200 rad s–1
As, I1W1 = I2W2 = Constant
∴ Wf = I1W1 + I2W2 / I1 + I2 
= [(1 / 2 m1r1w)+ (1 / 2 m2r2w2 )] / (1 / 2 m1r12 + 1 / 2 m2r22)​
By putting the value of m1, m2, r1, r2 and solving we get final angular velocity = 100 rad s–1

Test: System of Particles and Rotational Motion (September 10) - Question 10

The ratio of radii of gyration of a circular ring and a circular disc, of the same mass and radius, about an axis passing through their centres and perpendicular to their planes are:
[2013]

Detailed Solution for Test: System of Particles and Rotational Motion (September 10) - Question 10

I = MK2  ⇒ K = √I / M
Iring = MR2 and Idisc = 1 / 2 MR2
► K1 / K2 = (I1 / I2 )1/2 = √MR2 / (MR2 / 2) = √2 : 1

Test: System of Particles and Rotational Motion (September 10) - Question 11

A wheel of radius 1m rolls for war d half a revolution on horizontal ground. The magnitude of the displacement of the point of the wheel initially in contact with the ground is:
[2002]

Detailed Solution for Test: System of Particles and Rotational Motion (September 10) - Question 11

Linear distance moved by the wheel in half revolution = πr
Point P1 after half revolution reaches at P2 vertically 2m above the ground.
∴ Displacement P1P= (π2r2 + 22)1/2 = 2 + 4)1/2     [∵ r = 1m]

Test: System of Particles and Rotational Motion (September 10) - Question 12

A disc is rotating with angular velocity ω. If a child sits on it, what is conserved?
[2002]

Detailed Solution for Test: System of Particles and Rotational Motion (September 10) - Question 12

If external torque is zero, angular momentum remains conserved (External torque is zero because the weight of the child acts downward).

L = Iω = constant

Test: System of Particles and Rotational Motion (September 10) - Question 13

When a mass is rotating in a plane about a fixed point, its angular momentum is directed along:
[2012]

Detailed Solution for Test: System of Particles and Rotational Motion (September 10) - Question 13

  • By right-hand screw rule, the direction of L is ⊥ to the plane containing .
  • The mass is rotating in the plane, about a fixed point, thus this plane will contain  and the direction of  will be perpendicular to this plane.
Test: System of Particles and Rotational Motion (September 10) - Question 14

ABC is an equilateral triangle with O as its centre. represent three forces acting along the sides AB, BC and AC respectively. If the total torque about O is zero the magnitude of

Detailed Solution for Test: System of Particles and Rotational Motion (September 10) - Question 14

The torques F1​d and F2​d of F1​ and F2​respectively are counterclockwise. The torque F3​d is clockwise. Applying condition for rotational equilibrium,

F1 d + F2 d = F3 d
F3 = F1 + F2

Test: System of Particles and Rotational Motion (September 10) - Question 15

Three masses are placed on the x-axis: 300g at origin, 500g at x = 40cm and 400g at x = 70 cm.The distance of the centre of mass from the origin is:
[2012M]

Detailed Solution for Test: System of Particles and Rotational Motion (September 10) - Question 15

► Xcm = (m1x1 + m2x2 + m3x3) / (m1 + m2 + m3)
► Xcm = (300 x (0) + 500(40) + 400 x 70) / (300 + 500 + 400)
► Xcm = 500 x 40 + 400 x 70 / 1200
► Xcm = 50 + 70 / 3 = 120 / 3 = 40 cm

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