A thin circular ring of mass M and radius R is rotating in a horizontal plane about an axis vertical to its plane with a constant angular velocity ω. If two objects each of mass m be attached gently to the opposite ends of a diameter of the ring. the ring will then rotate with an angular velocity

  • fraction numerator straight omega space left parenthesis straight M space minus 2 straight m right parenthesis over denominator straight M space plus space 2 straight m end fraction
  • fraction numerator ωM over denominator straight M space plus space 2 straight m end fraction
  • fraction numerator straight omega left parenthesis straight M space plus 2 straight m right parenthesis over denominator straight M end fraction
  • fraction numerator straight omega left parenthesis straight M space plus 2 straight m right parenthesis over denominator straight M end fraction

B.

fraction numerator ωM over denominator straight M space plus space 2 straight m end fraction

Applying law of conservation of angular momentum.
I1ω1 = I2ω2

In the given case
I1 = MR2
I2 = MR2 + 2mR2

ω1

then ω2 = I1ω /I2 = Mω / M +2m

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Under the influence of a uniform magnetic field, a charged particle moves with constant speed v in a circle of radius R. The time period of rotation of the particle

  • depends on v and not on R

  • depends on R and not on v

  • is independent of both v and R

  • is independent of both v and R


C.

is independent of both v and R

The time period of circular motion of their charged particle is given by 



Hence, time period of rotation of the charged particle in uniform magnetic field is independent of both v and R.

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Four identical thin rods each of mass M and length l , inertial form a square frame. moment of inertia of this frame about an axis through the centre of the square and perpendicular to this plane is 

  • 4Ml2/3

  • 2Ml2/3

  • 13Ml2/3

  • 13Ml2/3


A.

4Ml2/3

Apply theorem of the parallel axis and the total moment of inertia will be the sum of the moment of inertia of each rod.
Moment of inertia of rod about an axis through its centre of mass and perpendicular to rod + (mass) of rod) x (perpendicular distance between two axes)


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A closely wound solenoid of 2000 turns and area of cross -section 1.5 x 10-4 m2 carries a current of 2.0 A. It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field 5 x 10-2 T making an angle of 30o with axis of  the solenoid. The torque on the solenoid will be

  • 3 x 10-3 N-m

  • 1.5 x 10-3 N-m

  • 1.5 x 10-2 N-m

  • 1.5 x 10-2 N-m


A.

3 x 10-3 N-m

Given, N = 2000,
A = 1.5 x 10-4 m2
i = 2.0 A B = 5 x 10-2 T
and

θ = 30o

Torque = NiBA sin θ

2000 x 2 x 5 x 10-2 x 1.5 x 10-14 x sin 30o

= 2000 x 50 x 10-6 x (1/2)

= 1.5 x 10-2 Nm



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A thin circular ring of mass M and radius r is rotating about its axis with constant angular velocity ω. Two objects each of mass m are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with angular velocity given by

  • (M +2m)ω /2m

  • 2Mω / M + 2m

  • (M +2m)ω / M

  • (M +2m)ω / M


D.

(M +2m)ω / M

In the absence of external torque, angular momentum remain constant

L = Iω = I'ω'
therefore,
MR2ω = (M + 2m) R2ω'


ω' = Mω / (M + 2m)

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