A rectangular coil of length 0.12 m and width 0.1 m having turns of wire is suspended vertically in a uniform magnetic field of strength 0.2 Wb/m2.The coil carries a current of 2 A. If the plane of the coil is inclined at an angle of 30o with the direction of the field, the torque required to keep the coil in stable equilibrium will be

  • 0.15 Nm

  • 0.20 Nm

  • 0.24 Nm

  • 0.24 Nm


B.

0.20 Nm

Given N = 50

B = 0.2 Wb/m2, I = 2A

θ = 60o, A = 0.12 x 0.1 = 0.012 m2


Thus, torque required to keep the coil in stable equilibrium i.e.,

 = NIAB sin θ = 50 x 2 x 0.012 x0.2 sin 60o


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straight A space force space straight F space equals space straight alpha stack space straight i with hat on top space plus space 3 space straight j with hat on top space plus space 6 space straight k with hat on top space is space acting space at space straight a space point space straight r equals space 2 straight i with hat on top space minus space 6 space straight j with hat on top space minus 12 space straight k with hat on top. The value of α for which angular momentum about origin is conserved is
  • -1

  • 2

  • zero

  • zero


A.

-1

When the resultant external torque acting on a system is zero, the total angular momentum of a system is zero, the total angular momentum of a system remains constant.This is the principle of the conservation of angular momentum.

Given,

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An automobile moves on a road with a speed of 54 km h-1. The radius of its wheels is 0.45 m and the moment of inertia of the wheel about its axis of rotation is 3 kg m2 . If the vehicle is brought to rest in 15 s, the magnitude of average torque transmitted by its breaks to the wheel is

  • 6.66 kg m2s-2

  • 8.58 kg m2s-2

  • 10.86 kg m2s-2

  • 10.86 kg m2s-2


A.

6.66 kg m2s-2

As velocity of an automobile vehicle

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A body of mass M hits normally a rigid wall with velocity v and bounces back with the same velocity. The impulse experienced by the body is 

  • 1.5 Mv

  • 2 Mv

  • zero

  • zero


B.

2 Mv

Impulse |J| = |Δp|

= Mv - (- Mv)

 = 2 Mv

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Point masses m1 and m2 are placed at the opposite ends of a rigid rod of length L and negligible mass. The rod of length L and negligible mass. The rod is to be set rotating about an axis perpendicular to it. The position of point P on this rod through which the axis should pass so that the work required to set the rod rotating with angular velocity ω0 is minimum is given by 


D.

As two point masses m1 and m2 are placed at opposite ends of a rigid rod of length L and negligible as shown in the figure. 


Total moment of inertia of the rod

 I = m1x2  + m2(L-x)2

I = m1x2 + m2L2 +m2x2 - 2m2Lx

As I is minimum i.e 


When I is minimum, the work done by rotating a rod will be minimum.

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