A screw gauge has a least count of 0.01 mm, there are 50 divisions in its circular scale, then the pitch of the screw is…………

  • A)     0.01 mm
  • B)     0.25mm
  • C)     0.5 mm
  • D)     1.0 mm

SOLUTION

As per question the least count of the screw gauge is 0.01 mm, total number of divisions on its circular scale is 50

 Pitch = Least count X total number of divisions in the circular scale

 Pitch of the screw = 0.01 mm X 50

  Pitch of the screw = 0.5 mm

Two particles of mass 5 Kg and 10 Kg respectively are attached to the two ends of of a rigid rod of length 1 m with negligible mass. The center of mass of the system at a distance from the 5 Kg particle is nearly at a distance of ……………..

  • A) 33 cm
  • B) 50 cm
  • C) 67 cm
  • D) 80 cm

The center of mass of a body or a system of particles is defined as the point where the whole of the mass of the body or all the masses of a set of particles appeared to be concentrated.

In a certain region of space with volume 0.2 cubic meter, the electrical potential is found to be 5 V throughout.The magnitude of the electric field in this region is———–

  • A) zero
  • B) O.5 newton per coulomb
  • C) 1 newton per coulomb
  • D) 5 newton per coulomb

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