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 Multiple Choice QuestionsMultiple Choice Questions


A square loop ABCD, carrying a current I2, is placed near and coplanar with a long straight conductor XY carrying a current I1, as shown in figure. The net force on the loop will be


  • μ0 I1 I22 π

  • μ0 I1I2 L2 π

  • 2 μ0 I1I2 L3 π

  • 2μ0 I1 I23 π


2μ0 I1 I23 π

Force on arm AB due to current in conductor XY is

  F1μ04π  2 I1I2 LL2

  F1μ0I1I23 π

This force acting towards the wire in the plane of loop.

Force on arm CD due to current in conductor XY is

  F2μ04π 2 I1 I2L4 π 3L 2

  F2μ0 I1I23 π

F2 is away from the wire the plane loop.

Ner force on the loop is

   F =  F1 - F2

      = μ0 I1 I2π - μ0 I1I23 π

      = μ0 I1 I2π 1 - 13 

  F = 23 μ0 I1 I2π          


An electron falls from rest through a vertical distance h in a uniform and vertically upward directed electric field E. The direction of electric field is now reversed, keeping its magnitude the same. A proton is allowed to fall from rest in it through the same vertical distance h. The time of fall of the electron, in comparison to the time of fall of the proton is

  • Smaller

  • 5 times greater

  • Equal

  • 10 times greater



As we know, F = qE = ma
 a = qEmh = IqE2mt2t = 2hmqE

i.e., time tm as 'q' is the same for electron and proton. Since the electron has a smaller mass so it will take a smaller time.


Two concentric conducting spherical shells A and B having radii rA and rB (rB > rA) are charged to QA and -QB QB > QA . The electric field along a line passing through the centre is

  • QB > QA



Inside the inner spherical shell A, E = 0. At r = rA, the field is constant. As QB is − and  QB > QA, the field in the space between A and B decreases with the increase in the value of x and becomes constant at the surface of B. Outside the shell B, the field E decreases from a negative value to zero. Hence, option (a) is correct.