Q79 / 90 · JEE Main 2021 (26 Feb Shift 2) · Physics

An aeroplane, with its wings spread 10 m, is flying at a speed of 180 km h-1 in a horizontal direction. The total intensity of earth's field at that part is 2.5×10-4 Wb m-2 and the angle of dip is 60°. The emf induced between the tips of the plane wings will be

  • Correct answer
    A
    108.25 mV
  • B
    54.125 mV
  • C
    88.37 mV
  • D
    62.50 mV

Solution

∈=B→V→L→=BVLsinθ

=2.5×10-4 T180×518 m s-110 msin60°

=108.25×10-3 V

Hence, the correct answer is option: 'A'.

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