JEE Main 2021 · PhysicsMultiple choiceSingle correctMediumCalculation

JEE Main 26 February 2021, Shift 2, Physics Q7

Question 7 of 90 in this shift, Physics question 7 of 30, Section A.

An aeroplane, with its wings spread 10 m, is flying at a speed of 180 km/h in a horizontal direction. The total intensity of earth’s field at that part is

2.5 \times 10^{-4}

Wb/m

^2

and the angle of dip is

60^\circ

. The emf induced between the tips of the plane wings will be ________.

(1)108.25 mVOfficial answer
(2)62.50 mV
(3)88.37 mV
(4)54.125 mV

Official answer

Option 1

NTA final key.

Chapter: Electromagnetic Induction
Topic: Faraday's Law
Idea tested: Linear Motional EMF

Same idea in other shifts

Asked 7× in all

8 Apr 2023, Shift 2 · Q48An emf of 0.08 V is induced in a metal rod of length 10 cm held normal to a uniform magnetic field of 0.4 T, when moves with a velocity of:EasySingle correct
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11 Apr 2023, Shift 2 · Q58A metallic cube of side $15\ \mathrm{cm}$ moving along y-axis at a uniform velocity of $2\ \mathrm{ms^{-1}}$ . In a region of uniform…MediumNumerical valueHas a figure
27 Jun 2022, Shift 2 · Q59A metallic rod of length 20 cm is placed in North-South direction and is moved at a constant speed of 20 m/s towards East. The horizontal…MediumNumerical value
4 Aug 2021, Shift 1 · Q10A metal rod of 10 cm length is moving with a velocity 5 m/s at the centre of a current carrying circular coil of radius 4 cm and 2A…MediumSingle correct
4 Aug 2021, Shift 1 · Q14The magnetic field in any region is given by $\vec{B}=B_0\left(2+\frac{x}{a}\right)\hat{k}$ . A square loop of edge length ‘l’ is placed…MediumSingle correct

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.