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Saturday, 31 October

JEE Main 2020 · PhysicsMultiple choiceDiagram basedHardMulti-step

JEE Main 9 January 2020, Shift 1, Physics Q10

Question 10 of 75 in this shift, Physics question 10 of 25, Section A.

Consider a sphere of radius R which carries a uniform charge density ρ\rho. If a sphere of radius R2\frac{R}{2} is carved out of it, as shown, the ratio ∣E⃗A∣∣E⃗B∣\frac{|\vec{E}_A|}{|\vec{E}_B|} of magnitude of electric field E⃗A\vec{E}_A and E⃗B\vec{E}_B, respectively, at points A and B due to the remaining portion is :

The figure, in words

A large shaded circle with a smaller unshaded (white) circular cavity inside it at the top, touching the large circle's top edge. A vertical dashed line passes through the centre of the cavity (marked with a dot), then point A (a dot at the bottom of the cavity, labelled A), down to point B (a dot on the bottom edge of the large circle, labelled B). A horizontal arrow from the cavity centre to the cavity's left edge is labelled R/2; an arrow from point A to the lower-left edge of the large circle is labelled R.
  1. (1)1834\frac{18}{34}Official answer
  2. (2)2134\frac{21}{34}
  3. (3)1754\frac{17}{54}
  4. (4)1854\frac{18}{54}

Official answer

Option 1

NTA final key (Jan 2020).

Chapter
Electric Charges and Fields
Topic
Electric Field
Idea tested
Field Inside a Spherical Cavity

Same topic in other shifts

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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. This question needs a figure we do not reproduce; it is described in words above. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.