JEE Main 2021 · PhysicsMultiple choiceDiagram basedHardMulti-step

JEE Main 3 August 2021, Shift 1, Physics Q10

Question 10 of 90 in this shift, Physics question 10 of 30, Section A.

The diagram shows a scheme of two unequal masses,

m_1 = 2m

and

m_2 = m

having unequal positive charge on them. They are suspended by two mass-less threads of unequal lengths from a common point such that, during equilibrium,

m_1

and

m_2

are on same horizontal level. Angle between two strings is

\theta

= 45° in this position. Electrostatic force of interaction between the masses will be -

The figure, in words

A horizontal line labelled 'Fixed Top Head' with two threads from a common point on it going down to the left and right, forming an angle marked 'θ=45°' with a curved arrow between them. The left thread ends in a dot labelled 'mass 1 (m1)' and the right thread ends in a dot labelled 'mass 2 (m2)'; the two dots are at the same horizontal level and the left thread is drawn longer/steeper than the right.

(1) $F = \left(\frac{\sqrt{17}-3}{2}\right) mg$ Official answer
(2) $F = \left(\sqrt{17}-3\right) mg$
(3) $F = mg$
(4) $F = \left(\frac{\sqrt{11}-3}{2}\right) mg$

Official answer

Option 1

NTA final key.

Chapter: Electric Charges and Fields
Topic: Coulomb's Law
Idea tested: Suspended Charge Mechanics

JEE Main 3 August 2021, Shift 1, Physics Q10

Same idea in other shifts