Physics Circle home Student Login

Home›Chapters›Laws of Motion PYQs›Equilibrium, Third Law & Frames›Question 7

JEE Main 2024 (6 Apr, Shift 2) · Equilibrium, Third Law & Frames

Equilibrium, Third Law & Frames question from JEE Main 2024 (6 Apr, Shift 2)

Q7JEE Main 2024 · 6 APR · SHIFT 2Easy

Equilibrium, Third Law & Frames

A body of weight 200 N200\text{ N} is suspended from a tree branch through a chain of mass 10 kg10\text{ kg}. The branch pulls the chain by a force equal to (if g=10 m/s2g = 10\text{ m/s}^2):

Show answer and solution

Answer: (a)

  1. The branch supports the chain and the body hanging from it.
  2. F=200+10×10=300 NF = 200 + 10 \times 10 = 300\text{ N}
  3. So the answer is (a) 300 N.
Report a mistake

More questions on Equilibrium, Third Law & Frames