Easy MCQ +4 / -1 PYQ · JEE Mains 2024

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

  1. A 300 N Correct answer
  2. B 100 N
  3. C 150 N
  4. D 200 N

Solution

<p>To determine the force that the branch pulls on the chain, we need to consider the combined weight of the body and the chain, as this is the total force that the branch must support due to gravity.</p> <p>The weight of the body is given as $200\, \mathrm{N}$.</p> <p>To find the weight of the chain, we use the formula for weight, which is the mass of an object multiplied by the acceleration due to gravity ($g$). The mass of the chain is given as $10\, \mathrm{kg}$ and $g = 10\, \mathrm{m/s}^2$, so:</p> <p>$\text{Weight of the chain} = \text{Mass of the chain} \times g$ <p>$= 10\, \mathrm{kg} \times 10\, \mathrm{m/s}^2$</p> <p>$= 100\, \mathrm{N}$</p></p> <p>The total force that the branch must support is the sum of the weight of the body and the weight of the chain:</p> <p>$\text{Total force} = \text{Weight of the body} + \text{Weight of the chain}$ <p>$= 200\, \mathrm{N} + 100\, \mathrm{N}$</p> <p>$= 300\, \mathrm{N}$</p></p> <p>Therefore, the branch pulls the chain by a force of $300\, \mathrm{N}$, which corresponds to Option A.</p>

About this question

Subject: Physics · Chapter: Laws of Motion · Topic: Newton's First Law

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