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Laws of Motion question

2024 · 31 Jan · Shift 2 · Q70
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  5. /2024 · 31 Jan · Shift 2 · Q70

Laws of Motion question

2024 · 31 Jan · Shift 2 · Q70

JEE MainPhysicsLaws of MotionMCQ+4 / −1
A light string passing over a smooth light fixed pulley connects two blocks of masses m1m_1m1​ and m2m_2m2​. If the acceleration of the system is g/8g / 8g/8, then the ratio of masses is: JEE Main 2024 (Online) 31st January Evening Shift Physics - Laws of Motion Question 27 English
  1. A
    81\frac{8}{1}18​
  2. B
    97\frac{9}{7}79​
  3. C
    53\frac{5}{3}35​
  4. D
    43\frac{4}{3}34​
View written solutionFree

Correct answer: B

  1. Use the formula for an Atwood machine

For two masses m1m_1m1​ and m2m_2m2​ connected by a light string over a smooth light pulley, the acceleration is

a=∣m1−m2∣m1+m2ga = \frac{|m_1-m_2|}{m_1+m_2}ga=m1​+m2​∣m1​−m2​∣​g

Given:

a=g8a=\frac{g}{8}a=8g​

So,

∣m1−m2∣m1+m2g=g8\frac{|m_1-m_2|}{m_1+m_2}g=\frac{g}{8}m1​+m2​∣m1​−m2​∣​g=8g​

Cancelling ggg:

∣m1−m2∣m1+m2=18\frac{|m_1-m_2|}{m_1+m_2}=\frac{1}{8}m1​+m2​∣m1​−m2​∣​=81​
  1. Assume the heavier mass is m1m_1m1​

Then,

m1−m2m1+m2=18\frac{m_1-m_2}{m_1+m_2}=\frac{1}{8}m1​+m2​m1​−m2​​=81​

Cross-multiplying:

8(m1−m2)=m1+m28(m_1-m_2)=m_1+m_28(m1​−m2​)=m1​+m2​ 8m1−8m2=m1+m28m_1-8m_2=m_1+m_28m1​−8m2​=m1​+m2​ 7m1=9m27m_1=9m_27m1​=9m2​

Therefore,

m1m2=97\frac{m_1}{m_2}=\frac{9}{7}m2​m1​​=79​
  1. Match with the options
m1m2=97\frac{m_1}{m_2}=\frac{9}{7}m2​m1​​=79​

So the correct option is B.

  1. Comparison with stored answer

Stored correct answer: B

This matches our derived answer.

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