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Laws of Motion question

2002 · Shift 0 · Q174
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Laws of Motion question

2002 · Shift 0 · Q174

JEE MainPhysicsLaws of MotionMCQ+4 / −1
A light string passing over a smooth light pulley connects two blocks of masses m1{m_1}m1​ and m2{m_2}m2​(vertically). If the acceleration of the system is g/8g/8g/8, then the ratio of the masses is
  1. A
    8:18:18:1
  2. B
    9:79:79:7
  3. C
    4:34:34:3
  4. D
    5:35:35:3
View written solutionFree

Correct answer: B

  1. Identify the system

This is an Atwood machine with two masses m1m_1m1​ and m2m_2m2​ connected by a light string over a smooth light pulley.

The acceleration of such a system is

a=(m1−m2)gm1+m2a = \frac{(m_1-m_2)g}{m_1+m_2}a=m1​+m2​(m1​−m2​)g​

assuming m1>m2m_1>m_2m1​>m2​ so that m1m_1m1​ moves downward.

  1. Use the given acceleration

Given:

a=g8a = \frac{g}{8}a=8g​

So,

(m1−m2)gm1+m2=g8\frac{(m_1-m_2)g}{m_1+m_2} = \frac{g}{8}m1​+m2​(m1​−m2​)g​=8g​

Cancel ggg from both sides:

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

  1. Solve for the ratio

Cross-multiply:

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​

So the ratio of the masses is

m1:m2=9:7m_1:m_2 = 9:7m1​:m2​=9:7

  1. Check options
  • A: 8:18:18:1 ❌
  • B: 9:79:79:7 ✅
  • C: 4:34:34:3 ❌
  • D: 5:35:35:3 ❌

Hence, the correct option is B.

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