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Laws of Motion question

2004 · Shift 0 · Q182
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Laws of Motion question

2004 · Shift 0 · Q182

JEE MainPhysicsLaws of MotionMCQ+4 / −1
Two masses m1=5kg{m_1} = 5kgm1​=5kg and m2=4.8kg{m_2} = 4.8kgm2​=4.8kg tied to a string are hanging over a light frictionless pulley. What is the acceleration of the masses when left free to move? (g=9.8m/s2)\left( {g = 9.8m/{s^2}} \right)(g=9.8m/s2) AIEEE 2004 Physics - Laws of Motion Question 130 English
  1. A
    5  m/s25\,\,m/{s^2}5m/s2
  2. B
    9.8  m/s29.8\,\,m/{s^2}9.8m/s2
  3. C
    0.2  m/s20.2\,\,m/{s^2}0.2m/s2
  4. D
    4.8  m/s24.8\,\,m/{s^2}4.8m/s2
View written solutionFree

Correct answer: C

  1. Identify the system

This is an Atwood machine: two masses m1=5 kgm_1=5\,\text{kg}m1​=5kg and m2=4.8 kgm_2=4.8\,\text{kg}m2​=4.8kg are connected by a light string over a frictionless pulley.

Since m1>m2m_1>m_2m1​>m2​, the 5 kg5\,\text{kg}5kg mass moves downward and the 4.8 kg4.8\,\text{kg}4.8kg mass moves upward.

  1. Write the formula for acceleration

For an Atwood machine,

a=(m1−m2)gm1+m2a=\frac{(m_1-m_2)g}{m_1+m_2}a=m1​+m2​(m1​−m2​)g​
  1. Substitute the values
a=(5−4.8)×9.85+4.8a=\frac{(5-4.8)\times 9.8}{5+4.8}a=5+4.8(5−4.8)×9.8​ a=0.2×9.89.8a=\frac{0.2\times 9.8}{9.8}a=9.80.2×9.8​ a=0.2 m/s2a=0.2\,\text{m/s}^2a=0.2m/s2
  1. Check the options
  • A: 5 m/s25\,\text{m/s}^25m/s2 ✗
  • B: 9.8 m/s29.8\,\text{m/s}^29.8m/s2 ✗
  • C: 0.2 m/s20.2\,\text{m/s}^20.2m/s2 ✓
  • D: 4.8 m/s24.8\,\text{m/s}^24.8m/s2 ✗

Therefore, the correct option is C.

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