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Laws of Motion question

2022 · 26 Jul · Shift 2 · Q56
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  5. /2022 · 26 Jul · Shift 2 · Q56

Laws of Motion question

2022 · 26 Jul · Shift 2 · Q56

JEE MainPhysicsLaws of MotionMCQ+4 / −1
Two masses M1M_{1}M1​ and M2M_{2}M2​ are tied together at the two ends of a light inextensible string that passes over a frictionless pulley. When the mass M2M_{2}M2​ is twice that of M1M_{1}M1​, the acceleration of the system is a1a_{1}a1​. When the mass M2M_{2}M2​ is thrice that of M1M_{1}M1​, the acceleration of the system is a2a_{2}a2​. The ratio a1a2\frac{a_{1}}{a_{2}}a2​a1​​ will be : JEE Main 2022 (Online) 26th July Evening Shift Physics - Laws of Motion Question 54 English
  1. A
    13\frac{1}{3}31​
  2. B
    23\frac{2}{3}32​
  3. C
    32\frac{3}{2}23​
  4. D
    12\frac{1}{2}21​
View written solutionFree

Correct answer: B

  1. Use the acceleration formula for an Atwood machine

For two masses m1m_1m1​ and m2m_2m2​ connected by a light string over a frictionless pulley,

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

assuming m2>m1m_2 > m_1m2​>m1​.

  1. Case 1: M2=2M1M_2 = 2M_1M2​=2M1​

Let M1=MM_1 = MM1​=M. Then M2=2MM_2 = 2MM2​=2M.

So,

a1=(2M−M)gM+2M=Mg3M=g3a_1 = \frac{(2M-M)g}{M+2M} = \frac{Mg}{3M} = \frac{g}{3}a1​=M+2M(2M−M)g​=3MMg​=3g​

  1. Case 2: M2=3M1M_2 = 3M_1M2​=3M1​

Again let M1=MM_1 = MM1​=M. Then M2=3MM_2 = 3MM2​=3M.

So,

a2=(3M−M)gM+3M=2Mg4M=g2a_2 = \frac{(3M-M)g}{M+3M} = \frac{2Mg}{4M} = \frac{g}{2}a2​=M+3M(3M−M)g​=4M2Mg​=2g​

  1. Find the ratio a1a2\dfrac{a_1}{a_2}a2​a1​​

a1a2=g/3g/2=g3⋅2g=23\frac{a_1}{a_2} = \frac{g/3}{g/2} = \frac{g}{3}\cdot\frac{2}{g} = \frac{2}{3}a2​a1​​=g/2g/3​=3g​⋅g2​=32​

  1. Match with the options

a1a2=23\frac{a_1}{a_2} = \frac{2}{3}a2​a1​​=32​

So the correct option is B.

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