Two masses M 1 and M 2 are tied together at the two ends of a light inextensible string that passes over a…

Two masses M1 and M2 are tied together at the two ends of a light inextensible string that passes over a frictionless pulley. When the mass M2 is twice that of M1. the acceleration of the system is a1. When the mass M2 is thrice that of M1. The acceleration of The system is a2. The ratio a1a2 will be

  1. 13
  2. 23
  3. 32
  4. 12

Solution

In both cases m1>m2. Therefore, m1 will accelerate downward and m2 will accelerate upward.

For m1,

m1g-T=m1a

For m2,

T-m2g=m2a

Adding the above equations we get,

a=m1-m2m1+m2g

a1=m1-m2m1+m2g=2m2-m22m2+m2g=g3

 and

a2=m1-m2m1+m2g=3m2-m23m2+m2g=g2

a1a2=23

Asked in: JEE Main 2022 (26 Jul Shift 2)

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