Two bodies of masses m1 and m2 are connected by a light string which passes over a frictionless massless pulley . If the pulley is moving upward with uniform acceleration g/2 , then tension in the string is - 1) 3 m1 m2 g / m1 + m2 ..............2)m1 + m2 g / 4m1 m2 ..........3)2 m1 m2 g / m1 + m2 ..............4)m1 m2 g / m1 + m2 ..............

When the system is accelerating upward with acceleration (g/2), the net downward acceleration experienced by the masses is = g + g/2 = 3g/2

Now,

M(3g/2) – T = Ma ……..(1)

T – m(3g/2) = ma ………(2)

Adding the above two equations we get,

(3g/2)(M – m) = a(M + m)

=> a = (3g/2)(M – m)/(M + m)

So,

(2) => T = m(3g/2) + ma

=> T = m(3g/2) + (m)(3g/2)(M – m)/(M + m)

=> T = 3mgM/(M + m)

This is the tension in the string.

Correct option is (1).

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