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In the system shown in the adjoining fig...

In the system shown in the adjoining figure, the acceleration of the 1 kg mass is `m//s^(2)`. (take `g = 10 m//s^(2)`)

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The correct Answer is:
5
Suppose a be the downward acceleration of the 4 kg mass, therefore, 2a is the upward acceleration of the 1 kg mass. Hence, equations of motion are :
`1 xx 2a = T - 1g` …(i)
`4a = 4g - 2T`…..(ii)
Adding, after multiplying the equation (i) by 2
8a = 2g or `a = (2g)/(8) = (g)/(4)`
or `2a = (g)/(2)`
Thus, the acceleration of the mass 1 kg is `(g)/(2)` upwards
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