Two situations are shown in fig. (a) and (b)
In each case `m_(1) = 3 kg and m_(2) = 4 kg ` . If `a_(1), a_(2) ` are the respective accelerations of the blocks in these situations, then the values of `a_(1) and a_(2)` are respectively `[g = 10 ms^(-2)]`

Options

A. `(20)/(7)ms^(-2),(10)/(7)ms^(-2)`
B. `(10)/(7)ms^(-2),(25)/(7)ms^(-2)`
C. `(40)/(7)ms^(-2),(10)/(7)ms^(-2)`
D. `(30)/(7)ms^(-2),(5)/(7)ms^(-2)`

Answer

C

Explanation

For fig (a)
`(m_(1) +m_(2)) a_(1) = m_(2) g`
`a_(1) = (m_(2) g)/(m_(1) + m_(2))`
Given `m_(1) = 3kg`
`m_(2) = 4 kg`
`g = 10 m//s^(2)`
`a_(1) = (4 xx 10)/(3 + 4)`
`= (40)/(7) m//s^(2)`
For fig. (b)
`(m_(1)+m_(2)) a = (m_(2) - m_(1)) g`
`a = ((m_(2) - m_(1))/((m_(1) +m_(2))))g`
`= ((4-3)/(4+3)) xx 10 `
`=(10)/(7) m//s^(2)`