For the given figure the acceleration of 1 kg block if string is massless and mass of pulley is 2 kg and diameter of pulley is 0.2 m :-

Options

A. `2m//s^(2)`
B. `2.5m//s^(2)`
C. `0.2m//s^(2)`
D. `1m//s^(2)`

Answer

A

Explanation

`m_(1)g-T_(1)=m_(1)a " " ` …(i)
`T_(2) -m_(2)g=m_(2)a " " `… (ii)
`T_(1)R-T_(2)R=I alpha = I (a)/(R )`
`T_(1)=T_(2)=(MR^(2))/(2)(a)/(R^(2))`
`T_(1)-T_(2)=(Ma)/(2) " " ` ....(iii)
`m_(1)g-m_(2)g +T_(2)-T_(1) = (m_(1)a+m_(2)a)`
By adding (i), (ii) and (iii)
`10-5=(2a)/(2)=(a+a)/(2)`
`5=(5a)/(2) rArr a =2m//s^(2)`