A block of mass `m_(1)=5` kg on a smooth table is pulled by a block of mass `m_(2)=2` kg through a unifrom rope ABC of length 2m and mass 1kg. The pulley is smooth and massless As the block of mass `m_(2)` falls from BC=0 to BC=2m its acceleration `("in" m//s^(2))` changes from: `(Take g=10m//s^(2))`

Options

A. `20/6 "to " 30/5`
B. `20/8 "to" 30/8`
C. `20/5 "to" 30/6`
D. `30/5 "to"20/6`

Answer

B

Explanation

(i) When `BC=0` pulling force is `F=m_(2)g`
`:.a=(m_(2)g)/((m_(2)+m_(R)+m_(1)))`
`=(2xx10)/((2+1+5))=20/8 m//s^(2)`
(ii) When `BC=2m` pulling force is `F'=m_(2)g+m_(R)g`
`:.a'=(m_(2)g+m_(R)g)/((m_(2)+m_(R)+m_(1)))`
`=(2xx10+1xx10)/(2+1+5)=30/8 m//s^(2)`