A trolley of mass 5 kg on a horizontal smooth surface is pulled by a load of mass 2 kg by means of uniform rope ABC of length 2 m and mass 1kg. As the load falls from BC=0 to BC=2m. Its accerleration in `m//s^(2)` change-

Options

A. `(20)/(6)` to `(30)/(6)`
B. `(20)/(8)` to `(30)/(8)`
C. `(20)/(5)` to `(30)/(6)`
D. None of these

Answer

A

Explanation

Initial force =`2g = 20 N`
Initial acceleration `= ("Force")/("Mass")=(20)/(5+1)=(20)/(6) ms^(-2)`
Final force `= ("Load + mass of thread" ) xx g`
`=(2+1) xx 10 = 30 N`
Therefore, final acceleration `= (30)/(6) ms^(-2)`.