The ratio of tensions in the string connected to the block of mass `m_(2)` in figure, respectively, is (friction is absent everywhere):`[m_(1)=50 kg, m_(2)=80kg` and `F=1000N]`

Options

A. `7:2`
B. `2:7`
C. `3:4`
D. `4:3`

Answer

C

Explanation

In figure (i): Let tension in the string is `T_(1)`
`F-T_(1)=m_(1)a`
implies `1000-T_(1)=50a` (1)
`T_(1)-m_(2)g=m_(2)a`
implies `T_(1)-80g=80a` (2)
From (1) and (2) `T_(1)=(12,000)/(13)N`
In figure (ii): Let tension in the string is `T_(2)` .
`F+m_(1)g-T_(2)=m_(1)a`
`implies 1000+50g-T_(2)=50a` (3)
`T_(2)-80g=80a` (4)
From (3) and (4)
`T_(2)=(16000)/(13)N implies (T_(1))/(T_(2))=(12000)/(16000)=(3)/(4)` .