Three blocks of masses `m_(1),m_(2)`and `m_(3)` are connected by massless strings as shown on a frictionless table. They are pulled with a force `T_(3)=40N`. If `m_(1)=10kg,m_(2)=6kg` and `m_(3)=4kg` the tension `T_(2)` will be

Options

A. 20 N
B. 40 N
C. 10 N
D. 32 N

Answer

D

Explanation

`a = (40)/(10+6+4) = 2m//s^(2)`
`40- T_(2) = 4a`
`rArr T_(2) = 40 - 4 xx 2 = 32 N`
Aliter
Direct formula
`T_(2) = ((m_(1) + m_(2)))/(m_(1) + m_(2) +m_(3)) xx T_(3)`
` = (16)/(20) xx 40 = 32 N`