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. 13 N
B. 32 N
C. 25 N
D. 35 N

Answer

B

Explanation

Let a be the acceleration of each block. Then
`T_(3)=(m_(1)+m_(2)+m_(3))a " "` …(i)
`and T_(2)=(m_(1)+m_(2))a " " ` …(ii)
From Eqs. (i) and (ii) , we get
`T_(2)=((m_(1)+m_(2))/(m_(1)+m_(2)+m_(3)))xxT_(3)`
`=((10+6)/(10+6+4))xx40 =32 N`