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A 40N force pulls a system of three mass...

A 40N force pulls a system of three masses on a horizontal frictionless surface the value of tension `T_(1)` is

A

10 N

B

20 N

C

30 N

D

40 N

Text Solution

Verified by Experts

The correct Answer is:
b

Acceleration of the system
`"a"= ("Force" )/( "Total mass" )`
`= (40)/( 10 + 6 + 4) =(40)/(20) =2ms^(-2)`
`F- T_(1) ma`
`40 - T_(1) = 10 xx 2`
`40 - T_(1) = 20`
`:. T_(1) = 20 - 40 =20`
`T_(1) = 20N`
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