In the arrangement shown tension in the string connecting $4 k g$ and $6 k g$ masses is

8 N

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12 N

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6 N

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4 N

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Solution

The correct option is A $8 N$
Maximum frictional force=12+8+4=24N
$∴$ Using 20N we can't accelerate the system
Applying equilibrium condition on 2kg mass,
$T_{1} + 4 = 20 \Rightarrow T_{1} = 16 N$
Similarly on, mass 4kg
$T_{2} + 8 = T_{1} \Rightarrow T_{2} = 8 N$

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In the arrangement shown tension in the string connecting 4kg and 6kg masses is

The correct option is A 8NMaximum frictional force=12+8+4=24N∴Using 20N we can't accelerate the systemApplying equilibrium condition on 2kg mass,T1+4=20⇒T1=16NSimilarly on, mass 4kgT2+8=T1⇒T2=8N

In the arrangement shown tension in the string connecting $4 k g$ and $6 k g$ masses is

The correct option is A $8 N$
Maximum frictional force=12+8+4=24N
$∴$ Using 20N we can't accelerate the system
Applying equilibrium condition on 2kg mass,
$T_{1} + 4 = 20 \Rightarrow T_{1} = 16 N$
Similarly on, mass 4kg
$T_{2} + 8 = T_{1} \Rightarrow T_{2} = 8 N$