A chain of 5 link, each of mass 0.1 kg , is lifted vertically with a constant acceleration 1.2m/s^2 . The force of interaction between the top link and the one immediately below it is

a) 1.10 N b) 0.88N c) 0.66N d) 0.44N

Answer is option A

Explanation:

Mass (m)=0.1kg

Acceleration (a)= 1.2m/s²

We know,

Force(F)=n×m×(g+a)

=n×0.1(9.8+1.2)

=1.1×n N

Where, n= number of links

g= acceleration due to gravity

We have n =1 [from question]

So, F=1×1.1N= 1.1N