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 isa) 1.10 N b) 0...
Answer is option A
Explanation:
Mass (m)=0.1kg
Acceleration (a)= 1.2m/s²
We know,
Force(F)=nm(g+a)
=n0.1(9.8+1.2)
=1.1n N
Where, n= number of links
g= acceleration due to gravity
We have n =1 [from question]
So, F=11.1N= 1.1N