Class 11 NEET Physics Laws of Motion Questions

Let M is the mass of the chain of length L. If y is the maximum length of chain which can hang outside the table without sliding, then for equilibrium of the chain, the weight of hanging part must be balanced by the force of friction on the portion of the table.W=fL ....(i)
But from figure
W=

M

L

yg and R=W′=

M

L

(L−y)g
So that fL=µR =µ

M

L

(L−y)g
Substituting these values of W and fL in eqn.(i),we get
µ

M

L

(L−y)g=

M

L

yg
or m(L−y)=y or y=

µL

µ+1

=

0.25L

1.25

=

L

5

or

y

L

=

1

5

=

1

5

×100%=20%