question_answer

A uniform chain of mass m hangs from a light pulley, with unequal lengths of the chain hanging from the two sides of the pulley. The force exerted by the moving chain on the pulley is

A)  mg                

B)  <mg  

C)       >mg               

D)       variable

Correct Answer: B

Solution :

Let mass per unit length of the chain be\[\lambda \]and acceleration of the chain a Invoking Newton?s second law on the left and right side of the chain, we get \[T-\lambda xg=\lambda xa\]                            (1) \[\lambda yg-T=\lambda ya\]                                              (2) Adding equation (1) and (2), we get: \[a=\frac{\lambda yg-\lambda xg}{\lambda x+\lambda y}=\frac{g(y-x)}{y+x}\] Inserting this value of a in equation (1), we get? \[T-\lambda xg=\lambda x\frac{g(y-x)}{y+x}\] \[T=\lambda xg+\frac{\lambda xg(y-x)}{(y+x)}=\lambda xg\left[ 1+\frac{y-x}{y+x} \right]\] \[\therefore \]      \[T=\lambda xg\frac{2y}{(y+x)}=\frac{2\lambda xyg}{(x+y)}\] or, \[T=\frac{2mxy}{{{(x+y)}^{2}}}g\,\left[ \because \,\,\lambda =\frac{m}{x+y} \right]\] Total force acting on the pulley\[=2T\] \[2T=\frac{4mg\,xy}{{{(x+y)}^{2}}}=\frac{mg(4xy)}{{{(x-y)}^{2}}+4xy}\] \[\therefore \]      \[2T=\frac{mg}{\frac{{{(x-y)}^{2}}}{4xy}+1}\Rightarrow 2T<mg\] Note-It has been assumed that the mass of the small  part of the chain wrapped over the pulley is negligible Hence, the correction option is [b].