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A uniform chain of length 2 m is kept on a table such that a length of 60 cm hangs freely from the edge of the table. The total mass of the chain is 4 kg. What is the work done in pulling the entire chain on the table?     AIEEE  Solved  Paper-2003

A) 7.2 J         

B)                        3.6 J                      

C)        120 J                     

D)        1200 J

Correct Answer: B

Solution :

Mass per unit length\[=\frac{M}{L}=\frac{4}{2}=2\text{ }kg/m\] The mass of 0.6 m of chain\[=0.6\times 2=1.2\text{ }kg\] The centre of mass of hanging part \[h=\frac{0.6+0}{2}=0.3\,m\] Hence, work done in pulling the chain on the table = work done against gravity force \[W=mgh=1.2\times 10\times 0.3=3.6J\] Alternative Method As\[\frac{60\,cm}{200\,cm}=\frac{3}{10}\] part of chain is hanging from the table, the change in potential energy in pulling hanging part onto the table. \[\Delta {{U}_{1}}=-\int\limits_{0}^{\frac{3}{10}l}{\frac{m}{l}g}\,xdx\]\[(\because \Delta U=W=F.dx=mg.dx)\]\[=\frac{m}{l}\,g\,\int_{0}^{\frac{3}{10}l}{xdx=\frac{m}{l}g\left[ \frac{{{x}^{2}}}{2} \right]_{0}^{\frac{3}{10}l}=\frac{m}{l}\frac{g}{2}{{\left( \frac{3}{10}l \right)}^{2}}}\]         \[=-\frac{9mgl}{200}=-3.6J\] Now, work done\[=-\Delta U=-(-3.6J)\]                                 \[=3.6\,J\] (suppose potential energy level at table surface is zero)