A) \[MgL\]
B) \[\frac{1}{3}MgL\]
C) \[\frac{1}{9}MgL\]
D) \[\frac{1}{18}MgL\]
Correct Answer: D
Solution :
If\[m\]is the mass per unit length of the chain, the mass of length\[y\]will be \[m\]and the force acting on it due to gravity will be\[mgy\](assuming that\[y\]is the length of the chain hanging over the edge). So, the work done in pulling the\[dy\]length of the chain on the table \[dW=F(-dy)\](as\[y\]is decreasing) \[i.e.,\] \[dW=(mgy)(-dy)\](as\[F=mgy)\] So, the work done in pulling the hanging portion on the table \[W=-\int_{L/g}^{0}{mgy\,\,dy=\frac{mg{{L}^{2}}}{2{{(3)}^{2}}}}\]You need to login to perform this action.
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