Question

A chain of mass m and length L is over hanging from the edge of a smooth horizontal table such that ${\frac{1}{n}}^{th}$ of its length is lying on the table. The work done in pulling the chain completely on to the table is

• A. $\frac{mgL}{2n^2}$
• B. $\frac{mgL(n-1{)}^2}{2n^2}$
• C. $\frac{mgL(n-1{)}^2}{n^2}$
• D. $\frac{mgL}{n^2}$

Answer 1

Answer: (B)

$\frac{mgL(n-1{)}^2}{2n^2}$

Mass of the chain lying below the table $=M\left(\frac{L(1-\frac{1}{n})}{L}\right)$ $=M(1-\frac{1}{n})$

The center of mass of the part of chain lying below the table is at a distance of $\frac{1}{2}\times L(1-\frac{1}{n})$

Thus, work done to pull the chain up $=M^{\prime }g\Delta H$

$=M(1-\frac{1}{n})\times g\times \frac{L}{2}(1-\frac{1}{n})$

$=\frac{MgL(n-1{)}^2}{2n^2}$