A chain is lying on a smooth table with half its length hanging over the edge of the table [fig(i)]. If the chain is released it slips off the table in time t₁. Now, two identical small balls are attached to the two ends of the chain and the system is released [fig(ii)]. This time the chain took t₂ time to slip off the table. Which of the following is true?
 

 

Let the mass of the chain be m and its length be L
The mass of unit length of the chain= m/L
When a part of the chain of length x is hanging, the acceleration of the chain will equal to
$\mathrm{a}=\frac{(\mathrm{m}/\mathrm{L})\mathrm{xg}}{\mathrm{m}}=\frac{\mathrm{gx}}{\mathrm{L}}$
Initially, when x = L/2 , the acceleration equals g/2, but then it increases. After identical masses M are attached to the ends of the chain, then acceleration at the moment when a length x of the chain is hanging will be
${\mathrm{a}}^{\mathrm{\prime }}=\frac{(\mathrm{m}/\mathrm{L})\mathrm{xg}+\mathrm{Mg}}{\mathrm{m}+2\mathrm{M}}=\frac{\mathrm{gx}}{\mathrm{L}}\left[\frac{\left(\mathrm{m}\right)+\frac{\mathrm{ML}}{\mathrm{x}}}{\mathrm{m}+2\mathrm{M}}\right]$
To decide whether a is larger or a’ let us assume $\mathrm{a}>{\mathrm{a}}^{\mathrm{\prime }}$
$1>\left[\frac{\left(\mathrm{m}\right)+\frac{\mathrm{ML}}{\mathrm{x}}}{\mathrm{m}+2\mathrm{M}}\right]\Rightarrow \mathrm{m}+2\mathrm{M}>\mathrm{m}+\frac{\mathrm{ML}}{\mathrm{x}}$
$\Rightarrow 2\mathrm{x}>\mathrm{L}\text{, which is true. }$
Hence, the chain without the balls will slip off faster.