Maximum Length Of Hung Chain MCQ - Practice Questions with Answers

A uniform chain of length l is placed on the table in such a manner that its l' part is hanging over the edge of the table without sliding.

$
\text { As } \mu=\frac{m_2}{m_1}=\frac{\text { mass hanging from table }}{\text { mass on table }}
$

The chain will have uniform linear density.
So the ratio of mass and ratio of length for any part of the chain will be equal.

$
\begin{aligned}
& \mu=\frac{\text { length of part hanging from table }}{\text { length of part on table }}=\frac{l^{\prime}}{l-l^{\prime}} \\
& l^{\prime}=\frac{\mu l}{(\mu+l)}
\end{aligned}
$

Where $l=$ length of chain

$
\begin{aligned}
& l^{\prime}=\text { chain hanging } \\
& \left(l-l^{\prime}\right)=\text { chain lying on table }
\end{aligned}
$