Question 91: Laws of Motion

Question

A uniform metal chain of mass m and length 'L' passes over a massless and frictionless pulley. It is released from rest with a part of its length 'l' is hanging on one side and rest of its length '$$\mathrm{L}-l$$' is hanging on the other side of the pully. At a certain point of time, when $$l=\frac{L}{x}$$, the acceleration of the chain is $$\frac{g}{2}$$. The value of x is __________.

Accepted Answer

Mass of length $l, m_1=\frac{M}{L} l$
Mass of length $L-l, m_2=\frac{M}{L}(L-l)$
Total mass of chain $\mathbf{M}=m_1+m_2$
From newton's laws of motion
$$
\begin{aligned}
& m_2 g-m_1 g=\left(m_2+m_1ight) \frac{g}{2} \\
& \Rightarrow \frac{M}{L}(L-l) g-\frac{M}{L}(l) g=M \frac{g}{2} \\
& \Rightarrow L-l-l=\frac{L}{2} \\
& \Rightarrow l=\frac{L}{4}
\end{aligned}
$$
Comparing with given equation, $l=\frac{L}{x}$, we get $x=4$

Answer

6

Answer

2

Answer

1.5

Answer

4

Laws of Motion

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