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 pulley. At a certain point of time, when \(l=\frac{\mathrm{L}}{\mathrm{x}}\), the acceleration of the chain is \(\frac{g}{2}\). The value of \(x\) is

JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Laws of Motion

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 '

L - l

' is hanging on the other side of the pulley. At a certain point of time, when

l = \frac{L}{x}

, the acceleration of the chain is

\frac{g}{2}

. The value of

x

Answer: (D)

For a chain over a pulley with unequal hanging lengths, the acceleration is

a = \frac{( m _{2} - m _{1} )}{( m _{2} + m _{1} )} g

. Here

m_{1} = λ ℓ

and

m_{2} = λ (L - ℓ)

, where

λ = m / L

. Substituting

a = g /2

\frac{g}{2} = \frac{( λ ( L - ℓ ) - λ ℓ ) g}{λ L} = \frac{( L - 2 ℓ ) g}{L}

. Solving:

L /2 = L - 2 ℓ

, so

ℓ = L /4

. Given

ℓ = L / x

, we get

x = 4

JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Single correct option