A uniform chain of length \(\mathrm{L}\) and mass \(\mathrm{M}\) is lying on a smooth table and \((1 / 4)^{\text {th }}\) of its length is hanging vertically down over the edge of the table. If \(g\) is acceleration due to gravity, the work required to pull the hanging part on to the table is (A) MgL (B) \(\mathrm{MgL} / 9\) (C) \(\mathrm{MgL} / 18\) (D) \(\mathrm{MgL} / 32\)

We are given the following variables: - Length of the chain (L) - Mass of the chain (M) - Acceleration due to gravity (g) We need to find the work required to pull 1/4 of the chain hanging vertically back onto the table.

First, let's find the mass of the hanging portion of the chain which is 1/4 of the chain. Since the chain is uniform, the mass of a portion of chain can be determined by the length of that portion. Mass per unit length will be \(\frac{M}{L}\). Therefore, mass of the hanging 1/4 portion will be: \[ m = \frac{M}{4} \]

Now we'll find the work done to lift the hanging part of the chain onto the table. As the chain is uniform, the mass per unit length is evenly distributed along its length. We can find the work done for a small length (dx) for the hanging part and then integrate to find the total work done. Let x be the length from the bottom end of the hanging part. The work done to bring this small length dx on to the table would be: \[ dw = m \cdot g \cdot x \cdot dx. \] Here, m is mass per unit length, which is \(\frac{M}{4L}\). Now, to find the total work done, we'll integrate from the bottom end (x = 0) to the top end (x = L/4) of the hanging part: \[ W = \int_{0}^{L/4} \frac{M}{4L} g x \, dx = \frac{Mg}{4L} \int_{0}^{L/4} x \, dx. \]

Now, we'll compute the integral: \[ W = \frac{Mg}{4L} \left[\frac{x^2}{2} \right]_{0}^{L/4} = \frac{Mg}{4L} \cdot \frac{(L/4)^2}{2} \] Simplifying the equation, we get: \[ W = \frac{MgL}{32} \] So the correct answer is (D), the work required to pull the hanging part on to the table is \(\mathrm{MgL} / 32\).