A uniform chain of mass 4 kg and legth 2 m overhags a smooth table with its one third part lying on the table Find the speed of chain as it c ompletely slips of the table. (Take `g=10m//s^(2)`)

Options

A. `(10)/(3)m//s`
B. `2m//s`
C. `3/2m//s`
D. `4/2m//s`

Answer

To solve the problem of finding the speed of the chain as it completely slips off the table, we can use the principles of energy conservation. Here is the step-by-step solution:

### Step 1: Understand the setup
The chain has a total length of 2 m and a mass of 4 kg. One-third of the chain (which is \( \frac{2}{3} \) m) is hanging off the table, while the remaining one-third (which is \( \frac{2}{3} \) m) is on the table.

### Step 2: Calculate the mass of the hanging part
Since the chain is uniform, the mass per unit length is given by:
\[ \text{Mass per unit length} = \frac{\text{Total mass}}{\text{Total length}} = \frac{4 \, \text{kg}}{2 \, \text{m}} = 2 \, \text{kg/m} \]
The length of the hanging part is \( \frac{2}{3} \) m, so the mass of the hanging part is:
\[ \text{Mass of hanging part} = \text{Mass per unit length} \times \text{Length of hanging part} = 2 \, \text{kg/m} \times \frac{2}{3} \, \text{m} = \frac{4}{3} \, \text{kg} \]

### Step 3: Calculate the potential energy of the hanging part
The center of mass of the hanging part is at a distance of \( \frac{1}{3} \) m from the table edge (since it is hanging down). The potential energy (PE) of the hanging part when it is at rest is given by:
\[ \text{PE} = mgh = \left(\frac{4}{3} \, \text{kg}\right) \times (10 \, \text{m/s}^2) \times \left(\frac{1}{3} \, \text{m}\right) = \frac{40}{9} \, \text{J} \]

### Step 4: Apply conservation of energy
When the chain slips off the table, the potential energy will convert into kinetic energy (KE). The total mass of the chain is 4 kg, and the kinetic energy when the chain has completely slipped off the table is given by:
\[ \text{KE} = \frac{1}{2} mv^2 \]
Setting the potential energy equal to the kinetic energy gives:
\[ \frac{40}{9} \, \text{J} = \frac{1}{2} \times 4 \, \text{kg} \times v^2 \]

### Step 5: Solve for the speed \( v \)
Rearranging the equation to solve for \( v^2 \):
\[ \frac{40}{9} = 2v^2 \implies v^2 = \frac{40}{18} = \frac{20}{9} \]
Taking the square root gives:
\[ v = \sqrt{\frac{20}{9}} = \frac{\sqrt{20}}{3} = \frac{2\sqrt{5}}{3} \, \text{m/s} \]

### Final Answer
The speed of the chain as it completely slips off the table is:
\[ v \approx 1.49 \, \text{m/s} \]

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