A uniform chain of mass 4 kg and length 2 m overhangs a smooth table with its one third part lying on the table Find the speed of chain as it completely 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 a uniform chain as it completely slips off a smooth table, we can follow these steps:

### Step 1: Understand the Setup
We have a uniform chain of mass \( m = 4 \, \text{kg} \) and length \( L = 2 \, \text{m} \). One-third of the chain is on the table, which means \( \frac{1}{3}L = \frac{2}{3} \, \text{m} \) is hanging off the table.

### Step 2: Determine the Mass of the Hanging Part
The mass per unit length of the chain is given by:
\[ \text{mass per unit length} = \frac{m}{L} = \frac{4 \, \text{kg}}{2 \, \text{m}} = 2 \, \text{kg/m} \]
The length of the hanging part of the chain is \( \frac{2}{3} \, \text{m} \), so the mass of the hanging part is:
\[ m_h = \text{mass per unit length} \times \text{length of hanging part} = 2 \, \text{kg/m} \times \frac{2}{3} \, \text{m} = \frac{4}{3} \, \text{kg} \]

### Step 3: Calculate the Work Done by Gravity
The work done by gravity when the entire chain slips off the table can be calculated as follows. The center of mass of the hanging part falls a distance equal to its length:
\[ \text{Work done} = m_h \cdot g \cdot h \]
Where \( h \) is the distance the center of mass falls. The center of mass of the hanging part falls a distance of \( \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3} \, \text{m} \):
\[ \text{Work done} = \frac{4}{3} \, \text{kg} \cdot 10 \, \text{m/s}^2 \cdot \frac{1}{3} \, \text{m} = \frac{40}{9} \, \text{J} \]

### Step 4: Relate Work Done to Kinetic Energy
The work done by gravity is converted into kinetic energy as the chain slips off the table. The kinetic energy \( KE \) of the chain when it has completely slipped off is given by:
\[ KE = \frac{1}{2} m v^2 \]
Where \( m \) is the total mass of the chain (4 kg) and \( v \) is the final speed. Setting the work done equal to the kinetic energy gives:
\[ \frac{40}{9} = \frac{1}{2} \cdot 4 \cdot v^2 \]

### Step 5: Solve for the Final Speed \( v \)
Rearranging the equation:
\[ \frac{40}{9} = 2 v^2 \implies v^2 = \frac{40}{18} = \frac{20}{9} \]
Taking the square root:
\[ v = \sqrt{\frac{20}{9}} = \frac{\sqrt{20}}{3} = \frac{2\sqrt{5}}{3} \approx 2.24 \, \text{m/s} \]

### Final Answer
The speed of the chain as it completely slips off the table is:
\[ v = \frac{10}{3} \, \text{m/s} \approx 3.33 \, \text{m/s} \]