Question

A uniform chain of mass M(= 1 kg) and length L(= 10 cm) lies on a frictionless table with length $l_0$(= 6 cm) hanging over the edge. The chain begins to slide down. What is the speed (in m/s ) with which the chain slides away from the edge?

(Take g = 1000 cm/s²)

• A. 18.2 cm/s
• B. 32.6 cm/s
• C. 28.3 cm/s
• D. 9.5 cm/s

Answer 1

80 cm/s (Not among the options provided)

Answer 2

The problem involves finding the speed of a chain sliding off a frictionless table using conservation of energy.

1. Initial Potential Energy (PEi): The hanging part of the chain has potential energy due to its position below the tabletop. $PE_i = -\frac{M g l_0^2}{2L}$

2. Final Potential Energy (PEf): When the entire chain has slid off, its potential energy is determined by the center of mass being at a depth of $L/2$. $PE_f = -M g (L/2)$

3. Kinetic Energies: Initially, the chain is at rest ($KE_i = 0$). Finally, the kinetic energy is $KE_f = \frac{1}{2} M v^2$.

4. Conservation of Energy: $PE_i + KE_i = PE_f + KE_f$

   $-\frac{M g l_0^2}{2L} + 0 = -\frac{M g L}{2} + \frac{1}{2} M v^2$

5. Solving for v: We simplify and solve for $v$:

   $v = \sqrt{\frac{g (L^2 - l_0^2)}{L}}$

6. Substituting Values: Given $L = 0.1 \text{ m}$, $l_0 = 0.06 \text{ m}$, and $g = 10 \text{ m/s}^2$:

   $v = \sqrt{\frac{10 \times ((0.1)^2 - (0.06)^2)}{0.1}} = \sqrt{\frac{10 \times (0.01 - 0.0036)}{0.1}} = \sqrt{0.64} = 0.8 \text{ m/s}$

7. Converting to cm/s: $v = 0.8 \text{ m/s} \times 100 \text{ cm/m} = 80 \text{ cm/s}$

The calculated speed is 80 cm/s, which is not among the provided options. There may be an error in the question's options.