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(A) Let the mass per unit length be $\lambda = \frac{m}{l}$.Initial state: The length hanging is $h_1 = \frac{l}{n}$. The mass of this part is $m_1 =

Vedclass · Accepted Answer

$\sqrt{g l\left(1-\frac{1}{n^2}\right)}$

Vedclass · Answer

(A) Let the mass per unit length be $\lambda = \frac{m}{l}$.Initial state: The length hanging is $h_1 = \frac{l}{n}$. The mass of this part is $m_1 = \lambda \cdot \frac{l}{n} = \frac{m}{n}$. The center of mass of the hanging part is at a distance $\frac{l}{2n}$ below the table surface. Taking the table surface as the reference level $(U=0)$,the initial potential energy is $U_i = -m_1 g \left(\frac{l}{2n}\right) = -\left(\frac{m}{n}\right) g \left(\frac{l}{2n}\right) = -\frac{mgl}{2n^2}$.Final state: When the chain completely slips off,its center of mass is at a distance $\frac{l}{2}$ below the table surface. The final potential energy is $U_f = -mg \left(\frac{l}{2}\right) = -\frac{mgl}{2}$.By the law of conservation of energy,the loss in potential energy equals the gain in kinetic energy:$K_f - K_i = U_i - U_f$$\frac{1}{2}mv^2 - 0 = -\frac{mgl}{2n^2} - \left(-\frac{mgl}{2}\right)$$\frac{1}{2}mv^2 = \frac{mgl}{2} \left(1 - \frac{1}{n^2}\right)$$v^2 = gl \left(1 - \frac{1}{n^2}\right)$$v = \sqrt{gl \left(1 - \frac{1}{n^2}\right)}$Wait,checking the calculation: $\frac{1}{2}mv^2 = \frac{mgl}{2}(1 - 1/n^2) \implies v = \sqrt{gl(1 - 1/n^2)}$. Re-evaluating the options,the correct expression is $\sqrt{gl(1 - 1/n^2)}$. However,if the question implies the standard result,let's re-check the factor of $2$. The kinetic energy is $\frac{1}{2}mv^2$. The potential energy change is $\frac{mgl}{2}(1 - 1/n^2)$. Thus $v = \sqrt{gl(1 - 1/n^2)}$. Option $A$ matches this.

$A$ uniform chain of mass $m$ and length $l$ is on a smooth horizontal table with $\left(\frac{1}{n}\right)^{\text{th}}$ part of its length hanging from one end of the table. The velocity of the chain,when it completely slips off the table is

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