A uniform metal chain of mass m and length L | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $\lambda$ be the linear mass density of the chain,so $\lambda = \frac{m}{L}$.The mass of the chain on one side is $m_1 = \lambda l$ and on the

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) Let $\lambda$ be the linear mass density of the chain,so $\lambda = \frac{m}{L}$.The mass of the chain on one side is $m_1 = \lambda l$ and on the other side is $m_2 = \lambda (L - l)$.The net force on the chain is $F_{net} = (m_2 - m_1)g = \lambda(L - l - l)g = \lambda(L - 2l)g$.The total mass of the chain is $m = \lambda L$.The acceleration $a$ of the chain is given by $a = \frac{F_{net}}{m} = \frac{\lambda(L - 2l)g}{\lambda L} = \frac{(L - 2l)g}{L}$.Given that $a = \frac{g}{2}$ when $l = \frac{L}{x}$,we substitute these values:$\frac{g}{2} = \frac{(L - 2(\frac{L}{x}))g}{L}$$\frac{1}{2} = 1 - \frac{2}{x}$$\frac{2}{x} = 1 - \frac{1}{2} = \frac{1}{2}$$x = 4$.

Similar Questions

$A$ particle of small mass $m$ is joined to a very heavy body of mass $M$ by a light string passing over a light pulley. Both bodies are free to move. The total downward force on the pulley is

In the figure given below,what is the reading of the spring balance (in $, N$)? (Take $g = 10\, N \,kg^{-1}$)

The ends of a rope are held by two men who pull on it with equal and opposite forces of magnitude $F$ each. Then the tension in the rope is

$A$ $2 \, kg$ block is lying on a smooth table and is connected to a body of mass $1 \, kg$ by a string passing over a pulley. The $1 \, kg$ mass is hanging vertically. Find the acceleration of the block and the tension in the string.

Vedclass Test Series

Exam Paper Generator

Online Exam Module