A homogeneous chain of length $L$ lies on a table. Some part of chain is hanging down the table. The coefficient of friction between the chain and the table is $μ$ . The maximum length which can hangover the table in equilibrium is :

\frac{μ L}{1 + μ}

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(\frac{1 - μ}{μ}) L

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{(\frac{1 - μ}{1 + μ})}^{2}

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(\frac{2 μ}{1 + 2 μ}) L

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Solution

The correct option is A $\frac{μ L}{1 + μ}$
Let $m$ be the mass of the chain.
Force pulling the chain down $= \frac{m}{L} x \cdot g$
Friction Force $= μ \frac{m}{L} (L - x) g$
$\Rightarrow \frac{m}{L} x g = μ \frac{m}{L} (L - x) g$ .
$\Rightarrow x = \frac{μ L}{1 + μ}$

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A homogeneous chain of length L lies on a table. Some part of chain is hanging down the table. The coefficient of friction between the chain and the table is μ. The maximum length which can hangover the table in equilibrium is :

The correct option is A μL1+μLet m be the mass of the chain.Force pulling the chain down=mLx⋅gFriction Force=μmL(L−x)g⇒mLxg=μmL(L−x)g.⇒x=μL1+μ

A homogeneous chain of length $L$ lies on a table. Some part of chain is hanging down the table. The coefficient of friction between the chain and the table is $μ$ . The maximum length which can hangover the table in equilibrium is :

The correct option is A $\frac{μ L}{1 + μ}$
Let $m$ be the mass of the chain.
Force pulling the chain down $= \frac{m}{L} x \cdot g$
Friction Force $= μ \frac{m}{L} (L - x) g$
$\Rightarrow \frac{m}{L} x g = μ \frac{m}{L} (L - x) g$ .
$\Rightarrow x = \frac{μ L}{1 + μ}$