JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Laws of Motion

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A uniform chain of 6 m length is placed on a table such that a part of its length is hanging over the edge of the table. The system is at rest. The co-efficient of static friction between the chain and the surface of the table is 0.5, the maximum length of the chain hanging from the table is ______ m.

Answer: 2

Solution

Let

λ

be the mass per unit length,

x

be the hanging length, and

L = 6

m be the total length.

The normal force on the table is

N = λ (L - x) g

.

The maximum static friction is

f_{s, m a x} = μ_{s} N = 0.5 λ (L - x) g

.

For equilibrium, this friction must balance the weight of the hanging part:

0.5 λ (L - x) g = λ xg

Simplifying:

0.5 (L - x) = x

, which gives

0.5 L = 1.5 x

.

Therefore,

x = \frac{L}{3} = \frac{6}{3} = 2

JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Integer answer

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