The density and breaking stress of a wire are 6 × 10⁴ kg / m³ and 1.2 × 10⁸ N / m² respectively. The wire is suspended from a rigid support on a planet where acceleration due to gravity is 1/3ʳᵈ of the value on the surface of earth. The maximum length of the wire with breaking is _______ m (take, g=10 m / s²).

Solution

Physics sir · 4 steps

The top point of the wire holds up the whole wire, so set the stress there equal to the breaking stress.

1. Where it breaks. Each cross-section of a hanging wire supports only the wire below it. The top section supports the entire length, so the stress is largest there and the wire breaks there first.
2. Stress at the top. A wire of length L and cross-section A has weight ρ A L g'. Dividing by A, the area cancels, and the stress at the top is ρ L g'.

   ρ g' Lₘₐₓ = σ_(breaking) ⇒ Lₘₐₓ = σ_(breaking)/(ρ g')

3. Put in numbers. On this planet g' = 10/3 m s⁻², so ρ g' = 6 × 10⁴ × 10/3 = 2 × 10⁵ N m⁻³. Then Lₘₐₓ = (1.2 × 10⁸)/(2 × 10⁵) = 600 m.
4. So. The longest wire that can hang without breaking is 600 m.

The common slip: 200. Using Earth's g = 10 m s⁻² instead of 10/3 m s⁻² gives 200 m, three times too short.

The idea to keep. For a wire hanging under its own weight, the maximum length is σ/(ρ g). It does not depend on the thickness, and a weaker g allows a proportionally longer wire.

Answer from the NTA final answer key. Solution by PrepWiser.