A uniform heavy rod of mass 20 kg, cross sectional area 0.4 m² and length 20 m is hanging from a fixed support. Neglecting the lateral contraction, the elongation in the rod due to its own weight is x × 10⁻⁹ m. The value of x is _______________.
(Given, young modulus Y = 2 × 10¹¹ Nm⁻² and g = 10 ms⁻²)

Solution

Physics sir · 4 steps

Tension grows from zero at the bottom to Mg at the top, so use half the weight as the average load.

1. Tension varies. Take a thin slice dy at height y above the free bottom end. It holds up only the rod below it, which has weight M/Lgy. So the tension rises from 0 at the bottom to Mg at the support.
2. Add up stretches. That slice stretches by (T dy)/AY = (Mg y dy)/LAY. Integrating from y = 0 to y = L gives Mg/LAY × L²/2.

   Δ L = MgL/2AY

3. Put in numbers. Δ L = (20 × 10 × 20)/(2 × 0.4 × 2 × 10¹¹) = 4000/(1.6 × 10¹¹) = 2.5 × 10⁻⁸ m.
4. So. Δ L = 25 × 10⁻⁹ m, so x = 25.

The common slip: 50. Using the full weight Mg as the tension everywhere drops the factor of 2 and gives 50.

The idea to keep. A rod hanging under its own weight stretches by MgL/2AY. That is half of what the same weight would do if hung from its end, because the average tension is Mg/2.

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