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’s modulus Y = 2 × 10¹¹ Nm^–2 and g = 10 ms^–2)

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answer is 25.

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Detailed Solution

$\begin{array}{l} Y = \frac{T}{A} \frac{dx}{dy} \\ m = 20 kg \\ A = 0.4 m^{2} \\ 1 = 20 m \end{array}$

let extension is dy in length dx

$\begin{array}{l} Y = \frac{stress}{strain} \\ Y = \frac{\frac{T}{dy}}{\frac{d}{dx}} = \frac{T}{A} \cdot \frac{dx}{dy} \\ dy = \frac{Tdx}{AY} \end{array}$

Tension at a distance x from lower end $= \frac{mg}{ℓ} x$

So. $\int_{0}^{Δl} dy = \int_{0}^{ℓ} \frac{mg}{ℓ} x \frac{dx}{AY}$

$\begin{array}{l} Δℓ = \frac{mg}{ℓAY} {[\frac{x^{2}}{2}]}_{0}^{ℓ} \\ Δℓ = \frac{mgℓ}{2 AY} \\ Δℓ = \frac{20 \times 10 \times 20}{2 \times 0 .4 \times 2 \times 10^{11}} \\ = 2500 \times 10^{- 11} \\ Δℓ = 25 \times 10^{- 9} \\ = x \times 10^{- 9} \\ x = 25 \end{array}$

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