Explanation
$$\dfrac{\Delta V}{V}=\dfrac{h\rho g}{k}=\dfrac{3\times 10^3\times 9.8 \times 10^3}{22\times 10^8} $$
$$=\dfrac{3}{22}\times 10^{-3}\times 98\\ =\dfrac{147}{11}\times 10^{-3}\\=1.34\times 10^{-2}m$$
An iron rod of length 2m and cross- sectional area of $$50mm^{2}$$ stretched by 0.5mm, when a mass of 250 kg is hung from its lower end. Young's modulus of iron rod is
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