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← JEE Main 2022 · 26 Jul · Shift 2
PhysicsInteger+4 / -1Question 55 of 90

JEE Main 2022 · 26 Jul · Shift 2, Physics question 55

A uniform heavy rod of mass 20 kg, cross sectional area 0.4 m2^2 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×10−9x\times10^{-9} m. The value of xx is ______.

(Given, young modulus Y = 2×10112\times10^{11} Nm−2^{-2} and g = 10 ms−2^{-2})

Answer and worked solution

Official answer: 25

Own weight stretches the rod by MgL2AY=20×10×202×0.4×2×1011=2.5×10−8\frac{MgL}{2AY}=\frac{20\times10\times20}{2\times0.4\times2\times10^{11}}=2.5\times10^{-8} m.

Answer: 25