Physics · 2022
JEE Main · 26 July 2022, Shift 2 · Q55
A uniform heavy rod of mass 20 kg, cross sectional area 0.4 m^2 and length 20 m is hanging from a fixed support. Neglecting the lateral contraction,…
A uniform heavy rod of mass $\displaystyle 20$ kg , cross sectional area $\displaystyle 0.4 \mathrm{~m}^2$ and length $\displaystyle 20$ m is hanging from a fixed support. Neglecting the lateral contraction, the elongation in the rod due to its own weight is $\displaystyle x \times 10^{-9} \mathrm{~m}$. The value of $\displaystyle x$ is \_\_\_\_.
(Given, young modulus $\displaystyle \mathrm{Y}=2 \times 10^{11} \mathrm{Nm}^{-2}$ and $\displaystyle g=10 \mathrm{~ms}^{-2}$ )
Official answer
25
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JEE Main 2022 Physics question, with the answer from NTA’s final answer key. Where our answers come from.