Physics · 2020
JEE Main · 9 January 2020, Shift 2 · Q5
A rod of length L has non-uniform linear mass density given by ρ(x)= a + b (x/L)^2, where a and b are constants and 0 ≤ x ≤ L. The value of x for the…
A rod of length L has non-uniform linear mass density given by $\displaystyle \rho(x)=\mathrm{a}+\mathrm{b}\left(\frac{x}{\mathrm{~L}}\right)^2$, where a and b are constants and $\displaystyle 0 \leq x \leq \mathrm{L}$. The value of $\displaystyle x$ for the centre of mass of the rod is at:
Official answer
(1)
$\displaystyle \frac{3}{4}\left(\frac{2 a+b}{3 a+b}\right) L$
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JEE Main 2020 Physics question, with the answer from NTA’s final answer key. Where our answers come from.