Question bank/JEE Main/Physics

A thin rod of length 1 m is kept vertical and is hinged from the
lower end Its linear mass density varies with the distance (x)
from the end $H^{'}$ as $\lambda = (12x)kg/m$, where x is distance along
rod from point H. The potential energy (in Joules) of the rod is 5 N.
Find N (at H the potential energy is zero)
$\left( g = 10m/s^{2} \right)$.

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Lets take an element of dx width at a distance x from the lowest
end.
$dU = (dm)gx,U = 12g\int_{x = 0}^{x - 1}\mspace{2mu} x^{2}dx = 40J$

Source: Grass-G / P-06 Physics paper 14 October.docx