Solve the following :

Question:

A rod of length $\mathrm{L}$ is placed along the $\mathrm{X}$-axis between $\mathrm{x}=0$ and $\mathrm{X}=\mathrm{L}$. The linear density (mass/length) $p$ of the rod varies with the distance $\mathrm{x}$ from the origin as $p=\alpha+b \mathrm{~b}$. (a) Find the SI units of $\mathrm{a}$ and $\mathrm{b}$. (b) Find the mass of the rod in terms of $a, b$ and $L$.

Solution:

It is given that linear density (mass per unit length) $p=\mathrm{a}+\mathrm{bx}$, where $\mathrm{x}$ is distance from origin.

a) S.I. unit of $a=\mathrm{kg} / \mathrm{m}$

S.I. unit of $b=\mathrm{kg} / \mathrm{m}^{2}$

b) To find the mass of the rod, we have to take a small segment $d x$ placed at a distance $x$ from the origin.

or, $\mathrm{dm}=p \mathrm{dx}$

or, $m=\int_{0}^{L} \rho d x=\int_{0}^{L}(a+b x) d x=\left[a x+b x^{2} / 2\right]_{0}^{L_{0}}=a L+0.5 b^{2}$

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