Question
A uniform rod of length $L$ has a mass per unit length $\lambda$ and area of cross-section $A .$ If the young's modulus of the rod is $Y$, then the elongation in the rod due to its own weight is(a) $\frac{2 \lambda g L^{2}}{\Lambda Y}$(b) $\frac{\lambda g L^{2}}{\Lambda Y}$(c) $\frac{\lambda g L^{2}}{4 \Lambda Y}$(d) $\frac{\lambda g L^{2}}{2 \Lambda Y}$
Step 1
This is given by the formula $\Delta L = \frac{F}{A} \cdot \frac{L}{Y}$, where $F$ is the force acting on the rod, $A$ is the cross-sectional area of the rod, $L$ is the length of the rod, and $Y$ is the Young's modulus of the rod. Show more…
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