Question

A uniform dense rod with non uniform young's modulus is hanging from ceiling under gravity. If elastic energy density at every point is same then young's modulus with x will change as which of the shown graph :-

• A. A
• B. B
• C. C
• D. D

Answer 1

Answer: (C)

C

Answer 2

Let the rod have length $l$, mass $m$, and constant cross-sectional area $A$. Let $x$ be the distance measured from the top end of the rod (hanging from the ceiling), so $x=0$ at the top and $x=l$ at the bottom. The rod has a uniform density $\rho$.

The tension $T(x)$ at a point $x$ is the weight of the rod below that point. The mass of the segment from $x$ to $l$ is $dm = \rho A dx$. The weight of this segment is $dF = dm \cdot g = \rho A g dx$. The total tension at $x$ is $T(x) = \int_x^l \rho A g dx = \rho A g (l-x)$.

The stress $\sigma(x)$ at position $x$ is $\sigma(x) = \frac{T(x)}{A} = \rho g (l-x)$. The strain $\epsilon(x)$ at position $x$ is given by $\epsilon(x) = \frac{\sigma(x)}{Y(x)}$, where $Y(x)$ is the Young's modulus at position $x$. So, $\epsilon(x) = \frac{\rho g (l-x)}{Y(x)}$.

The elastic energy density $u(x)$ at position $x$ is given by $u(x) = \frac{1}{2} \sigma(x) \epsilon(x)$. Substituting the expressions for $\sigma(x)$ and $\epsilon(x)$: $u(x) = \frac{1}{2} \left(\rho g (l-x)\right) \left(\frac{\rho g (l-x)}{Y(x)}\right) = \frac{(\rho g (l-x))^2}{2 Y(x)}$.

We are given that the elastic energy density $u(x)$ is the same at every point, so $u(x) = u_0$, where $u_0$ is a constant. $u_0 = \frac{\rho^2 g^2 (l-x)^2}{2 Y(x)}$.

We need to find how $Y(x)$ changes with $x$. Rearranging the equation for $Y(x)$: $Y(x) = \frac{\rho^2 g^2 (l-x)^2}{2 u_0}$.

Since $\rho$, $g$, and $u_0$ are constants, we can write $Y(x) = C(l-x)^2$, where $C = \frac{\rho^2 g^2}{2 u_0}$ is a positive constant.

We need to find the graph that represents $Y(x) = C(l-x)^2$ for $x$ in the range $[0, l]$.

1. At $x=0$ (top of the rod), $Y(0) = C(l-0)^2 = Cl^2$. This is the maximum value of Young's modulus.
2. At $x=l$ (bottom of the rod), $Y(l) = C(l-l)^2 = 0$. This is the minimum value of Young's modulus.
3. The function $Y(x)$ is a decreasing function of $x$ in the range $[0, l]$.
4. To determine the concavity, we find the second derivative of $Y(x)$ with respect to $x$: $\frac{dY}{dx} = C \cdot 2(l-x) \cdot (-1) = -2C(l-x)$. $\frac{d^2Y}{dx^2} = -2C(-1) = 2C$. Since $C > 0$, $\frac{d^2Y}{dx^2} > 0$. This means the graph of $Y(x)$ versus $x$ is concave up.

Observing the given graphs:

• Graph (A) shows a linear increase.
• Graph (B) shows a rapid decrease, resembling an inverse relationship.
• Graph (C) shows a decrease from a high value to a low value, and the curve is bending upwards (concave up).
• Graph (D) shows a decrease from a high value to a low value, and the curve is bending downwards (concave down).

Graph (C) correctly represents a function that is decreasing and concave up, matching our derived function $Y(x) = C(l-x)^2$.