Elasticity

$ \text { } \begin{aligned} \Delta l_1 & =\frac{w L}{\left(4 \pi R^2\right) y}, \Delta l_2=\frac{w L}{\pi R^2 Y} \\ \therefore U & =\frac{1}{2} K_1\left(\Delta l_1\right)^2+\frac{1}{2} K_2\left(\Delta l_2\right)^2\left(K=\frac{Y A}{L}\right) \\ & =\frac{1}{2} \times \frac{Y\left(4 \pi R^2\right)}{L} \times\left[\frac{w L}{4 \pi R^2 Y}\right]^2+\frac{1}{2} \times \frac{Y\left(\pi R^2\right)}{L} \times\left[\frac{w L}{\pi R^2 Y}\right]^2\\ & =\frac{5 w^2 L}{8 \pi R^2 Y} \end{aligned} $

To calculate the elastic potential energy in the system, we consider two wires of identical material (both having Young's modulus $ Y $) and length $ L $, but with different radii: $ R $ and $ 2R $. These wires are sequentially connected, and a weight $ w $ is hung from them.

First, determine the elongation of each wire due to the weight $ w $:

For the first wire with radius $ R $:

$ \Delta l_1 = \frac{wL}{4\pi R^2 Y} $

For the second wire with radius $ 2R $:

$ \Delta l_2 = \frac{wL}{\pi R^2 Y} $

The elastic potential energy stored in a wire is given by:

$ U = \frac{1}{2} K (\Delta l)^2 $

where $ K = \frac{YA}{L} $ is the spring constant and $ A $ is the cross-sectional area of the wire.

Calculate the spring constants for both wires:

For the first wire $ (K_1) $:

$ K_1 = \frac{Y \times \pi R^2}{L} $

For the second wire $ (K_2) $:

$ K_2 = \frac{Y \times 4 \pi R^2}{L} $

Now compute the potential energy for each wire and sum them up:

$ U = \frac{1}{2} \times \frac{Y \times 4 \pi R^2}{L} \times \left(\frac{wL}{4\pi R^2 Y}\right)^2 + \frac{1}{2} \times \frac{Y \times \pi R^2}{L} \times \left(\frac{wL}{\pi R^2 Y}\right)^2 $

After performing the calculations, the total elastic potential energy in the system is:

$ U = \frac{5 w^{2} L}{8 \pi R^{2} Y} $

To understand the relationship between the fractional change in volume of an object $(\Delta V / V)$ and its bulk modulus $(B)$ under a constant hydraulic stress, we need to recall the definition and formula of bulk modulus.

The bulk modulus $(B)$ is defined as the ratio of the hydraulic stress (pressure change) to the corresponding relative change in volume (fractional change in volume). Mathematically, it can be expressed as:

$B = -\frac{\Delta P}{\frac{\Delta V}{V}}$

where:

• $B$ is the bulk modulus
• $(\Delta P)$ is the change in pressure (hydraulic stress)
• $(\Delta V / V)$ is the fractional change in volume

From the above equation, we can solve for the fractional change in volume $(\Delta V / V)$:

$\frac{\Delta V}{V} = -\frac{\Delta P}{B}$

Here, it becomes clear that the fractional change in volume $(\Delta V / V)$ is inversely proportional to the bulk modulus $(B)$, assuming the change in pressure $(\Delta P)$ remains constant. This means that if the bulk modulus increases, the fractional change in volume decreases, and vice versa.

Thus, the correct option is:

Option C

$\frac{\Delta V}{V} \propto \frac{1}{B}$

Bulk modulus $ = {{Hydraulic\,stress} \over {Volumetric\,strain}}$

$ \Rightarrow B = {{Hydraulic\,stress} \over {{{\Delta V} \over V}}}$

or, ${{\Delta V} \over V} = $ Hydraulic stress $ \times {1 \over B}$

$\therefore$ For constant hydraulic stress, ${{\Delta V} \over V} \propto {1 \over B}$

Given, linear dimensions increase by a factor of 8.

So, volume will become (8)³ times and area of cross-section will become (8)² times.

Now, $stress = {{weight} \over {area}}$

Weight = (volume $\times$ density $\times$ g) will also become (8)³ times.

$\therefore$ New stress $ = {{{{(8)}^3} \times {w_0}} \over {{{(8)}^2} \times {A_0}}} = 8\left( {{{{w_0}} \over {{A_0}}}} \right)$

Hence, the stress increases by a factor of 8.