An indian ruber cord L meter long and area of cross-secion A metre is suspended vertically. Density of rubber is `rho` kg/ `"metre"^(3)` and Young's modulus of rubber is Y newton/ `"metre"^(2)`. IF the cord extends by l metre under its own. Weight, then extension l is

Options

A. `L^(2) rho g//Y`
B. `L^(2) rho g//2y`
C. `L^(2)) rho g//4 Y`
D. `Y//L^(2) rho g`.

Explanation

To solve the problem of finding the extension \( l \) of a rubber cord under its own weight, we will follow these steps:

### Step 1: Understand the Problem
We have a rubber cord of length \( L \) and cross-sectional area \( A \). The density of rubber is \( \rho \) kg/m³, and Young's modulus is \( Y \) N/m². The cord extends by \( l \) meters due to its own weight.

### Step 2: Consider a Differential Element
Consider a small differential element of the cord at a distance \( x \) from the bottom of the cord. The length of this element is \( dx \).

### Step 3: Calculate the Weight of the Element
The mass of the differential element can be expressed as:
\[ dm = \rho \cdot A \cdot dx \]
The weight \( dF \) of this differential element is given by:
\[ dF = dm \cdot g = \rho \cdot A \cdot dx \cdot g \]
where \( g \) is the acceleration due to gravity.

### Step 4: Apply the Concept of Stress and Strain
The force acting on the differential element causes it to stretch. The extension \( dL \) of the element can be related to the stress and strain as follows:
\[ \text{Stress} = \frac{F}{A} = \frac{dF}{A} = \frac{\rho \cdot A \cdot g \cdot dx}{A} = \rho \cdot g \cdot dx \]
\[ \text{Strain} = \frac{dL}{dx} \]
Using Young's modulus \( Y \):
\[ Y = \frac{\text{Stress}}{\text{Strain}} \implies Y = \frac{\rho \cdot g \cdot dx}{\frac{dL}{dx}} \]
Rearranging gives:
\[ dL = \frac{\rho \cdot g \cdot dx}{Y} \]

### Step 5: Integrate to Find Total Extension
To find the total extension \( l \) of the cord, we integrate \( dL \) from \( 0 \) to \( L \):
\[ l = \int_0^L \frac{\rho \cdot g \cdot dx}{Y} \]
This simplifies to:
\[ l = \frac{\rho \cdot g}{Y} \int_0^L dx = \frac{\rho \cdot g}{Y} \cdot L \]

### Step 6: Final Expression for Extension
Thus, the total extension \( l \) of the rubber cord under its own weight is:
\[ l = \frac{\rho \cdot g \cdot L^2}{2Y} \]

### Conclusion
The extension \( l \) of the rubber cord is given by:
\[ l = \frac{\rho g L^2}{2Y} \]