A metal wire of length `L_1` and area of cross section A is attached to a rigid support. Another metal wire of length `L_2` and of the same cross sectional area is attached to the free end of the first wire. A body of mass M is then suspended from the free end of the second wire, if `Y_1` and `Y_2` are the Young's moduli of the wires respectively the effective force constant of the system of two wires is

Options

A. `([(Y_(1)Y_(2))A])/([2(Y_(1)L_(2) +Y_(2)L_(1))])`
B. `([(Y_(1)Y_(2))A])/((L_(1)L_(2))^(2))`
C. `([(Y_(1)Y_(2))A])/((Y_(1)L_(2) +Y_(2)L_(1)))`
D. `([(Y_(1) Y_(2))^(1//2)A])/((L_(2)+L_(1))^(1//2))`

Explanation

To find the effective force constant of the system of two metal wires, we will follow these steps:

### Step 1: Understand the System
We have two metal wires attached in series. The first wire has length \( L_1 \) and Young's modulus \( Y_1 \), and the second wire has length \( L_2 \) and Young's modulus \( Y_2 \). Both wires have the same cross-sectional area \( A \).

### Step 2: Use Young's Modulus to Relate Force and Elongation
Young's modulus \( Y \) is defined as the ratio of stress to strain:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} \]
From this, we can derive the relationship for force \( F \) and elongation \( \Delta L \):
\[ Y = \frac{F L}{A \Delta L} \implies F = \frac{Y A \Delta L}{L} \]

### Step 3: Define the Force Constants for Each Wire
The force constant \( k \) for a wire can be defined as:
\[ k = \frac{F}{\Delta L} \]
From the previous equation, we can express \( k \) in terms of Young's modulus:
\[ k = \frac{Y A}{L} \]
Thus, for the first wire:
\[ k_1 = \frac{Y_1 A}{L_1} \]
And for the second wire:
\[ k_2 = \frac{Y_2 A}{L_2} \]

### Step 4: Calculate the Effective Force Constant
When two springs (or wires in this case) are connected in series, the effective force constant \( k_{\text{equiv}} \) is given by:
\[ \frac{1}{k_{\text{equiv}}} = \frac{1}{k_1} + \frac{1}{k_2} \]
Substituting the expressions for \( k_1 \) and \( k_2 \):
\[ \frac{1}{k_{\text{equiv}}} = \frac{L_1}{Y_1 A} + \frac{L_2}{Y_2 A} \]
This can be simplified to:
\[ \frac{1}{k_{\text{equiv}}} = \frac{L_1 Y_2 + L_2 Y_1}{Y_1 Y_2 A} \]

### Step 5: Invert to Find \( k_{\text{equiv}} \)
Taking the reciprocal gives us:
\[ k_{\text{equiv}} = \frac{Y_1 Y_2 A}{L_1 Y_2 + L_2 Y_1} \]

### Final Result
Thus, the effective force constant of the system of two wires is:
\[ k_{\text{equiv}} = \frac{Y_1 Y_2 A}{L_1 Y_2 + L_2 Y_1} \]