A wire loaded by a weight of density `7.8 g//c c` is found to be of length 100cm. On immersing the weight in water. The length decrease by 0.20 cm find the original length of the wire.

Explanation

To solve the problem step by step, we will follow the principles of elasticity and the relationship between stress, strain, and Young's modulus.

### Step 1: Define the Variables
Let:
- \( L \) = original length of the wire (in cm)
- \( \Delta L \) = change in length of the wire when immersed in water = 0.20 cm
- \( \rho \) = density of the weight = 7.8 g/cm³
- \( g \) = acceleration due to gravity = 9.8 m/s² (or 980 cm/s² for consistency in units)

### Step 2: Write the Stress-Strain Relationship
According to the stress-strain relationship, we have:
\[ \text{Stress} = \text{Strain} \times Y \]
Where:
- Stress \( \sigma = \frac{F}{A} \)
- Strain \( \epsilon = \frac{\Delta L}{L} \)
- \( Y \) = Young's modulus

### Step 3: Calculate the Force
The force \( F \) acting on the wire due to the weight is given by:
\[ F = mg \]
Where:
- \( m \) = mass of the weight
- \( g \) = acceleration due to gravity

The mass \( m \) can be expressed in terms of density and volume:
\[ m = \rho \cdot V \]
Assuming the volume \( V \) can be expressed as \( A \cdot L \) (where \( A \) is the cross-sectional area and \( L \) is the length of the wire):
\[ F = \rho \cdot (A \cdot L) \cdot g \]

### Step 4: Substitute into the Stress Equation
Substituting for \( F \) in the stress equation:
\[ \sigma = \frac{\rho \cdot (A \cdot L) \cdot g}{A} = \rho \cdot L \cdot g \]
Thus,
\[ \sigma = \rho \cdot g \cdot L \]

### Step 5: Relate to Young's Modulus
From the stress-strain relationship:
\[ \rho \cdot g \cdot L = Y \cdot \frac{\Delta L}{L} \]
Rearranging gives:
\[ Y = \frac{\rho \cdot g \cdot L^2}{\Delta L} \]

### Step 6: Set Up the Equations
1. For the original length of the wire:
\[ Y = \frac{7.8 \cdot 980 \cdot L^2}{\Delta L} \]
where \( \Delta L \) is the decrease in length when the weight is immersed in water.

2. For the decrease in length when immersed in water:
\[ Y = \frac{1 \cdot 980 \cdot L^2}{0.20} \]

### Step 7: Equate the Two Expressions for Y
Equating the two expressions for \( Y \):
\[ \frac{7.8 \cdot 980 \cdot L^2}{0.20} = \frac{1 \cdot 980 \cdot L^2}{0.20} \]
This simplifies to:
\[ 7.8 = \frac{100 - L}{0.20} \]

### Step 8: Solve for L
Rearranging gives:
\[ 7.8 \cdot 0.20 = 100 - L \]
\[ 1.56 = 100 - L \]
Thus,
\[ L = 100 - 1.56 = 98.44 \text{ cm} \]

### Final Answer
The original length of the wire is:
\[ \boxed{98.44 \text{ cm}} \]