A wire of length `3 m` diameter `0.4 mm` and young's modulus `8xx10^(10) N//m^(2)` is suspended from a point and supports a heavy cylinder of volume `10^(-3) m^(3)` at its lower end . Find the decrease in length when the metal cylinder is immersed in a liquid of density `800 kg //m^(3)`.

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Calculate the weight of the cylinder
The weight of the cylinder can be calculated using the formula:
\[ W = V \cdot \rho \cdot g \]
where:
- \( V = 10^{-3} \, m^3 \) (volume of the cylinder)
- \( \rho = 800 \, kg/m^3 \) (density of the liquid)
- \( g = 9.8 \, m/s^2 \) (acceleration due to gravity)

Calculating the weight:
\[ W = 10^{-3} \cdot 800 \cdot 9.8 = 7.84 \, N \]

### Step 2: Calculate the area of cross-section of the wire
The area of cross-section \( A \) can be calculated using the formula for the area of a circle:
\[ A = \frac{\pi}{4} \cdot d^2 \]
where:
- \( d = 0.4 \, mm = 0.4 \times 10^{-3} \, m \)

Calculating the area:
\[ A = \frac{\pi}{4} \cdot (0.4 \times 10^{-3})^2 \]
\[ A = \frac{\pi}{4} \cdot 0.16 \times 10^{-6} \]
\[ A \approx 0.1256 \times 10^{-6} \, m^2 \]

### Step 3: Calculate the decrease in length of the wire
The decrease in length \( \Delta L \) can be calculated using the formula:
\[ \Delta L = \frac{F \cdot L}{Y \cdot A} \]
where:
- \( F = 7.84 \, N \) (upthrust force)
- \( L = 3 \, m \) (length of the wire)
- \( Y = 8 \times 10^{10} \, N/m^2 \) (Young's modulus)
- \( A \) is calculated from Step 2.

Substituting the values:
\[ \Delta L = \frac{7.84 \cdot 3}{8 \times 10^{10} \cdot 0.1256 \times 10^{-6}} \]

Calculating:
\[ \Delta L = \frac{23.52}{10.048 \times 10^{4}} \]
\[ \Delta L \approx 2.34 \times 10^{-3} \, m \]

### Final Answer
The decrease in length of the wire when the metal cylinder is immersed in the liquid is approximately:
\[ \Delta L \approx 2.34 \, mm \]

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