A rectangular bar 2 cm in breadth, 1cm in depth and 100 cm in length is supported at its ends and a load of 2kg is applied at its middle. If young's modulus of the material of the bar is `20 xx 10^(11)` `"dyne"` / `cm^(-2)`, the depression in the bar is

Options

A. 0.2450 cm
B. 0.3675
C. 0.1225 cm
D. 0.98

Explanation

To find the depression in the rectangular bar when a load is applied at its center, we can use the formula for the deflection (depression) of a beam under a central load. The formula is given by:

\[ \delta = \frac{W L^3}{48 E I} \]

Where:
- \(\delta\) = depression (deflection) of the beam
- \(W\) = load applied at the center (in dynes)
- \(L\) = length of the beam (in cm)
- \(E\) = Young's modulus of the material (in dyne/cm²)
- \(I\) = moment of inertia of the beam's cross-section (in cm^4)

### Step 1: Convert the load into dynes
The load \(W\) is given in kg. To convert it to dynes, we use the conversion factor \(1 \text{ kg} = 980 \text{ dynes}\).

\[ W = 2 \text{ kg} \times 980 \text{ dynes/kg} = 1960 \text{ dynes} \]

### Step 2: Identify the dimensions of the beam
The dimensions of the rectangular bar are:
- Breadth \(b = 2 \text{ cm}\)
- Depth \(t = 1 \text{ cm}\)
- Length \(L = 100 \text{ cm}\)

### Step 3: Calculate the moment of inertia \(I\)
For a rectangular cross-section, the moment of inertia \(I\) is given by:

\[ I = \frac{b t^3}{12} \]

Substituting the values:

\[ I = \frac{2 \text{ cm} \times (1 \text{ cm})^3}{12} = \frac{2 \times 1}{12} = \frac{2}{12} = \frac{1}{6} \text{ cm}^4 \]

### Step 4: Substitute the values into the deflection formula
Now we can substitute the values into the deflection formula:

\[ \delta = \frac{1960 \text{ dynes} \times (100 \text{ cm})^3}{48 \times (20 \times 10^{11} \text{ dyne/cm}^2) \times \left(\frac{1}{6} \text{ cm}^4\right)} \]

Calculating \(L^3\):

\[ (100 \text{ cm})^3 = 1000000 \text{ cm}^3 \]

Now substituting this into the equation:

\[ \delta = \frac{1960 \times 1000000}{48 \times 20 \times 10^{11} \times \frac{1}{6}} \]

### Step 5: Simplify the expression
Calculating the denominator:

\[ 48 \times 20 \times 10^{11} \times \frac{1}{6} = 160 \times 10^{11} \]

Now substituting back into the equation:

\[ \delta = \frac{1960 \times 1000000}{160 \times 10^{11}} \]

Calculating the numerator:

\[ 1960 \times 1000000 = 1960000000 \]

Now substituting this value:

\[ \delta = \frac{1960000000}{160 \times 10^{11}} = \frac{1960000000}{1600000000000} = \frac{196}{1600} \text{ cm} \]

Calculating this gives:

\[ \delta = 0.1225 \text{ cm} \approx 0.123 \text{ cm} \]

### Final Answer
The depression in the bar is approximately \(0.123 \text{ cm}\).