When a tensile or compressive load `'P'` is applied to rod or cable, its length changes. The change length `x` which, for an elastic material is proportional to the force Hook' law).
`P alpha x` or `P = kx`
The above equation is similar to the equation of spring. For a rod of length `L`, area `A` and young modulue `Y`. the extension `x`can be expressed as.
`x = (PL)/(AY)` or `P = (AY)/(x)`, hence `K = (AY)/(L)`
Thus rod or cables attached to lift can be treated as springs. The energy stored in rod is called strain energy & equal to `(1)/(2)Px`. The loads placed or dropped on the floor of lift cause stresses in the cable and can be evaluted by spring analogy. If the cable of lift cause stresses and load is placed dropped, then maximum extension in cable can be calculated by energy conservation.
In above problem if mass of `10 kg` falls on the massless collar attached to rod from the height of `99cm` then maximum extension in the rod is equal (rod of length 4m, area `4cm^(2)` and young modules `2 xx 10^(10)N//m^(2)` `g = 10m//sec^(2)`)-

Options

A. `9.9 cm`
B. `10 cm`
C. `0.99 cm`
D. `1 cm`

Explanation

To solve the problem of maximum extension in a rod when a mass falls on it, we can follow these steps:

### Step 1: Identify the Given Values
- Mass of the object (m) = 10 kg
- Height from which the mass falls (h) = 99 cm = 0.99 m
- Length of the rod (L) = 4 m
- Cross-sectional area of the rod (A) = 4 cm² = 4 × 10⁻⁴ m²
- Young's modulus (Y) = 2 × 10¹⁰ N/m²
- Acceleration due to gravity (g) = 10 m/s²

### Step 2: Calculate the Spring Constant (K)
Using the formula for the spring constant:
\[ K = \frac{AY}{L} \]

Substituting the values:
\[ A = 4 \times 10^{-4} \, \text{m}^2 \]
\[ Y = 2 \times 10^{10} \, \text{N/m}^2 \]
\[ L = 4 \, \text{m} \]

Now substituting these into the formula:
\[ K = \frac{(4 \times 10^{-4}) \times (2 \times 10^{10})}{4} \]
\[ K = \frac{8 \times 10^{6}}{4} \]
\[ K = 2 \times 10^{6} \, \text{N/m} \]

### Step 3: Calculate the Potential Energy (PE) of the Falling Mass
The potential energy when the mass is at height \( h \) is given by:
\[ PE = mgh \]
Substituting the values:
\[ PE = 10 \times 10 \times 0.99 \]
\[ PE = 99 \, \text{J} \]

### Step 4: Use Energy Conservation to Find Maximum Extension
At maximum extension, all potential energy is converted into strain energy stored in the rod:
\[ PE = \frac{1}{2} K x_{\text{max}}^2 \]

Setting the two energies equal:
\[ 99 = \frac{1}{2} (2 \times 10^{6}) x_{\text{max}}^2 \]
\[ 99 = 10^{6} x_{\text{max}}^2 \]

### Step 5: Solve for Maximum Extension \( x_{\text{max}} \)
Rearranging the equation:
\[ x_{\text{max}}^2 = \frac{99}{10^{6}} \]
\[ x_{\text{max}} = \sqrt{\frac{99}{10^{6}}} \]
\[ x_{\text{max}} = \sqrt{0.000099} \]
\[ x_{\text{max}} = 0.00995 \, \text{m} \]
Converting to centimeters:
\[ x_{\text{max}} \approx 1 \, \text{cm} \]

### Final Answer
The maximum extension in the rod is approximately **1 cm**.

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