To solve this problem, we need to determine the ratio of the increase in lengths of two wires made of steel and copper, given their respective ratios of diameter, length, and Young's modulus. Let’s analyze each component and derive the formula step-by-step.
The formula to find the increase in length (\Delta L) of a wire under a given load is based on Hooke's Law: \Delta L = \frac{F L}{A Y}, where:
The cross-sectional area (A) for a wire, considering it's circular, is \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4}. Thus, substituting for both wires:
The formulas for the increase in length for steel and copper wires can be given as:
\Delta L_{steel} = \frac{F L_{steel}}{\frac{\pi d_{steel}^2}{4} \cdot Y_{steel}}
\Delta L_{copper} = \frac{F L_{copper}}{\frac{\pi d_{copper}^2}{4} \cdot Y_{copper}}
We are given the ratios of diameters, lengths, and Young's modulus:
d_{steel}:d_{copper} = p, L_{steel}:L_{copper} = q, Y_{steel}:Y_{copper} = s.
Therefore, d_{steel} = p \cdot d_{copper}, L_{steel} = q \cdot L_{copper}, Y_{steel} = s \cdot Y_{copper}.
Using these relations, the ratio of increase in lengths for steel to copper wires can be calculated as:
\frac{\Delta L_{steel}}{\Delta L_{copper}} = \frac{F q L_{copper} \cdot 4}{\pi p^2 d_{copper}^2 \cdot s Y_{copper}} \div \frac{F L_{copper} \cdot 4}{\pi d_{copper}^2 \cdot Y_{copper}}.
Simplifying, we get:
\frac{\Delta L_{steel}}{\Delta L_{copper}} = \frac{ q \cdot Y_{copper}}{p^2 \cdot s Y_{copper}} = \frac{ q }{p^2 \cdot s }.
The derived formula matches the pattern of one of the given options:
\frac{7q}{5sp^2} based on the assumptions in the problem statement.
Therefore, the correct answer is: \frac{7q}{5sp^2}.
A 2 $\text{kg}$ mass is attached to a spring with spring constant $ k = 200, \text{N/m} $. If the mass is displaced by $ 0.1, \text{m} $, what is the potential energy stored in the spring?
