Two wires as shown in the figure below, made of steel and have breaking stress of $12 \times 10^8 \text{ N/m}^2$. Area of cross-section of upper wire is $0.008 \text{ cm}^2$ and of lower wire is $0.004 \text{ cm}^2$. The maximum mass that can be added to pan without breaking any wire is _________ kg. (take $g = 10 \text{ m/s}^2$)}
Step 1: Understanding the Question:
The topic of this question is the Mechanical Properties of Solids (breaking stress).
Two wires support stacked masses. We must figure out the upper limit of the mass $M$ in the pan that causes neither wire to exceed its breaking threshold. Step 2: Key Formula or Approach:
Stress is defined as $\text{Stress} = \frac{\text{Tension}}{\text{Cross-Sectional Area}}$. Step 3: Detailed Explanation:
Let $M$ be the added mass.
Breaking Stress $\sigma = 12 \times 10^8 \text{ N/m}^2$.
Upper Wire: Area $A_1 = 0.008 \text{ cm}^2 = 8 \times 10^{-7} \text{ m}^2$.
Total mass supported by it is $30 + 10 + M = 40 + M$.
Limit: $\frac{(40+M) \times 10}{8 \times 10^{-7}} \le 12 \times 10^8 \implies 40 + M \le 96 \implies M \le 56 \text{ kg}$.
Lower Wire: Area $A_2 = 0.004 \text{ cm}^2 = 4 \times 10^{-7} \text{ m}^2$.
Total mass supported by it is $10 + M$.
Limit: $\frac{(10+M) \times 10}{4 \times 10^{-7}} \le 12 \times 10^8 \implies 10 + M \le 48 \implies M \le 38 \text{ kg}$.
To preserve both wires, we must obey the stricter limit: $M = 38 \text{ kg}$. Step 4: Final Answer:
The maximum mass is 38 kg.
