Question

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 $g = 10 \text{ m/s}^2$)}

• A. 56
• B. 38
• C. 96
• D. 5.6

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We have two wires in series supporting different loads. We need to find the limit of the additional mass $M$ such that neither wire exceeds its breaking stress.
Step 2: Detailed Explanation:
Let $M$ be the mass added to the pan.
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 upper wire $= 30 \text{ kg} + 10 \text{ kg} + M = (40 + M)$ kg.
Stress in upper wire $\sigma_1 = \frac{(40+M)g}{A_1} \le 12 \times 10^8$.
\[ \frac{(40+M) \times 10}{8 \times 10^{-7}} \le 12 \times 10^8 \implies (40+M) \le 12 \times 0.08 \times 10^2 = 96 \implies M \le 56 \text{ kg} \]
Lower Wire: Area $A_2 = 0.004 \text{ cm}^2 = 4 \times 10^{-7} \text{ m}^2$.
Total mass supported by lower wire $= 10 \text{ kg} + M = (10 + M)$ kg.
Stress in lower wire $\sigma_2 = \frac{(10+M)g}{A_2} \le 12 \times 10^8$.
\[ \frac{(10+M) \times 10}{4 \times 10^{-7}} \le 12 \times 10^8 \implies (10+M) \le 12 \times 0.04 \times 10^2 = 48 \implies M \le 38 \text{ kg} \]
To ensure neither wire breaks, we take the minimum of the two limits:
$M_{max} = \min(56, 38) = 38$ kg.
Step 3: Final Answer:
The maximum mass is 38 kg.