Wires $$W_1$$ and $$W_2$$ are made of same material having the breaking stress of $$1.25 \times 10^9$$ N m$$^{-2}$$. $$W_1$$ and $$W_2$$ have cross-sectional area of $$8 \times 10^{-7}$$ m$$^2$$ and $$4 \times 10^{-7}$$ m$$^2$$, respectively. Masses of 20 kg and 10 kg hang from them as shown in the figure. The maximum mass that can be placed in the pan without breaking the wires is _________ kg (Use g = 10 m s$$^{-2}$$)

We need to find the maximum mass $$m_{\text{pan}}$$ that can be placed in the pan without breaking either of the two wires, $$W_1$$ or $$W_2$$.

1. Identify the Properties of the Wires

Both wires are made of the same material, meaning they share the same breaking stress limit:

• Breaking Stress ($$\sigma_{\text{max}}$$): $$1.25 \times 10^9\text{ N m}^{-2}$$
• Area of cross-section of $$W_1$$ ($$A_1$$): $$8 \times 10^{-7}\text{ m}^2$$
• Area of cross-section of $$W_2$$ ($$A_2$$): $$4 \times 10^{-7}\text{ m}^2$$

2. Calculate the Maximum Tension Capacity for Each Wire

The maximum tension ($$T_{\text{max}}$$) a wire can sustain before breaking is given by the formula:

$$T_{\text{max}} = \sigma_{\text{max}} \times A$$

• For Wire 1 ($$W_1$$):

  $$T_{1,\text{max}} = (1.25 \times 10^9) \times (8 \times 10^{-7}) = 10 \times 10^2 = 1000\text{ N}$$

• For Wire 2 ($$W_2$$):

  $$T_{2,\text{max}} = (1.25 \times 10^9) \times (4 \times 10^{-7}) = 5 \times 10^2 = 500\text{ N}$$

3. Formulate Tension Equations Based on the Masses

Let $$m$$ be the maximum mass added to the pan. The tensions experienced by the two wires under static equilibrium are determined by the hanging loads beneath them:

• Tension in Wire 2 ($$T_2$$): It supports the $$10\text{ kg}$$ mass and the pan with its added mass $$m$$.

  $$T_2 = (10 + m)g = (10 + m) \times 10 = 100 + 10m$$

• Tension in Wire 1 ($$T_1$$): It supports everything below it—the $$20\text{ kg}$$ mass, the $$10\text{ kg}$$ mass, and the pan with mass $$m$$.

  $$T_1 = (20 + 10 + m)g = (30 + m) \times 10 = 300 + 10m$$

4. Determine the Safe Mass Limit for Each Wire

To avoid breaking, the actual tension in each wire must not exceed its maximum capacity:

Condition for Wire 1 ($$W_1$$):

$$T_1 \le T_{1,\text{max}}$$

$$300 + 10m \le 1000 \implies 10m \le 700 \implies m \le 70\text{ kg}$$

Condition for Wire 2 ($$W_2$$):

$$T_2 \le T_{2,\text{max}}$$

$$100 + 10m \le 500 \implies 10m \le 400 \implies m \le 40\text{ kg}$$

5. Select the Limiting Constraint

To ensure neither wire breaks, we must choose the smaller value of the two mass constraints. If the mass exceeds $$40\text{ kg}$$, wire $$W_2$$ will snap first.

Therefore, the maximum safe mass that can be added to the pan is exactly $$40\text{ kg}$$.

Final Answer: 40