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PhyMedMain 2026 • 5th Apr - Shift 1

Maximum Mass Added to Pan Without Breaking Steel Wires

Two wires as shown in the figure below, made of steel and have breaking stress of 12×108 N/m212 \times 10^8\text{ N/m}^2. Area of cross-section of upper wire is 0.008 cm20.008\text{ cm}^2 and of lower wire is 0.004 cm20.004\text{ cm}^2. The maximum mass that can be added to pan without breaking any wire is ______ kg\text{kg}.

(take g=10 m/s2g = 10\text{ m/s}^2)

Question Diagram 1

Options

A

56

B

38

Correct
C

96

D

5.6

Topics & Concepts

Mechanical Properties of SolidsElasticity

Step-by-Step Solution

To find the maximum mass mm that can be added to the pan without breaking either wire, we must determine the maximum tension each wire can tolerate before reaching its breaking stress.

1. Calculation of Maximum Tolerable Tension

The breaking stress for steel is given as: σbreaking=12×108 N/m2\sigma_{\text{breaking}} = 12 \times 10^8 \text{ N/m}^2

For the Upper Wire:

  • Area of cross-section, A1=0.008 cm2=0.008×10−4 m2=8×10−7 m2A_1 = 0.008 \text{ cm}^2 = 0.008 \times 10^{-4} \text{ m}^2 = 8 \times 10^{-7} \text{ m}^2
  • Maximum tension allowed, T1,maxT_{1,\text{max}}: T1,max=σbreaking×A1=(12×108 N/m2)×(8×10−7 m2)=960 NT_{1,\text{max}} = \sigma_{\text{breaking}} \times A_1 = \left(12 \times 10^8 \text{ N/m}^2\right) \times \left(8 \times 10^{-7} \text{ m}^2\right) = 960 \text{ N}

For the Lower Wire:

  • Area of cross-section, A2=0.004 cm2=0.004×10−4 m2=4×10−7 m2A_2 = 0.004 \text{ cm}^2 = 0.004 \times 10^{-4} \text{ m}^2 = 4 \times 10^{-7} \text{ m}^2
  • Maximum tension allowed, T2,maxT_{2,\text{max}}: T2,max=σbreaking×A2=(12×108 N/m2)×(4×10−7 m2)=480 NT_{2,\text{max}} = \sigma_{\text{breaking}} \times A_2 = \left(12 \times 10^8 \text{ N/m}^2\right) \times \left(4 \times 10^{-7} \text{ m}^2\right) = 480 \text{ N}

2. Tension Equations in Terms of Added Mass (mm)

Let mm be the mass added to the pan (with g=10 m/s2g = 10 \text{ m/s}^2).

  • Lower Wire: Supports the 10 kg10 \text{ kg} block and the added mass mm in the pan. T2=(10+m)g=(10+m)×10T_2 = (10 + m)g = (10 + m) \times 10

  • Upper Wire: Supports the 30 kg30 \text{ kg} block, the lower wire, the 10 kg10 \text{ kg} block, and the added mass mm in the pan. T1=(30+10+m)g=(40+m)×10T_1 = (30 + 10 + m)g = (40 + m) \times 10

3. Conditions to Prevent Wire Failure

Condition 1: Upper wire does not break T1≤T1,maxT_1 \le T_{1,\text{max}} (40+m)×10≤960(40 + m) \times 10 \le 960 40+m≤9640 + m \le 96 m≤56 kgm \le 56 \text{ kg}

Condition 2: Lower wire does not break T2≤T2,maxT_2 \le T_{2,\text{max}} (10+m)×10≤480(10 + m) \times 10 \le 480 10+m≤4810 + m \le 48 m≤38 kgm \le 38 \text{ kg}

Conclusion

To ensure that neither wire breaks, mm must satisfy both constraints: m=min⁡(56 kg,38 kg)=38 kgm = \min(56 \text{ kg}, 38 \text{ kg}) = 38 \text{ kg}

Thus, the maximum mass that can be added to the pan without breaking any wire is 38 kg.

Correct Answer: B

Maximum Mass Added to Pan Without Breaking Steel Wires | Physics PYQ Solution - JEE Challenger