Mechanical Properties of Solids

JEE Main level, single correct, medium

11 of 118

Question 11 of 118

Two wires as shown in the figure below, made of steel and have breaking stress of 12×10812 \times 10^8 N/m2^2. Area of cross-section of upper wire is 0.0080.008 cm2^2 and of lower wire is 0.0040.004 cm2^2. The maximum mass that can be added to pan without breaking any wire is _____ kg.
(take g=10g = 10 m/s2^2)

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Answer: (B) 3838

Let the maximum mass that can be added to the pan be mm.

The breaking stress for both wires is σmax=12×108 N/m2\sigma_{max} = 12 \times 10^8 \text{ N/m}^2.

For the upper wire:
Area of cross-section, Au=0.008 cm2=8×10−7 m2A_u = 0.008 \text{ cm}^2 = 8 \times 10^{-7} \text{ m}^2
Maximum tension the upper wire can withstand is:
Tu,max=σmax×Au=(12×108)×(8×10−7)=960 NT_{u,max} = \sigma_{max} \times A_u = (12 \times 10^8) \times (8 \times 10^{-7}) = 960 \text{ N}
The total mass supported by the upper wire is (30+10+m)=(40+m) kg(30 + 10 + m) = (40 + m) \text{ kg}.
Tension in the upper wire, Tu=(40+m)g=(40+m)×10T_u = (40 + m)g = (40 + m) \times 10
To avoid breaking the upper wire:
(40+m)×10≤960(40 + m) \times 10 \le 960
40+m≤96  ⟹  m≤56 kg40 + m \le 96 \implies m \le 56 \text{ kg}

For the lower wire:
Area of cross-section, Al=0.004 cm2=4×10−7 m2A_l = 0.004 \text{ cm}^2 = 4 \times 10^{-7} \text{ m}^2
Maximum tension the lower wire can withstand is:
Tl,max=σmax×Al=(12×108)×(4×10−7)=480 NT_{l,max} = \sigma_{max} \times A_l = (12 \times 10^8) \times (4 \times 10^{-7}) = 480 \text{ N}
The total mass supported by the lower wire is (10+m) kg(10 + m) \text{ kg}.
Tension in the lower wire, Tl=(10+m)g=(10+m)×10T_l = (10 + m)g = (10 + m) \times 10
To avoid breaking the lower wire:
(10+m)×10≤480(10 + m) \times 10 \le 480
10+m≤48  ⟹  m≤38 kg10 + m \le 48 \implies m \le 38 \text{ kg}

To ensure neither wire breaks, the added mass mm must satisfy both conditions. Therefore, the maximum mass that can be added is the smaller of the two values:
m=38 kgm = 38 \text{ kg}

Answer: 3838

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