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Short Note
The two wires shown in figure are made of the same material which has a breaking stress of 8 × 108 N m−2. The area of cross section of the upper wire is 0.006 cm2 and that of the lower wire is 0.003 cm2. The mass m1 = 10 kg, m2 = 20 kg and the hanger is light. Repeat the above part if m1 = 10 kg and m2 = 36 kg.
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Solution
\[\text{If m}_1 = 10 \text{ kg and m}_2 = 36 \text{ kg }\]
Tension in lower wire \[ \text{T}_\text{l} = \text{m}_1 \text{ g + w} \]
Here: g is the acceleration due to gravity
w is the load
∴ Stress in lower wire:
w is the load
∴ Stress in lower wire:
\[\Rightarrow \frac{\text{T}_\text{l}}{\text{A}_\text{l}} = \frac{\text{m}_1\text{ g + w}}{\text{A}_\text{l}} = 8 \times {10}^5 \]
\[ \Rightarrow \text{ w = 140 N }\]
\[ \Rightarrow \text{ w = 140 N }\]
Now, tension in upper wire
\[\text{T}_2 = \text{m}_1 \text{ g + m}_2 \text{ g + w }\]
∴ Stress in upper wire:
\[\Rightarrow \frac{\text{T}_\text{u}}{\text{A}_\text{u}} = \frac{\text{m}_2 \text{g + m}_1 \text{g + w}}{\text{A}_\text{u}} = 8 \times {10}^5 \]
\[ \Rightarrow \text{w = 20 N}\]
\[ \Rightarrow \text{w = 20 N}\]
For the same breaking stress, the maximum load that can be put is 20 N or 2 kg. The upper wire will break first if the load is increased.
Concept: Elastic Moduli - Determination of Young’s Modulus of the Material of a Wire
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