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If the yield strength of steel is 2.5 × 108 Nm–2, what is the maximum weight that can be hung at the lower end of the wire?

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प्रश्न

If the yield strength of steel is 2.5 × 108 Nm–2, what is the maximum weight that can be hung at the lower end of the wire?

दीर्घउत्तर
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उत्तर

Clearly, tension will be maximum at x = L

∴ T = μgL + Mg = (m + M)g  ......[∵ m = μL]

The yield force = (Yield strength Y) area = 250 × 106 × π × (10–3)2 = 25 × πN

At the yield point, T = Yield force

⇒ (m + M)g = 250 × π

m = π × (10–3)2 × 10 × 7860 << M

∴ Mg = 250 × π

Hence, M = `(250 xx pi)/10` = 25 × π = 75 kg.

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अध्याय 9: Mechanical Properties of Solids - Exercises [पृष्ठ ७०]

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एनसीईआरटी एक्झांप्लर Physics Exemplar [English] Class 11
अध्याय 9 Mechanical Properties of Solids
Exercises | Q 9.25 (b) | पृष्ठ ७०

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