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The Ratio Stress/Strain Remain Constant for Small Deformation of a Metal Wire. When the Deformation is Made Larger, Will this Ratio Increase Or Decrease? - Physics

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प्रश्न

The ratio stress/strain remain constant for small deformation of a metal wire. When the deformation is made larger, will this ratio increase or decrease?

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उत्तर

The ratio of stress to strain will decrease. 

Beyond the elastic limit, the body loses its ability to restore completely when subjected to stress. Thus, there occurs more strain for a given stress. At some points, however, the body undergoes strain without the application of stress. So, the ratio of stress to strain decreases. 

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पाठ 14: Some Mechanical Properties of Matter - Short Answers [पृष्ठ २९७]

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एचसी वर्मा Concepts of Physics Vol. 1 [English] Class 11 and 12
पाठ 14 Some Mechanical Properties of Matter
Short Answers | Q 1 | पृष्ठ २९७

संबंधित प्रश्‍न

A rod of length 1.05 m having negligible mass is supported at its ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths as shown in Figure. The cross-sectional areas of wires A and B are 1.0 mm2 and 2.0 mm2, respectively. At what point along the rod should a mass be suspended in order to produce (a) equal stresses and (b) equal strains in both steel and aluminium wires.

 


Two strips of metal are riveted together at their ends by four rivets, each of diameter 6.0 mm. What is the maximum tension that can be exerted by the riveted strip if the shearing stress on the rivet is not to exceed 6.9 × 107 Pa? Assume that each rivet is to carry one-quarter of the load.


When a block a mass M is suspended by a long wire of length L, the elastic potential potential energy stored in the wire is `1/2`  × stress × strain × volume. Show that it is equal to `1/2`  Mgl, where l is the extension. The loss in gravitational potential energy of the mass earth system is Mgl. Where does the remaining `1/2` Mgl energy go ? 


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Answer in one sentence.

Define strain.


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A rod of length l and negligible mass is suspended at its two ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths (Figure). The cross-sectional areas of wires A and B are 1.0 mm2 and 2.0 mm2, respectively.

(YAl = 70 × 109 Nm−2 and Ysteel = 200 × 109 Nm–2)

  1. Mass m should be suspended close to wire A to have equal stresses in both the wires.
  2. Mass m should be suspended close to B to have equal stresses in both the wires.
  3. Mass m should be suspended at the middle of the wires to have equal stresses in both the wires.
  4. Mass m should be suspended close to wire A to have equal strain in both wires.

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Answer in one sentence:

What is an elastomer?


Strain is defined as the ratio of:


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