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JEE Mains Previous Paper 1 (Held On: 08 Apr 2019 Shift 1)

Option 3 : 3.1 × 10^{6} Nm^{-2}

JEE Mains Previous Paper 1 (Held On: 12 Apr 2019 Shift 2)

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90 Questions
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180 Mins

**Concept: **

**Stress:** Stress is defined as the **restoring force per unit area**. The restoring force is developed on the object depending on its deformation.

**Tensile stress:** When forces cause an **elongation** on an object, then the occurred stress is called tensile stress. For example, **stretching an elastic band**.

The tensile stress developed is given by the formula:

\({\rm{Stress}} = \frac{{{\rm{Mg}}}}{{{\rm{\pi }}{{\rm{r}}^2}}}\)

Where r is the radius of the object

M is the mass of the object

g is the acceleration due to gravity

**Calculation:**

Given:

Radius of steel wire, r = 2 × 10^{-3} m

Mass of the steel wire, M = 4 kg

Acceleration due to gravity, g = 3.1 × π ms^{-2}

\(\Rightarrow {\rm{Stress}} = \frac{{4 \times 3.1 \times {\rm{\pi }}}}{{{\rm{\pi }} \times {{\left( {2 \times {{10}^{ - 3}}} \right)}^2}}}\)

∴ Stress = 3.1 × 10