Exams › JEE Main › Physics

A uniform steel rod of cross-sectional area 1 m² is loaded along its length by a set of co-linear axial forces acting at intermediate sections, producing axial tensions of 50 kN, 30 kN and 70 kN over three successive segments of lengths 1.0 m, 1.0 m and 1.0 m respectively. Taking Young's modulus Y = 2.0*10¹¹ N/m², find the total elongation of the rod.

1. 1.3 micro-m
2. 1.7 micro-m
3. 0.9 micro-m
4. 1.2 micro-m

Correct answer: 1.7 micro-m

Solution

For an axially loaded bar with different internal forces in different segments, the total change in length is the sum of the individual segment elongations, each given by delta = F*L/(A*Y). Summing the three contributions for the given internal tensions and equal lengths, the total elongation works out to approximately 1.7 micro-m.

Related JEE Main Physics questions

• Select the incorrect statement.
• A steel wire has length l and cross-sectional area A. When a certain load is applied, it elongates by 1 cm. If the same load is now applied to another steel wire made of the same material, having twice the length and half the cross-sectional area, what will be its extension?
• For common metals such as aluminium and copper, which sequence correctly represents the relative magnitudes of the elastic moduli?
• A steel sheet fails under a shear stress of 3.5 × 10⁸ N m⁻². The approximate force required to punch a circular hole of diameter 1 cm through a steel plate 0.3 cm thick is:
• A copper wire of fixed volume V is stretched out to a length l. If a constant force F is applied to this wire and it elongates by Δl, which of the following plots would be a straight line?
• A material has Poisson’s ratio equal to 0.5. When a tensile force is applied to a wire made of this material, its cross-sectional area decreases by 4%. The corresponding percentage increase in its length is:

⚔️ Practice JEE Main Physics free + battle 1v1 →