Exams › JEE Main › Physics

A metal rod has initial length L at room temperature and a uniform cross-sectional area A. Its coefficient of linear expansion is α per °C. When the rod is heated through a temperature increase of ΔT K, equal compressive forces F are applied at both ends so that its length does not change at all. The Young’s modulus Y of the metal is:

1. F/(A α ΔT)
2. F/(A α (ΔT - 273))
3. F/(2A α ΔT)
4. 2F/(A α ΔT)

Correct answer: F/(A α ΔT)

Solution

Preventing expansion means the elastic compressive strain equals the thermal strain alpha*dT. With stress = F/A, Young's modulus Y = stress/strain = (F/A)/(alpha*dT) = F/(A alpha dT).

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 →