Two rods A and B are made of the same material. The diameter of both the rods are equal but the length of the rod A is more than rod B. If the tensile force applied on both the rods are equal, then which of the following statement is correct?

CONCEPT:

Stress:

• Stress is the ratio of the load or force to the cross-sectional area of the material to which the load is applied.
  • The standard unit of stress is N/m2.

Strain:

• Strain is a measure of the deformation of the material as a result of the force applied.
  • The strain is a unitless quantity.

Hooke's law:

• Hooke's law states that within the elastic limit the stress applied on a body is directly proportional to the strain produced.

⇒ Stress ∝ Strain

⇒  Stress = E × Strain (Where E = modulus of elasticity)

\( ⇒ σ=\frac{F(N)}{A(m^2)}\)

\( ⇒ strain=\frac{dl}{l}\)    

where σ = stress, F = applied force, A = cross-sectional area, dl = change in length, l = initial length and E = young's modulus of elasticity

EXPLANATION:

Given D_A = D_B = D, F_A = F_B = F, and l_A > l_B

• We know that when the tensile force F is applied on the rod of length l, the elongation of the rod is given as,

\(⇒ dl=\frac{Fl}{AE}\)       -----(1)

Where A = cross-sectional area, and E = Young's modulus of elasticity

• The cross-sectional area of the rod of diameter D is given as,

\(⇒ A=\frac{\pi}{4}D^2\)        -----(2)

​By equation 1 and equation 2, the elongation of the rod is given as,

\(⇒ dl=\frac{4Fl}{\pi D^2E}\)

⇒ dl ∝ l        -----(3)

• By equation 3, it is clear that the elongation of the rod is directly proportional to the length of the rod.
• Since the length of rod A is more than rod B, therefore the elongation of rod A will be more than rod B. Hence, option 1 is correct.