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?

  1. Elongation in rod A is more than rod B
  2. Elongation in rod A is less than rod B
  3. Elongation in rod A is equal to rod B
  4. None of these

Answer (Detailed Solution Below)

Option 1 : Elongation in rod A is more than rod B
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Detailed Solution

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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)

    

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 DA = DB = D, FA = FB = F, and lA > lB

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

       -----(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,

        -----(2)

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

⇒ 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.

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