AP EAMCET202119 Aug 2021Morning ShiftPhysicsMechanical Properties of SolidsActual
The Young's modulus of a rubber string of length 12 cm and density 1 . 5 kg m - 3 is 5 × 10 8 N m - 2 . When this string is suspended vertically, the increase in its length due to its own weight is (take g = 10 m s - 2 )
Options
- A2.16 × 10 − 10   m
- B9.6 × 10 − 11   m
- C9.6 × 10 − 3   m
- D2.16 × 10 − 3   m
Correct answer
A. 2.16 × 10 − 10   m
Step-by-step solution
Hooke's law for elasticity, σ = γ ε γ = σ ε = F A l ∆ l   here, l   is the length of string undergone through tension. Here, the string is suspended vertically and the string is elongated due to its own weight therefore, weight is acting at the centre of mass which in the given case is located at the half-length of string, therefore, l = 12   cm 2 = 6   cm   or  0 . 06   m γ = σ ε = F A l ∆ l γ = F A l l 2 ∆ l