BITSAT2019PhysicsMechanical Properties of SolidsActual
A rubber cord has a cross-sectional area 10 - 6 m 2 and total unstretched length 0 . 1 m . It is stretched to 0 . 125 m and then released to project a particle of mass 5 . 0 g . The velocity of projection is [Given, Young's modulus of rubber, Y = 5 × 10 8 N m - 2 ]
Options
- A45   m   s - 1
- B30   m   s - 1
- C25   m   s - 1
- D15   m   s - 1
Correct answer
C. 25   m   s - 1
Step-by-step solution
Force constant K of rubber is given by K = Y A l = 5 × 10 8 × 10 - 6 0 . 1 = 5 × 10 3   N   m - 1 Now, from conservation of energy, elastic potential energy of cord = kinetic energy of particle i.e. 1 2 K ∆ l 2 = 1 2 m v 2 ⇒ v = K m · ∆ l = 5 × 10 3 5 × 10 - 3 0 . 125 - 0 . 1 = 10 3 × 0 . 025 = 25   m   s - 1