JEE Main202127 Jul 2021Morning ShiftPhysicsMechanical Properties of SolidsActual
A stone of mass 20   g is projected from a rubber catapult of length 0 . 1   m and area of cross section 10 - 6   m 2 stretched by an amount 0 . 04   m . The velocity of the projected stone is m   s - 1 . (Young's modulus of rubber = 0 . 5 × 10 9   N   m - 2 )
Correct answer
20
Step-by-step solution
By energy conservation 1 2 · Y A L · x 2 = 1 2 m v 2 0 . 5 × 10 9 × 10 - 6 × ( 0 . 04 ) 2 0 . 1 = 20 1000 v 2 ∴   v 2 = 400 v = 20   m   s - 1