AP EAMCET201822 Apr 2018Evening ShiftPhysicsMechanical Properties of SolidsActual
A sphere of mass 2 ~kg and diameter 4.5 ~cm is attached to the lower end of a steel wire of 2 ~m length and area of cross-section 0.24 10⁻⁶ ~m ^2 . The wire is suspended from 205 ~cm high ceiling of a room. When the system is made to oscillate as a simple pendulum, the sphere just grazes the floor at its lowest position. The velocity of the sphere at the lowest position is (Young's modulus of steel =2 10¹¹ Nm ⁻² and
Options
- A10 ~ms ⁻¹
- B12 ~ms ⁻¹
- C15 ~ms ⁻¹
- D18 ~ms ⁻¹
Correct answer
A. 10 ~ms ⁻¹
Step-by-step solution
Extension in wire due to load and centrifugal force = l=(205-204.5) cm =0.5 ~cm If velocity of sphere at lowest point is v , then gathered Y= (M g+ M v^2 R ) L A l M g+ M v^2 R = Y A l L where, R=202.25 ~cm 2 10+ 2 v^2 202.25 10⁻² = 2 10¹¹ 0.24 10⁻⁶ 0.5 10⁻² 2 =0.12 10^3 2 v^2 202.25 10⁻² 202.00 v^2=101.125 v=10.05 ~ms ⁻¹ or v=10 ~ms ⁻¹ gathered