AIIMS2014PhysicsMotion in One Dimension
A steel wire with cross-section 3 ~cm ^2 has elastic limit 2.4 10^8 ~N ~m ⁻² . The maximum upward acceleration that can be given to a 1200 ~kg elevator supported by thiscable wire if the stress is not to exceed one-third of the elastic limit is (Take g=10 ~m ~s ⁻² )
Options
- A12 ~m ~s ⁻²
- B10 ~m ~s ⁻²
- C8 ~m ~s ⁻²
- D7 ~m ~s ⁻²
Correct answer
B. 10 ~m ~s ⁻²
Step-by-step solution
Maximum tension an elevator can tolerate is aligned T & = 1 3 stress area of cross-section & = 1 3 (2.4 10^8 ) (3 10⁻⁴ )=2.4 10^4 ~N aligned If a is the maximum upward acceleration of elevator then T=m(g+a) or 2.4 10^4=1200(10+a) On solving, a=10 ~m ~s ⁻² .