The maximum stress that can be applied to the material of a wire employed to suspend an elevator is 3 π × 10 8 N m - 2 . If the mass of the elevator is 900 kg , and it moves up with an acceleration of 2 . 2 m s - 2 then calculate the minimum radius of the wire. g = 10 m s - 2

Given, mass of the elevator m = 900 Kg , Acceleration a = 2 . 2 m s - 2 , Maximum stress S max = 3 π × 10 8 N m 2 Area A = π r min 2 , Tension T = S max A = S max π r min 2 , Weight = m g FBD equation of elevator: m a = T - m g ⇒ m a = S max π r min 2 - m g ⇒ r min = m a + g π S max = 900 2 . 2 + 10 π × 3 π × 10 8 ⇒ r min = 6 . 05 × 10 - 3 m Hence, minimum radius of the wire is ≅ 6 mm .

The maximum stress that can be applied to the material of a wire employed to suspend an elevator is 3 π × 10 8 N m - 2 . If the mass of the elevator is 900 kg , and it moves up with an acceleration of 2 . 2 m s - 2 then calculate the minimum radius of the wire. g = 10 m s - 2

Given, mass of the elevator m = 900 Kg , Acceleration a = 2 . 2 m s - 2 , Maximum stress S max = 3 π × 10 8 N m 2 Area A = π r min 2 , Tension T = S max A = S max π r min 2 , Weight = m g FBD equation of elevator: m a = T - m g ⇒ m a = S max π r min 2 - m g ⇒ r min = m a + g π S max = 900 2 . 2 + 10 π × 3 π × 10 8 ⇒ r min = 6 . 05 × 10 - 3 m Hence, minimum radius of the wire is ≅ 6 mm .

The bulk’s modulus of copper is 138 × 10 9 Pa . The additional pressure generated in an explosion chamber is 345 × 10 6 Pa . Then the percentage change in the volume of a piece of copper placed in this chamber will be

Given, bulk’s modulus of copper B = 138 × 10 9 Pa Excess pressure ∆ P = 345 × 10 6 Pa Bulk’s modulus, B = ∆ P ∆ V V Volume Strain is the ratio of the change in volume of a body to its original volume and Percentage change in the volume ∆ V V × 100 = ∆ P B × 100 ⇒ ∆ V V × 100 = 345 × 10 6 138 × 10 9 × 100 = 0 . 25 %