A `6 kg ` weight is fastened to the end of a steel wire of un stretched length `60 cm `. It is whirled in a vertical circle and has an angular velocity of ` 2 rev//s` at the bottom of the circle. The area of cross - section of the wire is `0.05 cm ^(2)` . Calculate the elongation of the wire when the weight is at the lowest point of the path . Young's modulus of steel `= 2xx10 ^(11) N //m^(2)` .

Answer

A, C, D

Explanation

` T - mg = ml omega ^(2) = ml (2pi f )^(2)`
` T = mg + 4 pi ^(2) mlf^(2) `
` 6xx 9.8 + 4 pi^(2)(6)(0.6)(2)^(2)`
` = 628 N`
` Delta l = (Tl)/ (Ay)`
` = ((628) (0.6))/(0.05xx10^(-4) (2xx10^(11)))`
`= 3.8 xx10^(-4) m `