A copper wire of negligilble mas, length `(l)` and cross-sectional area `(A)` is kept on a smooth horizontal table with one end fixed, a ball of mass `'m'` is attached at the other end. The wire and the ball are rotated with angular velocity `'omega'`. If wire elongates by `Delta l` then the Young's modulus of wire and if on incresing the angular velocity from `omega` to `omega^(1)` when the wire breaks-down, then the brekaing stress `(Deltal lt lt l )` are respectivley.

Options

A. `((m l^(2) omega^(2)))/(A Delta l), (ml omega^(2))/(A)`
B. `(ml)/(A Delta l omega^(2)), (ml omega^(2))/(A)`
C. `(ml omega^(2))/(A Delta l ), (m omega^(2))/(Al)`
D. `(ml omega^(2))/(A Delta l ), (ml omega^(2))/(Al)`

Answer

A

Explanation

(a) `r = l + Delta l, F = T = mr omega^(2) = m(l + Delta l) omega^(2)`
as `Delta l` in small, `F approx ml omega^(2), Y = ((ml omega^(2)))/(A Delta l) xx l`
(b) We know Breaking stress
`= ("Breaking force")/("Area of cross section") = (ml omega^(1^(2)))/(A)`