A block of mass 1 kg is fastended to one end of a copper wire of cross- sectional area 1` mm^(2)` and is rotated in a horizontal circle of radius 20 cm. If the breaking stress of copper is `5 xx 10^(8)Nm^(-2)`, find the maximum number of revolutions the block make in the minute without the string breaking.

Explanation

Maximum tension on the string = Breaking stress ` xx ` Area of cross- section
`= 5 xx 10^(8) xx 1 xx 10^(-6) = 500 N`.
When a body revolves in a horizontal circle,
Tension on the string = Centripetal force
` 500 = mr omega^(2)`
Where `m=1 kg, r= 0.20m, omega=?`
`500 = 1 xx 0.2 xx omega^(2)`
` omega^(2)= (500)/(0.2)=2500`
Maximum angular speed, ` omega = 50 rads^(-1)`.
`t = 60 s, n = ?`
`omega = (2pi n)/(t), 50= (2pi n)/(60)`
The maximum number of revolutions,
`n= (50 xx 60)/(2 pi) = (1500)/(pi)= 477.4 "rpm".`