A steel wire of length 5 m and diameter `10^(-3)` m is hanged vertically from a ceiling of 5.22 m height. A sphere of radius 0.1 m and mass 8tt kg is tied to the free end of the wire. When this sphere is oscillated like a simple pendulum, it touches the floor of the room in its velocity of the sphere in its lower most position. Calculate the velocity of the sphere in its lower Young's modulus for in its lower most position. Young's modulus for steel `=1.994 xx 10 ^(11) Nm ^(-2).`

Explanation

Radius of sphere `r = 0.1m`
Initial length `L = 5m`
Increase in the length of the wire
`Delta L = 5.22 - (L + 2r) = 5.22 - (5 + 2 xx 0.1) = 0.02 m`
Radius of the wire `r_(0) = 5 xx 10 ^(-4)m`
Radius of the wire `r _(o) = 5 xx 10 ^(-4) m`
If the tension produced in the wire at the lowest point is T, then
`Y = (T //A)/(Delta L//L) ` gives
`T = (YA Delta L )/(L ) = (Y (pi r _(0) ^(2) ) Delta L )/( L) = (1.994 xx 10 ^(11) xx pi ( 5 xx 10 ^(-4)) ^(2) xx 0.02)/( 5)`
`=199 .4pi N`
But, net force `T-Mg = (Mv ^(2))/(R )`
where R = radius of the path of oscillatins of the sphere `=5.22 =0.1 =5.12 m`
`therefore 199.4pi -8pi xx 9.8 = (8pi xx v ^(2))/(5.12 )`
`therefore 199.4 - 78.4 = (8v ^(2))/(5.12)`
`therefore 121 = (8v ^(2))/( 5.12 ) `
`therefore v ^(2)= ( 121 xx 5.12 )/( 8) = 77.44`
`therefore v = 8.8 ms ^(-1)`