A sphere of radius ` 0.1m` and mass `8 pi` `kg` is attached to the lower end of a steel wire of length `5.0 m` and diameter `10^(-3)m`. The wire is suspended from `5.22 m` high ceiling of a room . When the sphere is made to swing as a simple pendulum, it just grazes the floor at its lowest point. Calculate the velocity of the sphere at the lowest position . Young's modulus of steel is `(1.994xx10^(11) N//m^(2))`.

Options

A. `7.5ms`
B. `8.2ms`
C. `8.8ms^(-1)`
D. `6.5ms^(-1)`

Answer

C

Explanation

As the length of the wire is `5m` and diameter `2xx0.1 = 0.2m` and at lowest point it grazes the floor which is at a distance `5.22m` from the roof, the increase in the wire at the lowest point `Delta L = 5.22-(5+0.2) = 0.02m`
So tension in the wire (due to elasticity)
`T = (YA)/(L) Delta L = (1.994xx10^(11)xx pi (5xx10^(-4))^(2)xx0.02)/(5) = 199.4pi N`
and as equaction of circular motion of a mass `'m'` tied to a string in a vertical plane is
`(mv^(2)//r) = T - mg cos theta`
But herer `r = 5 + 0.02 + 0.1 = 5.12m`
So `(8pi^(2)//5.12) = (1.99.4pi - 8pi xx 9.8)`
`v^(2) = (121 xx 5.12//8) = 77.4`, So `v = 8.8m//s`