A sphere of radius 0. 1m and mass 8 π kg is attached to the lower...

As the length of the wire is $5m$ and diameter $2 \times 0 . 1 = 0 . 2m$ and at the lowest point, it grazes the floor which is at a distance $5 . 22m$ from the roof, the increase in the length of the wire at lowest point



$\Delta \text{L} = 5 \text{.} 2 2 - \left(5 + 0 \text{.}2\right)$

$= 0 \text{.}0 2 \text{m}$

So the tension in the wire (due to elasticity)

$T = \dfrac{YA}{L} \Delta$ L = \dfrac{1\text{.}994 \times 1{0}^{11} \times \pi\left(5 \times 1{0}^{- 4}\right)2 \times 0\text{.}02}{5} ==199\text{.}}4\pi\text{ N}

and as the equation of circular - motion of a mass $m$ tied to a string in a vertical plane is

$\dfrac{mv^{2}}{r} =$ T - m$\text{g cos}$\theta$<br /><br />So at the lowest point<br /><br />\dfrac{mv^{2}}{r} =$ T - m$\text{g} \left[\text{as} \theta = 0\right]$

But here $r = 5 + 0 . 02 + 0 . 1 = 5 . 12m$

So \left(\dfrac{8\piv^{2}}{5 . 12}\right) = 199\text{.}4\pi - 8\pi \times 9\text{.}8$<br /><br />i.e.,$v^{2} = \left(1 2 1 \times 5 \text{.} 1 2 / 8\right) = 7 7 \text{.} 4 4$<br /><br />So$v = 8 . 8ms^{- 1}