A sphere of radius `10 cm` and mass `25` kg is attached to the lower end of a steel wire of length `5 m ` and diameter `4 mm ` which is suspended from the ceiling of a room . The point of support is `521 cm ` above the floor. When the sphere is set swinging as a simple pendulum, its lowest point just grazes the floor. Calculate the velocity of the ball at its lowest position `(Y_(steel) = 2xx10^(11) N//m^(2))`.

Answer

A, B, C

Explanation

` Deltal = 521-500-20 = 1 cm = 0.01 m`
`T- mg = (mv^(2))/R`
`:. T = m(g+v^(2)/R) = m (g+ v^(2)/l)Delta l = (Tl)/(AY) = (m(g + (v^(2))/l)l)/((pid^(2)//4)Y )`
`= (mgl +mv^(2))/((pid^(2)//4)Y)`
`:. v = sqrt((pid^(2) DeltalY)/(4 m) - gl)`
`= sqrt (( (3.14) (4xx10^(-3))^(2) (0.01) (2xx10^(11)))/(4xx25) -9.8xx5)`
`~~ 31 m//s`