A sphere of mass 20 kg is suspended by a metal wire of unstretched length 4 m and diameter 1 mm. When in equilibrium there is a clear gap of 2 mm between the sphere and the floor. The sphere is gently pushed aside so that the wire makes an angle `theta` with the vertical and is released. Find the maximum value of `theta` so that the metal of the wire is `2.0xx10^11 Nm^-2`. Make appropriate approximation.

Answer

C, D

Explanation

`m=20kg`
L=4m
`2r=1mm, r=5xx10^-4m`
At equilibrium T=mg
When it moves at an angle `theta` and released the tension T at lowest point is
`T=mg+(mv^2)/r`
The change in tensiion is due to centrifugal force
`DeltaT=(mv^2)/r`........(1) Again by work energy principle
`1/2mv^2-0=mgr(1-costheta)`
`v^2=2gr(1-costheta)`.......2
So, `Delta T=(m[2 gr(1-costheta)])/r`
`=2mg(1-costheta)`
`rarr F=DeltaT`
`rarr F=(YADeltaL)/L`
`=2mg-2mgcostheta`
`rarr 2mgcostheta=2mg-(YADeltaL)/L`
`rarr cotheta=1-(YADeltaL)/(L(2mg))`
`rarr cos theta=1-[(2xx10^11xx4xx3.14xx(5)^2xx10^-8xx2xx10^-3)/(4x2xx20xx10)`
`rarr costheta=0.80`
`theta=36.4^@`