A sphere of mass 2 kg and diameter 4.5 cm is attached to the lower and of a steel wire of 2 m length and are of cross- section `0.24 xx 10^(-6)m^2` . The wire is suspended from 205 cm high ceilling of a room. When the system is made to oscillate as a simple pendulum, the sphere just grazes the floor at its lowest psition. The velocity of the sphere at the lowest position is (young's modulus of steel `=2xx 10^(11) Nm^(-2)` and acceleraiton due to gravity `=10 ms^(-2))`

Options

A. `10 ms^(-1)`
B. `12 ms^(-1)`
C. `15 ms^(-1)`
D. `18 ms^(-1)`

Answer

A

Explanation

Extension in wire due to the load and centrifugal force `= Delta l =(205-204.5) cm =0.5cm`
If velocity of sphere at lowest points is v, then `Y=((Mg+(Mv^2)/( R)) L)/(A Delta l)`
`rArr Mg+(Mv^2)/(R ) = (Y A Delta l)/(L)`
`rArr Mg+(Mv^2)/(R )= (YADeltal)/(L)`
where, R=202.25 cm
`rArr 2xx 10 + (2 xx v^2)/(202.25 xx 10^(-2))`
`(2 xx 10^11 xx 0.24 xx 10^(-6) xx 0.5 xx 10^(-2))/(2)`
`=0.12 xx 10^3`
`rArr (2xxv^2)/(202. 25 xx 10^(-2))=100 rArr v^2=101 .125`
`rArr v=10.05 ms^(-1) or v= 10ms^(-1)`