One end of a nylon roe of length 4.5 m and diameter 6 mm is fixed to a tree-limb. A monkey weighing 100N jumps to catch the free end and stays there. Find the correpsonding change in the diameter. Young's modulus of nylon `= 4.8 xx 10^(11) N//m^(2)` and Poisson's ratio of nylon `= 0.2`.

Options

A. `1.8 xx 10^(-5) m`
B. `3.32 xx 10^(-5) m`
C. `4.4 xx 10^(-9) m`
D. `8.8 xx 10^(-9) m`

Answer

D

Explanation

As the monkey stays in equilibrium the tension in the rope equals the weight of the mokey Hence,
`Y = ("Stress")/("Strain") = (T//A)/(DeltaL//L) rArr DeltaL = (TL)/(AY)`
`rArr "elongation" DeltaL`
`= ((100 N) xx (4.5 m))/((pi xx 9 xx 10^(-6) m^(2)) (4.8 xx 10^(11) N//m^(-2)))`
`= 3.32 xx 10^(-5) m`
Again, Poisson's ratio `(Delta d//d)/(DeltaL//L) = ((Deltad)L)/(DeltaLd)`
`rArr 0.2 = (Delta xx 4.5)/((3.32 xx 10^(-5)) xx (6 xx 10^(-3)))`
`rArr Delta d = (0.2 xx 6 xx 3.32 xx 10^(-8) m)/(4.5)`
`8.8 xx 10^(-9) m`