A horizontal aluminium rod of diameter 4 cm projected 6 cm from a wall . An object of mass `400 pi` kg is suspended from the end of the rod. The shearing modulus of aluminium is `3.0xx10^(10)N//m^(2)`. The vertical deflection of the end of the rod is ( `:' g = 10 m//s^(2)` )

Options

A. 0.01 mm
B. 0.02 mm
C. 0.03 mm
D. 0.04 mm

Answer

B

Explanation

Modulus of rigidity of a horizontal rod with hanging mass is given as
`mu=(E_(L))/(Ax)=(mgL)/(Ax)` ...(i)
Where, `x=` verticle deflection of the end of the rod.
``
Here, `L=6cm=6xx10^(-2)m` , radius `R=2cm=2xx10^(-2)m`
mass, `m=400 pi` kg and `eta=3xx10^(10)N/m^(2)`
Putting these values, in Eq. (i) we get
`3xx10^(10)=((400pi xx 10)xx6xx10^(-2))/(pi(2xx10^(-2))^(2)x)`
`rArr x = ( 4 pi xx 10^(3)xx6xx10^(-2))/(pi xx 4 xx 10^(-4)xx3xx10^(10)) rArr x = 0.02 mm `
Hence, the correct option is (b).