a. Shearing stress `=F/A=(Mg)/(pir^(2))=(1200xx9.8)/(3.14xx(0.024)^(2))`
`=6.5xx10^(6)Nm^(-2)`
b. Let `/_x` be the vertical defelection of the rod. Then shear modulus,
`G=(F/A)/(/_x/l)`
Thus `/_x=((F/A)l)/G=(6.5xx10^(6)xx0.53)/(3.0xx10^(10))=1.1xx10^(-5)M`