Two measure diameter of a wire a screwgauge is used The main scale division is of `1mm` In a complete rotation the screw advances by `1mm` and the circular scale has 100 devisions The reading of screwgauge is as shown in If there is no error is no error in mass measurement but error in length measurement is `1%` then find max Possible error in density
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Explanation

`rho=(m)/(((pid^(2))/(4))l)`
`((Deltarho)/(rho))=2(Deltad)/(d)+(Deltal)/(l)`
`Deltad =" least count of "= (1mm)/(100) = 0.1mm`
and `d = 3.07` mm from the
`so((Deltarho)/(rho))_(max)=(2xx(0.01)/(3.07)+(1)/(100))xx100%`
`((Deltarho)/(rho))_(max)=1.65%` .