In the above problem if the the yield strength of steel is `2.5xx10^(8) N//m^(2)`, then the maximum mass that can be hung at the lower end of the wire is

Options

A. `785kg`
B. `78.75kg`
C. `78.25kg`
D. `78.50kg`

Answer

C

Explanation

Yield force = (Yield strength `Y`) `xx` area
`= 250xx10^(6) pi xx (10^(-3))^(2) = 250 xx pi N = 785N`
At the yield point, `(m + M_("max"))g = 785`
`0.25+M_("max") = 78.5 rArr M_("max") = 78.25kg`