If the lengths of the arms of a physical balance are `l_(1)` and `l_(2)` and a salesman weighs out twice to a customer articles of mass M by putting weughts first in one pan and then in the other, shown that the salesman is a loser by `M[((l_(1)-l_(2))^(2))/(l_(1)l_(2))]`.

Explanation

For a physical balance of unequal arm, we have
`l_(1)M=l_(2)M_(1) rArr M_(2) = Ml_(1)/l_(2)`
and `l_(1)M_(2)=l_(2)M rArr M_(2) =M l_(2)/l_(1)`
With this false balance, the salesman gives `(M_(1)+M_(2))` instead of `M+M=2M`
Loss of the salesman `=M_(1)+M_(2)-2M`
or Loss `=M l_(1)/l_(2)+M l_(2)/l_(1)-2 M`
`=M[(l_(1)^(2)+l_(2)^(2)-2l_(1)l_(2))/(l_(1)l_(2))]`
`=M[((l_(1)-l_(2))^(2))/(l_(1)l_(2))]`