AL phy problem

Let the length of the left arm of the beam balance be L1, and the length of the right arm be L2
when the mass m is placed in the left pan, mass of m1 needs to be placed on the right pan for balancing,
thus equating the moments,
m.L1 = m1.L2
Now, when the mass m is placed on the right pan, mass of m2 needs to be placed on the left pan for balancing,
taking moments, m2.L1 = m.L2
dividing the two equations, mL1/m2.L1 = m1.L2/m.L2
we get, m/m2 = m1/m
i.e. m^2 = m1.m2
or m = square-root[m1.m2]

The two arms of a beam balance are of slightly different lengths.An unknown mass is weighted first on the left pan and then on the right pan.The two measurements are m1 and m2 respectively.What is the actual mass?
設天平的兩臂長 x(左) 及 y(右)，設這物體的質量為 m 則
放在左方時
mg x = m1g y ____(1)
放在右方時
m2g x = mg y ____(2)
兩式相除
m / m2 = m1 / m
m = √(m1m2)