Answer:
103 g
Step-by-step explanation:
given that
Mass of the attached mass, m = 40 g
Initial distance of balance, x1 = 49.7 cm = 0.497 m
Final distance of balance, x2 = 39.2 cm = 0.392 m
Equilibrium point of attachment, x3 = 14 cm = 0.14 m
To get the mass of the meter stick, we use the relation
M = m.[(x2 - x3) / (x1 - x2)]
And on solving in full, we gave
M = 40.[(0.392 - 0.14) / [0.497 - 0.392)]
M = 40(0.258 / 0.105)
M = 40 * 2.457
M = 102.8 g
Therefore the needed work good for mass of the stick j
Is 103 g