Answer :
Answer:1) 100 gm mass should be placed at 95 cm mark.
2) Mass of 112.5 gm should be placed at 90 cm mark.
Explanation:
For equilibrium of the meter stick the sum of the moment's generated by the masses should be equal and opposite
Answer to part b)
Since a meter stick is 100 cm long and it is pivoted at it's center i.e at 50 cm
Thus
1) Moment generated by 100 gm mass about center = [tex]M_{1}=m_{1}g\times r_{1}\\\\M_{1}=0.15\times 9.81\times 0.3=0.44145Nm[/tex]
Let a mass 'm' be placed at 90 cm mark thus moment it generates equals
[tex]M_{2}=m\times 9.81\times 0.4=3.924m[/tex]
Equating both the moments we get
[tex]0.44145=3.924m\\\\\therefore m=\frac{0.44145}{3.924}\times 1000grams\\\\\therefore m= 112.5grams[/tex]
Answer to part a)
Let the 100 grams weight be placed at a distance 'x' right of center
Moment generated by 100 grams weight equals
[tex]M_{1}=0.1\times 9.81\times x[/tex]
equating the moments of the forces we get
[tex]0.1\times 9.81\times x=0.15\times 0.3\times 9.81[/tex]
[tex]\therefore x=\frac{0.4415}{.981}=45centimeters[/tex]
thus the mass of 100 gm should be placed at 95 cm mark in the scale.