a uniform meter rule is pivoted at 20cm mark .
It is balanced when mass of 100gm on 10cm mark and mass of 50 gm on 30 cm mark are suspended. Find d mass of meter rule
By principle of moments.
(10*100)=(50*10)+W*30
So W=(1000-500)/30
=16.67
moments around pivot point at 20cm
m(50-20)+50(30-20)=100(20-10)
3m + 50 = 100
no idea
No Idea
To find the mass of the meter rule, we can use the principle of moments. The principle of moments states that for an object to be in equilibrium, the sum of the clockwise moments must be equal to the sum of the anticlockwise moments.
In this case, the meter rule is pivoted at the 20 cm mark, and we have two masses suspended, one at the 10 cm mark and the other at the 30 cm mark. Let's denote the mass of the meter rule as "x" grams.
Now, let's calculate the moments of each mass with respect to the pivot point:
Moment of 100 gm mass at 10 cm mark = Mass * Distance
= 100 gm * 10 cm = 1000 gm cm
Moment of 50 gm mass at 30 cm mark = Mass * Distance
= 50 gm * 30 cm = 1500 gm cm
According to the principle of moments, the sum of the clockwise moments (1500 gm cm) must be equal to the sum of the anticlockwise moments (1000 gm cm + moment of the meter rule).
So, sum of anticlockwise moments = 1000 gm cm + x gm * 20 cm = 1000 gm cm + 20x gm cm
1500 gm cm = 1000 gm cm + 20x gm cm
Simplifying the equation:
500 gm cm = 20x gm cm
Divide both sides by 20 cm:
25 gm = x
Therefore, the mass of the meter rule is 25 grams.