Answer:
2 kg
Explanation:
Note: For the meter stick to be balanced,
Sum of clock wise moment must be equal to sum of anti clock wise moment
Wd = W'd' ................ Equation 1
Where W = weight of the rock, d = distance of the meter stick from the point of support, W' = weight of the that must be suspended for the meter stick to be balanced, d' = distance of the mass to the point of support.
make W' the subject of the equation
W' = Wd/d'............... Equation 2
Taking our moment about the support,
Given: W = mg = 1 ×9.8 = 9.8 N, d = 50 cm, d' = (75-50) = 25 cm
Substitute into equation 2
W' = 9.8(50)/25
W' = 19.6 N.
But,
m = W'/g
m = 19.6/9.8
m = 2 kg.
The mass m suspended from the meter stick at the 75 cm mark if the system is balanced has the value of 2 kg.
Given data:
The mass of rock is, m' = 1 kg.
The distance of mark to ensure balance condition is, d'' = 75 cm.
As per the concept of equilibrium, the sum of clock wise moment must be equal to sum of anti clock wise moment
[tex]W \times d=W' \times d'[/tex] ................................................(1)
Here,
W is the weight of the rock.
d is distance of the meter stick from the point of support.
W' is the weight that must be suspended for the meter stick to be balanced.
d' is the distance of the mass to the point of support.
Solving as,
[tex]W'=\dfrac{W \times d}{d'}[/tex].........................................................(2)
Where,
W = mg = 1 ×9.8 = 9.8 N,
d = 50 cm
d' = (75-50) = 25 cm
Substitute the value in equation 2 as,
[tex]W'=\dfrac{9.8 \times 50}{25}\\\\W' = 19.6 \;\rm m[/tex]
Now, mass can be calculated as,
m = W'/g
m = 19.6/9.8
m = 2 kg.
Thus, we can conclude that the mass m suspended from the meter stick at the 75 cm mark if the system is balanced has the value of 2 kg.
Learn more about the equilibrium condition here:
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