Correct answer: B) at equator
Explanation:
We know that the value of acceleration due to gravity ,g is deduced to be: g = GM/R2 by the Law of Gravitational attraction.
According to this, the value of, g depends on the mass and the radius of the Earth (G being a constant).
However, as we know that the Earth is not perfectly spherical but bulged out (R is slightly more at the equator) at the equator on account of its rotation.
Thus, it is found that the value of g is 0.5% more at the poles than it is at the equator.
As a result the weight (W=mg) of the same amount of sugar must also be greater at the poles.
Weight of 1kg sugar at equator measured by the weighing machine=1X9.8=9.8N
Mass measured (as per the calibrated weighing machine) will be= (9.8/9,8)=1kg
Weight of 1kg sugar at poles=1X 9.85=9.85N
Mass measured (as per the calibrated weighing machine) will be= (9.85/9,8)=1.005kg
So at the poles, even 1kg sugar will be read as 1.005 kg by the vendor and you will get lesser sugar by him.
Thus, it is profitable to buy sugar at the equator than on poles.
KEY CONCEPT:
Distance from the centre of the Earth to the poles is 6,356 km whereas from any point on the equator is 6,378 km.
Acceleration due to gravity (g) is inversely proportional to the square of the distance between the body and the centre of the earth.
Due to lesser distance from the poles, the 'g' is around 0.5 % more at the poles when compared to 'g' at equators.
Increase in 'g' at poles will increase the weight (w=mg)
So 1 kg of your sugar at the equator will be weighed as more than 1 kg at the poles.
This will result in increasing the costing of the sugar which will apparently, be more than 1 kg at poles.
The sugar particles in the package won't increase, its measure on the weighing scale will, due to which it will be priced at a higher value at the poles.
Thus it is profitable to buy sugar or any other commodity at the equators as you will pay for its actual weight and not for it apparently increased weight due to variation of 'g' in 'w'.
