
Where will it be profitable to purchase one kilogram sugar?
A) At poles
B) At equator
C) At \[{{45}^{\circ }}\] latitude
D) At \[{{40}^{\circ }}\] latitude
Answer
126.6k+ views
Hint: Profit is paying less for more. One kilogram sugar will be more when the actual weight (\[mg\] ) of sugar will be more than the weight as weighed by any machine. Increase in \[g\] will increase the weight.
Complete step by step solution:
Distance from the center of the Earth to the poles is \[6356km\] and the distance from the center of Earth to any point on the equator is \[6378km\].
Acceleration due to gravity (\[g\] ) is inversely proportional to the square of the distance between the body and the center of the earth which is formulated below:
\[g=\dfrac{GM}{{{r}^{2}}}\]
Where \[g\] represents acceleration due to gravity
\[G\] Represents the universal Gravitational constant, \[G=6.674 \times {{10}^{-11}}{{m}^{3}}k{{g}^{-1}}{{s}^{-2}}\] ⋅
\[M\] Represents mass of Earth
\[r\] Represents the distance of a point on the surface from the center of Earth.
As \[g\] is inversely proportional to \[r\] hence value of \[g\] will be greater when r is smaller
Due to lesser distance from the poles, the ' \[g\] ' is around more at the poles when compared to ' \[g\]' at equators. Increase in ' \[g\]' at poles will increase the weight (\[W=mg\] ). So one kilogram of sugar at poles will weigh more than its actual weight.
One kilogram of your sugar at the equator will weigh less than its actual weight.
Note: The sugar particles in the package won't increase, its measure on the weighing scale will increase 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 the weight shown in the weighing machine will be relatively less.
Complete step by step solution:
Distance from the center of the Earth to the poles is \[6356km\] and the distance from the center of Earth to any point on the equator is \[6378km\].
Acceleration due to gravity (\[g\] ) is inversely proportional to the square of the distance between the body and the center of the earth which is formulated below:
\[g=\dfrac{GM}{{{r}^{2}}}\]
Where \[g\] represents acceleration due to gravity
\[G\] Represents the universal Gravitational constant, \[G=6.674 \times {{10}^{-11}}{{m}^{3}}k{{g}^{-1}}{{s}^{-2}}\] ⋅
\[M\] Represents mass of Earth
\[r\] Represents the distance of a point on the surface from the center of Earth.
As \[g\] is inversely proportional to \[r\] hence value of \[g\] will be greater when r is smaller
Due to lesser distance from the poles, the ' \[g\] ' is around more at the poles when compared to ' \[g\]' at equators. Increase in ' \[g\]' at poles will increase the weight (\[W=mg\] ). So one kilogram of sugar at poles will weigh more than its actual weight.
One kilogram of your sugar at the equator will weigh less than its actual weight.
Note: The sugar particles in the package won't increase, its measure on the weighing scale will increase 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 the weight shown in the weighing machine will be relatively less.
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