Answer
Verified
114.3k+ 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.
Recently Updated Pages
Uniform Acceleration - Definition, Equation, Examples, and FAQs
Class 11 JEE Main Physics Mock Test - FREE Online Mock Test Series 2025
JEE Colleges - Detailed Description of Top JEE Colleges
JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key
Geostationary Satellites and Geosynchronous Satellites for JEE
Complex Numbers - Important Concepts and Tips for JEE
Trending doubts
Learn About Angle Of Deviation In Prism: JEE Main Physics 2025
JEE Main Login 2045: Step-by-Step Instructions and Details
JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking
Collision - Important Concepts and Tips for JEE
Ideal and Non-Ideal Solutions Raoult's Law - JEE
Free Radical Substitution Mechanism of Alkanes for JEE Main 2025
Other Pages
Current Loop as Magnetic Dipole and Its Derivation for JEE
A particle performs SHM of amplitude A along a straight class 11 physics JEE_Main
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Keys & Solutions
JEE Main 2023 January 30 Shift 2 Question Paper with Answer Keys & Solutions
Inductive Effect and Acidic Strength - Types, Relation and Applications for JEE
NCERT Solutions for Class 11 Physics Chapter 4 Laws of Motion