Where will it be profitable to purchase 1 kil | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The weight of an object is given by $W = mg$,where $m$ is the mass and $g$ is the acceleration due to gravity.Since the Earth is not a perfect sph

Vedclass · Accepted Answer

At equator

Vedclass · Answer

(B) The weight of an object is given by $W = mg$,where $m$ is the mass and $g$ is the acceleration due to gravity.Since the Earth is not a perfect sphere and bulges at the equator,the radius of the Earth is maximum at the equator and minimum at the poles.The acceleration due to gravity $g$ is inversely proportional to the square of the radius $(g \propto 1/R^2)$.Therefore,$g$ is minimum at the equator and maximum at the poles.To get more sugar for the same weight (or to pay less for the same mass),one should purchase where $g$ is minimum,as the balance will show a smaller weight for the same mass $m$.Thus,it is most profitable to purchase sugar at the equator.

Where will it be profitable to purchase $1$ kilogram of sugar?

Similar Questions

Define acceleration due to gravity. Give the magnitude of $g$ on the surface of earth.

The length of a seconds pendulum at a height $h=2R$ from the earth's surface will be. (Given: $R =$ radius of earth and acceleration due to gravity at the surface of earth $g = \pi^{2} \ m/s^{2}$)

$A$ tunnel is dug along a diameter of the Earth. If $M_e$ and $R_e$ are the mass and radius of the Earth respectively,then the force on a particle of mass $m$ placed in the tunnel at a distance $r$ from the center is:

How does the acceleration due to gravity change with altitude?

The height $h$ at which the weight of a body will be the same as that at the same depth $h$ from the surface of the earth is (Radius of the earth is $R$ and effect of the rotation of the earth is neglected):

Vedclass Test Series

Exam Paper Generator

Online Exam Module