QJEEM22Ph6f82: JEE Main 2022 Physics - Class 11 JEEM22Ph6f82 Solution: An object is taken to a height above the surface of earth at a…

The weight of an object at a distance d from the center of the Earth is given by:

$$W^{\prime }=W{\left(\frac{R}{d}\right)}^2$$

where:

• W' is the weight at distance d,
• W is the weight at the Earth's surface,
• R is the radius of the Earth,
• d is the distance from the center of the Earth.

In this problem, we are asked to find the percentage decrease in weight, which can be calculated by:

$$\mathrm{\Delta }W=\frac{W-W^{\prime }}{W}\times 100\mathrm{\%}$$

where ΔW is the percentage change in weight. Substituting the weight formula into the percentage change formula gives:

$$\mathrm{\Delta }W=(1-{\left(\frac{R}{d}\right)}^2)\times 100\mathrm{\%}$$

The radius of the Earth R is 6400 km and the distance d from the center of the Earth is given as $\frac{5}{4}R$ . Substituting these values into the equation gives:

$$\mathrm{\Delta }W=(1-{\left(\frac{6400km}{\frac{5}{4}\cdot 6400km}\right)}^2)\times 100\mathrm{\%}=(1-{\left(\frac{4}{5}\right)}^2)\times 100\mathrm{\%}=36\mathrm{\%}$$

Therefore, the percentage decrease in the weight of the object when taken to a height of $\frac{1}{4}R$ above the surface of the Earth is 36%.