Question

Question

1.What will be the percentage decrease in weight of a body, When it is taken 32 km below the surface of earth.Take radius of earth 6400 kilom...,​

Answer 1

• A body is taken 32 km above the surface of the earth
• Radius of earth = 6400 km
• Percentage decrease in weight of the body = ?

$\begin{gathered}\sf :\implies g' = g\bigg\lgroup 1 - \dfrac{h}{R}\bigg\rgroup\;\;-eq(1)\\\end{gathered}$

• So then the % decrease in the value of acceleration due to gravity is gonna be,

$\begin{gathered}\sf :\implies \delta g = \dfrac{g - g'}{g} \times 100\\\end{gathered}$

$\begin{gathered}\sf :\implies \delta g = \dfrac{g - g\bigg\lgroup 1 - \dfrac{h}{R}\bigg\rgroup}{g} \times 100\\\end{gathered}$

$\begin{gathered}\sf :\implies \delta g = \dfrac{g\bigg\{ 1 - \bigg\lgroup 1 - \dfrac{h}{R}\bigg\rgroup\bigg\} }{g} \times 100\\\end{gathered}$

$\begin{gathered}\sf :\implies \delta g = \bigg\{ 1 - 1 + \dfrac{h}{R} \bigg\} 100\\\end{gathered}$

$\sf :\implies \delta g = \dfrac{100h}{R} \;\; -eq(2)$

• Finally, the % change in weight will be given by,

$\begin{gathered}\sf :\implies \delta W = \delta m + \delta g\\\end{gathered}$

$\begin{gathered}\sf :\implies \delta W = \dfrac{100h}{R}\\\end{gathered}$

$\displaystyle \underline{\bigstar\:\textsf{According to the Question :}}$

$\sf\dashrightarrow \delta W = \dfrac{100\times 32}{6400}$

$\dashrightarrow\underline{\boxed{\purple{\mathfrak {\delta w = 0.5 \%}}}}$

Answer 2

Answer:

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