Question

Find the percentage decrease in weight of the body, when taken 16km below the surface of the earth. Take radius of the earth as 6400km.

Answer 1

Answer:

Let be the decrease =. X

The taken below is =16km.

The radius of the earth =6400

X+16=6400.

X=(6400-16)km

X=6384.

x=6384,/100

X=63.84

Explanation:

Therefore the dec

Answer 2

The percentage decrease in weight of the body is 0.25%

Explanation:

Let radius of the earth be R

Let mass of the earth be M

Let mass of the object in earth be m

Now, the mass of the earth is given by the formula:

Mass = Volume × Density

⇒ Mass = 4/3πR³ × ρ

Weight of the body in the earth is given by the formula:

F = (GmM)/R² = Gm(4/3πR³ × ρ)/R² = Gm(4/3 πRρ)

Weight of the body at 16 km below from earth is given by the formula:

F' = Gm(4/3 π(R - 16)ρ)

Decrease in weight is given as:

F - F' = Gm(4/3 π(16)ρ)

The percentage decrease is given as:

(F - F')/F × 100 = (Gm(4/3 π(16)ρ))/(Gm(4/3 πRρ)) × 100

(F - F')/F × 100 = 1600/R

(F - F')/F × 100 = 1600/6400 = 1/4

∴ (F - F')/F × 100 = 0.25%