Question

When a metal sphere is suspended at the end of a metal wire its extension is 0.4mm. If another metal sphere of the same material with its radius half as the previous is suspended then extension would be

Answer 1

r₁ = radius of the initial sphere = r

V₁ = Volume of the initial sphere = 4πr₁³/3

ρ = density of material of sphere

m₁ = mass of the initial sphere = ρ V₁ = 4πr₁³ρ/3

r₂ = radius of the other sphere = r/2

V₂ = Volume of the other sphere = 4πr₂³/3

m₂ = mass of the other sphere = ρ V₂ = 4πr₂³ρ/3

k = spring constant of the wire

Δx₁ = stretch of the wire due to initial sphere = 0.4 mm = 0.0004 m

Δx₂ = stretch of the wire due to other sphere = ?

for the first sphere hanged :

weight of the sphere = spring force by wire

m₁ g = k Δx₁                    eq-1

for the other sphere hanged :

weight of the sphere = spring force by wire

m₂ g = k Δx₂                    eq-2

dividing eq-2 by eq-1

m₂ g /(m₁ g ) =  k Δx₂ /( k Δx₁ )

m₂  /m₁ = Δx₂ / Δx₁

(4πr₂³ρ/3) /(4πr₁³ρ/3) = Δx₂ / Δx₁

(r₂/r₁)³ =  Δx₂ / Δx₁

((r/2)/r)³ =  Δx₂ / 0.4

(0.5)³ =  Δx₂ / 0.4

Δx₂ = 0.05 mm