Question

The radius of a metal sphere at room temperature T is R and the coefficient of linear expansion of the metal is α. The sphere is heated a little by a temperature ΔT so that its new temperature is T + ΔT. The increase in the volume of the sphere is approximate.

Options

• 2πRαΔT

• 2πR²αΔT

• `4π"R"^3(Δ"T")/3`

• 4πR³αΔT

Answer 1

4πR³αΔT

Explanation:

Let original volume V₀ = `4/3π"R"^3`

We know that,

α = `γ/3`

⇒ γ = 3α

also, γ = `(Δ"V")/"V"_0Δ"T" - 3α`

∴ ΔV = `3αΔ"T" xx 4/3 π"R"^3`

= 4πR³αΔT