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 Δ T so that its new temperature is (T + Δ T). The increase in the volume of the sphere is approximately

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 Δ T so that its new temperature is (T + Δ T). The increase in the volume of the sphere is approximately (A) 2π RαΔ T (B) π RαΔ T (C) 4π R3 αΔ T/3 (D) 4π R3 αΔ T

Tardigrade Admin · Accepted Answer

As &gamma;=(Δ V/V × Δ T) and &gamma;=3α, ∴ 3α=(Δ V/((4π)3R3)Δ T) which gives, Δ V=4π R3 αΔ T

Q. 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 $Δ T$ so that its new temperature is $(T + Δ T)$ . The increase in the volume of the sphere is approximately

$2 π R α Δ T$

$π R α Δ T$

$4 π R^{3} α Δ T /3$

$4 π R^{3} α Δ T$

As $γ = \frac{Δ V}{V \times Δ T}$ and $γ = 3 α$ ,
$∴ 3 α = \frac{Δ V}{( \frac{4 π}{3} R ^{3} ) Δ T}$
which gives, $Δ V = 4 π R^{3} α Δ T$

Q. 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 approximately

2πRαΔT

πRαΔT

4πR3αΔT/3

4πR3αΔT