If the volume of a block of metal changes by 0.12 %, when it is heated through 20 °C, then the coefficient of linear expansion is

Question

If the volume of a block of metal changes by 0.12 %, when it is heated through 20 °C, then the coefficient of linear expansion is (A) 4 × 10-5 °C-1 (B) 2 × 10-5 °C-1 (C) 0.5 × 10-5 °C-1 (D) 4 × 10-4 °C-1

Tardigrade Admin · Accepted Answer

Here, (Δ V/V)=0.12 %, (0.12/100) Δ T=20 °C Now, &gamma;=(Δ V/VΔ T)=(0.12/100 × 20) =6 × 10-5 °C-1 ∴ α=(&gamma;/3)=2 ×10-5 °C-1

Q. If the volume of a block of metal changes by $0.12%$ , when it is heated through $20^{\circ} C$ , then the coefficient of linear expansion is

$4 \times 1 0^{- 5}^{\circ} C^{- 1}$

$2 \times 1 0^{- 5}^{\circ} C^{- 1}$

$0.5 \times 1 0^{- 5}^{\circ} C^{- 1}$

$4 \times 1 0^{- 4}^{\circ} C^{- 1}$

Here, $\frac{Δ V}{V} = 0.12%$ , $\frac{0.12}{100}$
$Δ T = 20^{\circ} C$
Now, $γ = \frac{Δ V}{V Δ T} = \frac{0.12}{100 \times 20}$
$= 6 \times 1 0^{- 5}^{\circ} C^{- 1}$
$∴ α = \frac{γ}{3} = 2 \times 1 0^{- 5}^{\circ} C^{- 1}$

Q. If the volume of a block of metal changes by 0.12%, when it is heated through 20∘C, then the coefficient of linear expansion is

4×10−5∘C−1

2×10−5∘C−1

0.5×10−5∘C−1

4×10−4∘C−1