A metallic bar of coefficient of linear expansion 10-5 K-1 is heated from 0°C to 100°C. The percentage increase in its length is

Question

A metallic bar of coefficient of linear expansion 10-5 K-1 is heated from 0°C to 100°C. The percentage increase in its length is (A) 0.1 % (B) 1% (C) 10% (D) 0.01 %

Tardigrade Admin · Accepted Answer

Given, α=10-5 K -1 and Δ T=100°-0°=100° C We know that, percentage increase in its length, (Δ l/l) × 100=α Δ T × 100 (Δ l/l)=10-5 × 100=10-3 ∴ (Δ l/l) × 100=0.1 %

Q. A metallic bar of coefficient of linear expansion $1 0^{- 5} K^{- 1}$ is heated from $0^{\circ} C$ to $10 0^{\circ} C$ . The percentage increase in its length is

0.1 %

1%

10%

0.01 %

0.001%

Given, $α = 1 0^{- 5} K^{- 1}$
and $Δ T = 10 0^{\circ} - 0^{\circ} = 10 0^{\circ} C$
We know that, percentage increase in its length,
$\frac{Δ l}{l} \times 100 = α Δ T \times 100$
$\frac{Δ l}{l} = 1 0^{- 5} \times 100 = 1 0^{- 3}$
$∴ \frac{Δ l}{l} \times 100 = 0.1%$

Q. A metallic bar of coefficient of linear expansion 10−5K−1 is heated from 0∘C to 100∘C. The percentage increase in its length is

0.1 %

1%

10%

0.01 %

0.001%

Given, α=10−5K−1 and ΔT=100∘−0∘=100∘C We know that, percentage increase in its length, lΔl​×100=αΔT×100 lΔl​=10−5×100=10−3 ∴lΔl​×100=0.1%

Q. A metallic bar of coefficient of linear expansion $1 0^{- 5} K^{- 1}$ is heated from $0^{\circ} C$ to $10 0^{\circ} C$ . The percentage increase in its length is

Given, $α = 1 0^{- 5} K^{- 1}$
and $Δ T = 10 0^{\circ} - 0^{\circ} = 10 0^{\circ} C$
We know that, percentage increase in its length,
$\frac{Δ l}{l} \times 100 = α Δ T \times 100$
$\frac{Δ l}{l} = 1 0^{- 5} \times 100 = 1 0^{- 3}$
$∴ \frac{Δ l}{l} \times 100 = 0.1%$