How much should the temperature of a brass rod be increased so as to increase its length by 1 % ? (Given, α for brass is 0.00002° C -1 ).

Question

How much should the temperature of a brass rod be increased so as to increase its length by 1 % ? (Given, α for brass is 0.00002° C -1 ). (A) 300° C (B) 400° C (C) 500° C (D) 550° C

Tardigrade Admin · Accepted Answer

Here, Δ T=?, (Δ L/L)=(1/100), α=0.00002° C -1 As, Δ L=α L Δ T ∴ α Δ T =(Δ L/L) or Δ T=(Δ L/L α)=(1/100 × 0.00002) Δ T =(105/2 × 102)=500° C So, 500° C of temperature of a brass rod should be increased, so as to increase its length by 1 %.

Q. How much should the temperature of a brass rod be increased so as to increase its length by $1%$ ? (Given, $α$ for brass is $0.0000 2^{\circ} C^{- 1}$ ).

$30 0^{\circ} C$

$40 0^{\circ} C$

$50 0^{\circ} C$

$55 0^{\circ} C$

Q. How much should the temperature of a brass rod be increased so as to increase its length by 1% ? (Given, α for brass is 0.00002∘C−1 ).

300∘C

400∘C

500∘C

550∘C

Here, ΔT=?,LΔL​=1001​,α=0.00002∘C−1 As, ΔL=αLΔT ∴αΔT=LΔL​ or ΔT=LαΔL​=100×0.000021​ ΔT=2×102105​=500∘C So, 500∘C of temperature of a brass rod should be increased, so as to increase its length by 1%.

Q. How much should the temperature of a brass rod be increased so as to increase its length by $1%$ ? (Given, $α$ for brass is $0.0000 2^{\circ} C^{- 1}$ ).

$30 0^{\circ} C$

$40 0^{\circ} C$

$50 0^{\circ} C$

$55 0^{\circ} C$