A blacksmith fixes iron ring on the rim of the wooden wheel of a bullock cart. The diameter of the rim and the iron ring are 5.243 m and 5.231 m respectively at 27° C. The αl of iron is 1.2 × 10-5 K -1 . To what temperature should the ring be heated so as to fit the rim of the wheel?

Question

A blacksmith fixes iron ring on the rim of the wooden wheel of a bullock cart. The diameter of the rim and the iron ring are 5.243 m and 5.231 m respectively at 27° C. The αl of iron is 1.2 × 10-5 K -1 . To what temperature should the ring be heated so as to fit the rim of the wheel? (A) 174° C (B) 192° C (C) 206° C (D) 218° C

Tardigrade Admin · Accepted Answer

L=L0(1+αl Δ T) Δ T=((L/L0)-1) (1/αl) =((5.243/5.231)-1) × (1/1.2 × 10-5)=191° C ⇒ T2=Δ T+T1=(191+27)=218° C

$17 4^{\circ} C$

$19 2^{\circ} C$

$20 6^{\circ} C$

$21 8^{\circ} C$

$L = L_{0} (1 + α_{l} Δ T)$
$Δ T = (\frac{L}{L _{0}} - 1) \frac{1}{α _{l}}$
$= (\frac{5.243}{5.231} - 1) \times \frac{1}{1.2 \times 1 0 ^{- 5}} = 19 1^{\circ} C$
$\Rightarrow T_{2} = Δ T + T_{1} = (191 + 27) = 21 8^{\circ} C$

Q. A blacksmith fixes iron ring on the rim of the wooden wheel of a bullock cart. The diameter of the rim and the iron ring are 5.243m and 5.231m respectively at 27∘C. The αl​ of iron is 1.2×10−5K−1. To what temperature should the ring be heated so as to fit the rim of the wheel?

174∘C

192∘C

206∘C

218∘C

L=L0​(1+αl​ΔT) ΔT=(L0​L​−1)αl​1​ =(5.2315.243​−1)×1.2×10−51​=191∘C ⇒T2​=ΔT+T1​=(191+27)=218∘C

$17 4^{\circ} C$

$19 2^{\circ} C$

$20 6^{\circ} C$

$21 8^{\circ} C$

$L = L_{0} (1 + α_{l} Δ T)$
$Δ T = (\frac{L}{L _{0}} - 1) \frac{1}{α _{l}}$
$= (\frac{5.243}{5.231} - 1) \times \frac{1}{1.2 \times 1 0 ^{- 5}} = 19 1^{\circ} C$
$\Rightarrow T_{2} = Δ T + T_{1} = (191 + 27) = 21 8^{\circ} C$