At what temperature a gold ring of diameter 6.230 cm be heated so that it can be fitted on a wooden bangle of diameter 6.241 cm ? Both the diameters have been measured at room temperature (27° C ). (Given: coefficient of linear thermal expansion of gold α L =1.4 × 10-5 K -1 )

Question

At what temperature a gold ring of diameter 6.230 cm be heated so that it can be fitted on a wooden bangle of diameter 6.241 cm ? Both the diameters have been measured at room temperature (27° C ). (Given: coefficient of linear thermal expansion of gold α L =1.4 × 10-5 K -1 ) (A) 125.7° C (B) 91.7° C (C) 425.7° (D) 152.7° C

Tardigrade Admin · Accepted Answer

Δ ℓ=6.241-6.230=0.011 cm Δ ℓ=ℓ α Δ θ 0.011=6.230 × 1.4 × 10-5(θ-27) θ-27=(0.011 × 105/6.230 × 1.4) θ ≈ 153.11 nearest is 152.7° C.

$125. 7^{\circ} C$

$91. 7^{\circ} C$

$425. 7^{\circ}$

$152. 7^{\circ} C$

$Δ ℓ = 6.241 - 6.230 = 0.011 c m$
$Δ ℓ = ℓ α Δ θ$
$0.011 = 6.230 \times 1.4 \times 1 0^{- 5} (θ - 27)$
$θ - 27 = \frac{0.011 \times 1 0 ^{5}}{6.230 \times 1.4}$
$θ \approx 153.11$ nearest is $152. 7^{\circ} C$ .

Q. At what temperature a gold ring of diameter 6.230cm be heated so that it can be fitted on a wooden bangle of diameter 6.241cm ? Both the diameters have been measured at room temperature (27∘C). (Given: coefficient of linear thermal expansion of gold αL​=1.4×10−5K−1 )

125.7∘C

91.7∘C

425.7∘

152.7∘C

Δℓ=6.241−6.230=0.011cm Δℓ=ℓαΔθ 0.011=6.230×1.4×10−5(θ−27) θ−27=6.230×1.40.011×105​ θ≈153.11 nearest is 152.7∘C.

$125. 7^{\circ} C$

$91. 7^{\circ} C$

$425. 7^{\circ}$

$152. 7^{\circ} C$

$Δ ℓ = 6.241 - 6.230 = 0.011 c m$
$Δ ℓ = ℓ α Δ θ$
$0.011 = 6.230 \times 1.4 \times 1 0^{- 5} (θ - 27)$
$θ - 27 = \frac{0.011 \times 1 0 ^{5}}{6.230 \times 1.4}$
$θ \approx 153.11$ nearest is $152. 7^{\circ} C$ .