NEET-XII-Physics
23: Heat and Temperature
- #15-aon a summer day when the temperature is 46°C and (b) on a winter day when the temperature is 6°C? Coefficient of linear expansion of steel = 11 × 10-6 °C-1.Ans : Let the correct length measured by a metre scale made up of steel 16 °C be L.
Initial temperature, t1 = 16 °C
Temperature on a hot summer day, t2 = 46 °C
So, change in temperature, Δθ = t2 `` -`` t1 = 30 °C
Coefficient of linear expansion of steel, `` \alpha `` = 1.1 × 10-5 °C-1
Therefore, change in length,
ΔL = L αΔθ = L × 1.1 × 10-5 × 30
`` \%\,\mathrm{\,of\,}\,\mathrm{\,error\,}=\left(\frac{∆L}{L}\times 100\right)\%``
`` =\left(\frac{L\alpha \Delta \theta }{L}\times 100\right)\%``
`` =\left[1.1\times {10}^{-5}\times 30\times 100\right]\%``
`` =3.3\times {10}^{-2}\%``
`` `` (b) Temperature on a winter day, t2 = 6 °C
So, change in temperature, Δθ = t1 `` -`` t2 = 10 °C
ΔL = L2 `` -`` L1 = L αΔθ = L × 1.1 × 10-5 × 10
`` \%\,\mathrm{\,of\,}\,\mathrm{\,error\,}=\left(\frac{∆L}{L}\times 100\right)\%``
`` =\left(\frac{L\alpha \Delta \theta }{L}\times 100\right)\%``
`` =1.1\times {10}^{-5}\times 10\times 100\%``
`` =1.1\times {10}^{-2}``
`` ``
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- #15-bon a winter day when the temperature is 6°C? Coefficient of linear expansion of steel = 11 × 10-6 °C-1.Ans : Temperature on a winter day, t2 = 6 °C
So, change in temperature, Δθ = t1 `` -`` t2 = 10 °C
ΔL = L2 `` -`` L1 = L αΔθ = L × 1.1 × 10-5 × 10
`` \%\,\mathrm{\,of\,}\,\mathrm{\,error\,}=\left(\frac{∆L}{L}\times 100\right)\%``
`` =\left(\frac{L\alpha \Delta \theta }{L}\times 100\right)\%``
`` =1.1\times {10}^{-5}\times 10\times 100\%``
`` =1.1\times {10}^{-2}``
`` ``
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