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NEET-XII-Physics

23: Heat and Temperature

with Solutions - page 4
Qstn# iv-15 Prvs-QstnNext-Qstn
  • #15
    A metre scale is made up of steel and measures correct length at 16°C. What will be the percentage error if this scale is used (a) on 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 : (a) 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 = L​2 `` -`` 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}``
    `` ``
    Page No 13:
  • #15-a
    on a summer day when the temperature is 46°C and
    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}\%``
    `` ``
  • #15-b
    on 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 = L​2 `` -`` 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}``
    `` ``
    Page No 13: