Qstnid -Iv-14-two Metre Scales, One Of Steel And The Other Of Aluminium, Agree At 20°c. Calculate The Ratio Aluminium-centimetre/steel-centimetre At ( Α For Steel = 1.1 &Times; 10-5 °C-1 And For Aluminium = 2.3 &Times; 10-5°c-1.) (A) 0°c, (B) 40°c And (C) 100°c. (A) 0°c, (B) 40°c And (C) 100°c. - 23: Heat And Temperature

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NEET-XII-Physics

23: Heat and Temperature

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#14
Two metre scales, one of steel and the other of aluminium, agree at 20°C. Calculate the ratio aluminium-centimetre/steel-centimetre at ( α for steel = 1.1 × 10^-5 °C^-1 and for aluminium = 2.3 × 10^-5°C^-1.) (a) 0°C, (b) 40°C and (c) 100°C. (a) 0°C, (b) 40°C and (c) 100°C.
Ans : Given:
At 20°C, length of the metre scale made up of steel, L_st= length of the metre scale made up of aluminium, L_al
Coefficient of linear expansion for aluminium, α_al = 2.3 × 10^-5°C^-1
Coefficient of linear expansion for steel, α_st = 1.1 × 10^-5°C^-1
Let the length of the aluminium scale at 0°C, 40°C and 100°C be L_0al,L₄_0aland L₁₀_0al.
And let the length of the steel scale at 0°C, 40°C and 100°C be L_0st,L₄_0stand L₁₀_0st. (a) So, L_0st(1 - α_st × 20) = L_0al(1 - α_al × 20)
`` \frac{{L}_{0st}}{{L}_{0al}}=\frac{\left(1-{\alpha }_{al}\times 20\right)}{\left(1-{\alpha }_{st}\times 20\right)}``
`` \Rightarrow \frac{{L}_{0st}}{{L}_{0al}}=\frac{1-2.3\times {{10}^{-}}^{5}\times 20}{1-1.1\times {10}^{-5}\times 20}``
`` \Rightarrow \frac{{L}_{0st}}{{L}_{0al}}=\frac{0.99954}{0.99978}``
`` \Rightarrow \frac{{L}_{0st}}{{L}_{0al}}=0.999759`` (b) `` \frac{{L}_{40al}}{{L}_{40st}}=\frac{{L}_{0al}\left(1+\alpha {}_{al}\times 40\right)}{{L}_{0st}\left(1+{\alpha }_{st}\times 40\right)}``
`` \Rightarrow \frac{{L}_{40al}}{{L}_{40st}}=\frac{{L}_{0al}}{{L}_{0st}}\times \frac{\left(1+2.3\times {10}^{-5}\times 40\right)}{\left(1+1.1\times {10}^{-5}\times 40\right)}``
`` \Rightarrow \frac{{L}_{40al}}{{L}_{40st}}=\frac{0.99977\times 1.00092}{1.00044}``
`` \Rightarrow \frac{{L}_{40al}}{{L}_{40st}}=1.0002496``
`` `` (c) `` \frac{{L}_{100al}}{{L}_{100st}}=\frac{{L}_{0al}\left(1+\alpha {}_{al}\times 100\right)}{{L}_{0st}\left(1+{\alpha }_{st}\times 100\right)}``
`` \Rightarrow \frac{{L}_{100al}}{{L}_{100st}}=\frac{0.99977\times 1.0023}{1.0023}``
`` \Rightarrow \frac{{L}_{100al}}{{L}_{100st}}=1.00096``
`` ``
Page No 13: (a) So, L_0st(1 - α_st × 20) = L_0al(1 - α_al × 20)
`` \frac{{L}_{0st}}{{L}_{0al}}=\frac{\left(1-{\alpha }_{al}\times 20\right)}{\left(1-{\alpha }_{st}\times 20\right)}``
`` \Rightarrow \frac{{L}_{0st}}{{L}_{0al}}=\frac{1-2.3\times {{10}^{-}}^{5}\times 20}{1-1.1\times {10}^{-5}\times 20}``
`` \Rightarrow \frac{{L}_{0st}}{{L}_{0al}}=\frac{0.99954}{0.99978}``
`` \Rightarrow \frac{{L}_{0st}}{{L}_{0al}}=0.999759`` (b) `` \frac{{L}_{40al}}{{L}_{40st}}=\frac{{L}_{0al}\left(1+\alpha {}_{al}\times 40\right)}{{L}_{0st}\left(1+{\alpha }_{st}\times 40\right)}``
`` \Rightarrow \frac{{L}_{40al}}{{L}_{40st}}=\frac{{L}_{0al}}{{L}_{0st}}\times \frac{\left(1+2.3\times {10}^{-5}\times 40\right)}{\left(1+1.1\times {10}^{-5}\times 40\right)}``
`` \Rightarrow \frac{{L}_{40al}}{{L}_{40st}}=\frac{0.99977\times 1.00092}{1.00044}``
`` \Rightarrow \frac{{L}_{40al}}{{L}_{40st}}=1.0002496``
`` `` (c) `` \frac{{L}_{100al}}{{L}_{100st}}=\frac{{L}_{0al}\left(1+\alpha {}_{al}\times 100\right)}{{L}_{0st}\left(1+{\alpha }_{st}\times 100\right)}``
`` \Rightarrow \frac{{L}_{100al}}{{L}_{100st}}=\frac{0.99977\times 1.0023}{1.0023}``
`` \Rightarrow \frac{{L}_{100al}}{{L}_{100st}}=1.00096``
`` ``
Page No 13:

#14-a
0°C,
Ans : So, L_0st(1 - α_st × 20) = L_0al(1 - α_al × 20)
`` \frac{{L}_{0st}}{{L}_{0al}}=\frac{\left(1-{\alpha }_{al}\times 20\right)}{\left(1-{\alpha }_{st}\times 20\right)}``
`` \Rightarrow \frac{{L}_{0st}}{{L}_{0al}}=\frac{1-2.3\times {{10}^{-}}^{5}\times 20}{1-1.1\times {10}^{-5}\times 20}``
`` \Rightarrow \frac{{L}_{0st}}{{L}_{0al}}=\frac{0.99954}{0.99978}``
`` \Rightarrow \frac{{L}_{0st}}{{L}_{0al}}=0.999759``

#14-b
40°C and
Ans : `` \frac{{L}_{40al}}{{L}_{40st}}=\frac{{L}_{0al}\left(1+\alpha {}_{al}\times 40\right)}{{L}_{0st}\left(1+{\alpha }_{st}\times 40\right)}``
`` \Rightarrow \frac{{L}_{40al}}{{L}_{40st}}=\frac{{L}_{0al}}{{L}_{0st}}\times \frac{\left(1+2.3\times {10}^{-5}\times 40\right)}{\left(1+1.1\times {10}^{-5}\times 40\right)}``
`` \Rightarrow \frac{{L}_{40al}}{{L}_{40st}}=\frac{0.99977\times 1.00092}{1.00044}``
`` \Rightarrow \frac{{L}_{40al}}{{L}_{40st}}=1.0002496``
`` ``

#14-c
100°C.
Ans : `` \frac{{L}_{100al}}{{L}_{100st}}=\frac{{L}_{0al}\left(1+\alpha {}_{al}\times 100\right)}{{L}_{0st}\left(1+{\alpha }_{st}\times 100\right)}``
`` \Rightarrow \frac{{L}_{100al}}{{L}_{100st}}=\frac{0.99977\times 1.0023}{1.0023}``
`` \Rightarrow \frac{{L}_{100al}}{{L}_{100st}}=1.00096``
`` ``
Page No 13: