NEET-XII-Physics
23: Heat and Temperature
- #14-a0°C, (b) 40°C and (c) 100°C.Ans : So, L0st(1 - αst × 20) = L0al(1 - αal × 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``
`` ``
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- #14-b40°C andAns : `` \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-c100°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``
`` ``
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