A) \[1.06\times {{10}^{-3}}{}^\circ {{C}^{-1}}\]
B) \[3.5\times {{10}^{-2}}\,{}^\circ {{C}^{-1}}\]
C) \[2.22\times {{10}^{-3}}\text{ }{}^\circ {{C}^{-1}}\]
D) \[3.95\times {{10}^{-2}}\text{ }{}^\circ {{C}^{-1}}\]
Correct Answer: C
Solution :
(c) \[2.22\times {{10}^{-3}}\,{}^\circ {{C}^{-1}}\] Here, \[{{R}_{1}}=2.5\,\Omega ,{{T}_{1}}=28{}^\circ C\] \[{{R}_{2}}=2.9\,\Omega \] and \[{{T}_{2}}=~100{}^\circ C\] As \[{{R}_{2}}={{R}_{1}}[1+\alpha ({{T}_{2}}-{{T}_{1}})]\] \[\therefore 2.9=2.5[1+\alpha (100-28)]\] \[\frac{2.9}{2.5}-1=72\alpha \] or \[\alpha =\frac{1}{72}\times \frac{2.9-2.5}{2.5}\] \[=\frac{1}{72}\times \frac{0.4}{2.5}=2.22\times {{10}^{-3}}\,{}^\circ {{C}^{-1}}\]
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