A) \[{{400}^{{}^\circ }}C\]
B) \[{{200}^{{}^\circ }}C\]
C) \[{{300}^{{}^\circ }}C\]
D) \[{{500}^{{}^\circ }}C\]
Correct Answer: A
Solution :
: As the resistance of the bulb filament at \[{{T}^{o}}C\]is \[{{R}_{T}}={{R}_{0}}(1+\alpha T)\] where \[{{R}_{0}}\] is its resistance at \[{{0}^{o}}C\]and a is the temperature coefficient of resistance. \[\therefore \] \[{{R}_{{{T}_{1}}}}={{R}_{0}}(1+\alpha {{T}_{1}})\] ?....(i) and \[{{R}_{{{T}_{2}}}}={{R}_{0}}(1+\alpha {{T}_{2}})\] ???(ii) Dividing eqn. (i) by eqn. (ii), we get \[\frac{{{R}_{{{T}_{1}}}}}{{{R}_{{{T}_{2}}}}}=\frac{1+\alpha {{T}_{1}}}{1+\alpha {{T}_{2}}}\] Here, \[{{R}_{{{T}_{1}}}}=100\Omega ,\,{{T}_{1}}={{100}^{o}}C\] \[{{R}_{{{T}_{2}}}}=200\Omega ,\,{{T}_{2}}=?\] \[\alpha ={{0.005}^{o}}{{C}^{-1}}\] \[\therefore \] \[\frac{100\Omega }{200\Omega }=\frac{1+({{0.005}^{o}}{{C}^{-1}})({{100}^{o}}C)}{1+({{0.005}^{o}}{{C}^{-1}}){{T}_{2}}}\] \[\frac{1}{2}=\frac{1+0.5}{1+({{0.005}^{o}}{{C}^{-1}}){{T}_{2}}}\] \[2(1+0.5)=1+({{0.005}^{o}}{{C}^{-1}}){{T}_{2}}\] \[3=1+({{0.005}^{o}}{{C}^{-1}}){{T}_{2}}\] or \[({{0.005}^{o}}{{C}^{-1}}){{T}_{2}}=3-1=2\] \[{{T}_{2}}=\frac{2}{({{0.005}^{o}}{{C}^{-1}})}={{400}^{o}}C\]
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