A) 1 ohm
B) 2 ohm
C) 3 ohm
D) 4 ohm
Correct Answer: D
Solution :
From the formula \[R={{R}_{0}}(1+\alpha t)\] \[\text{Ist}\] case at \[50{}^\circ C\], we have \[5={{R}_{0}}(1+50\alpha )\] ...(1) \[\text{IInd}\] case at \[100{}^\circ C\], we have \[6={{R}_{0}}(1+100\alpha )\] ...(2) Now dividing equation (2) by equation (1), we get \[\frac{6}{5}=\frac{1+100\alpha }{1+50\alpha }\] \[6+300\alpha =5+500\alpha \] so \[\alpha =\frac{1}{200}\] Again after putting the value of \[\alpha \] in equation (1), we get \[5={{R}_{0}}\left( 1+50\times \frac{1}{200} \right)\] \[{{R}_{0}}=\frac{20}{5}=4ohm\]
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