A) 1 ohm
B) 2 ohm
C) 3 ohm
D) 4 ohm
Correct Answer: D
Solution :
From the formula \[R={{R}_{0}}(1+\alpha t)\] Ist case at\[\text{50}{{\,}^{\text{o}}}\text{C,}\] we have \[5\,=\,{{R}_{0}}(1+50\,\alpha )\] ?(1) lInd case at \[100{{\,}^{o}}C,\] 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 a in equation (1), we get \[5={{R}_{0}}\left( 1+50\times \frac{1}{200} \right)\] \[{{R}_{0}}=\frac{20}{5}=4\,\text{ohm}\]You need to login to perform this action.
You will be redirected in
3 sec