
If the resistance of a conductor is $5\Omega $ at ${50^0}C$ and $7\Omega $ at ${100^0}C$, then mean temperature coefficient of resistance (of material) is
A. $0.013/{}^0C$
B. $0.004/{}^0C$
C. $0.006/{}^0C$
D. $0.008/{}^0C$
Answer
163.2k+ views
Hint: First try to find the relation between electric resistance and the temperature of conductor of the given material. Then find the ratio of resistance of the conductor in both the cases given in the question. Then apply the same ratio to find the temperature coefficient when the temperature in both cases is given.
Formula used:
Resistance, $R = 1 + \alpha t$
Where, $\alpha $ is temperature coefficient and t is temperature.
Complete step by step solution:
First start with the given information from the question:
Resistance of thermometer in normal case, ${R_1} = 5\Omega $.
Resistance of thermometer after dipped in liquid, ${R_2} = 7\Omega $.
Temperature in the first case, ${t_1} = {50^0}C$.
Temperature in the first case, ${t_1} = {100^0}C$.
Now, we know that resistance of conductor here is given by the following formula:
$R = 1 + \alpha t$
Then the ratio of both resistances will be given by:
$\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{\left( {1 + \alpha {t_1}} \right)}}{{\left( {1 + \alpha {t_2}} \right)}} \\ $
Putting the values in above ratio, we get:
$\dfrac{{{R_2}}}{{{R_1}}} = \dfrac{{\left( {1 + \alpha \times 100} \right)}}{{\left( {1 + \alpha \times 50} \right)}} = \dfrac{7}{5} \\ $
$\Rightarrow 7 + 350\alpha = 5 + 500\alpha \\ $
By solving, we get the following value:
$\therefore \alpha = \dfrac{2}{{150}} = 0.013/{}^0C$
Hence, the correct answer is option A.
Note: The temperature coefficient of resistance is commonly described as the change in electrical resistance of a material per degree change in temperature. So, when we look at the electrical resistance of conductors like gold, aluminium, silver, and copper, it all depends on the process of electron collision within the substance. The process of electron collision accelerates as temperature rises. As a result, as the conductor's temperature rises, so does its resistance.
Formula used:
Resistance, $R = 1 + \alpha t$
Where, $\alpha $ is temperature coefficient and t is temperature.
Complete step by step solution:
First start with the given information from the question:
Resistance of thermometer in normal case, ${R_1} = 5\Omega $.
Resistance of thermometer after dipped in liquid, ${R_2} = 7\Omega $.
Temperature in the first case, ${t_1} = {50^0}C$.
Temperature in the first case, ${t_1} = {100^0}C$.
Now, we know that resistance of conductor here is given by the following formula:
$R = 1 + \alpha t$
Then the ratio of both resistances will be given by:
$\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{\left( {1 + \alpha {t_1}} \right)}}{{\left( {1 + \alpha {t_2}} \right)}} \\ $
Putting the values in above ratio, we get:
$\dfrac{{{R_2}}}{{{R_1}}} = \dfrac{{\left( {1 + \alpha \times 100} \right)}}{{\left( {1 + \alpha \times 50} \right)}} = \dfrac{7}{5} \\ $
$\Rightarrow 7 + 350\alpha = 5 + 500\alpha \\ $
By solving, we get the following value:
$\therefore \alpha = \dfrac{2}{{150}} = 0.013/{}^0C$
Hence, the correct answer is option A.
Note: The temperature coefficient of resistance is commonly described as the change in electrical resistance of a material per degree change in temperature. So, when we look at the electrical resistance of conductors like gold, aluminium, silver, and copper, it all depends on the process of electron collision within the substance. The process of electron collision accelerates as temperature rises. As a result, as the conductor's temperature rises, so does its resistance.
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