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प्रश्न
The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law:
R = Ro [1 + α (T – To)]
The resistance is 101.6 Ω at the triple-point of water 273.16 K, and 165.5 Ω at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 Ω?
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उत्तर १
Here, R0 = 101.6 Ω; T0 = 273.16 K Case (i) R1= 165.5 Ω; T1 = 600.5 K, Case (ii) R2 = 123.4 , T2 = ?
Using the relation R = R0[1 + α (T – T0)]
Case (i) 165.5 = 101.6 [1 + α (600.5 – 273.16)]
`alpha= (165.5 - 101.6)/(101.6xx(600.5-273.16)) = 63.9/(101.6xx327xx37)`
Case II `123.4 = 101.6 [1 + alpha(T_2 - 273.16)]`
or `123.4 = 101.6 [1+ 63.9/(101.6xx327.34)(T_2-273.16)]`
`= 101.6 + 63.9/327.37 (T_2 - 273.16)`
or `T_2 = ((123.4-101.6)xx327.34)/63.9+ 273.16 = 111.67 + 273.16`
= 384.83 K
उत्तर २
It is given that:
R = R0 [1 + α (T – T0)] … (i)
Where,
R0 and T0 are the initial resistance and temperature respectively
R and T are the final resistance and temperature respectively
α is a constant
At the triple point of water, T0 = 273.15 K
Resistance of lead, R0 = 101.6 Ω
At normal melting point of lead, T = 600.5 K
Resistance of lead, R = 165.5 Ω
Substituting these values in equation (i), we get:
`R = R_0[1+alpha(T - T_0)]`
`165.5=101.6 [1+alpha(600.5 - 273.15)]`
`1.629 = 1 + alpha(327.35)`
`:.alpha = 0.629/327.35 = 1.92 xx 10^(-3) K^(-1)`
For resistance, `R_1= 123.4 Omega`
`R_1 = R_0[1+alpha(T-T_0)]`
Where T is the temperrature when the resistance of lead is `123.4 Omega`
`123.4 = 101.6[]1+1.92 xx 10^(-3)(T-273.15)]`
`1.214 = 1 + 1.92 xx 10^(-3)(T - 273.15)`
`0.214/(1.92xx10^(-3)) = T - 273.15`
:. T = 384.61 K
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