The electrical resistance in ohms of a certai | vedclass

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(C) The given relation is $R = R_{0} [1 + \alpha (T - T_{0})]$.At the triple point of water,$T_{0} = 273.16 \; K$ and $R_{0} = 101.6 \; \Omega$.At the

Vedclass · Accepted Answer

$384.61$

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(C) The given relation is $R = R_{0} [1 + \alpha (T - T_{0})]$.At the triple point of water,$T_{0} = 273.16 \; K$ and $R_{0} = 101.6 \; \Omega$.At the melting point of lead,$T = 600.5 \; K$ and $R = 165.5 \; \Omega$.Substituting these values into the equation:$165.5 = 101.6 [1 + \alpha (600.5 - 273.16)]$$1.629 = 1 + \alpha (327.34)$$\alpha = \frac{0.629}{327.34} \approx 1.9215 	imes 10^{-3} \; K^{-1}$.Now,for $R = 123.4 \; \Omega$,we find $T$:$123.4 = 101.6 [1 + 1.9215 	imes 10^{-3} (T - 273.16)]$$1.21457 = 1 + 1.9215 	imes 10^{-3} (T - 273.16)$$0.21457 = 1.9215 	imes 10^{-3} (T - 273.16)$$T - 273.16 = \frac{0.21457}{1.9215 	imes 10^{-3}} \approx 111.67$$T = 111.67 + 273.16 = 384.83 \; K$.Rounding to the nearest provided option,$T \approx 384.61 \; K$.

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