The electrical resistance of pure platinum increases linearly with increasingtemperature over a small range of temperature . This property is used in a Platinum resistance thermometer. The relation between R_(theta) (Resistance at theta K) and R_(0) (Resistance at theta_(0)K) is given by R_(theta)=R_(0)[1+a(theta-theta_(0))], where alpha= temperature coefficient of resistance. Now, if a Platinum resistance thermometer reads 0^(@)C when its resistance is 80Omega and 100^(@)C when its resistance is 90 Omega find the temperature at which its resistance is 86Omega.

The electrical resistance of pure platinum increases linearly with inc

1.	The electrical resistance of pure platinum increases linearly with increasingtemperature over a small range of temperature . This property is used in a Platinum resistance thermometer. The relation between R_(theta) (Resistance at theta K) and R_(0) (Resistance at theta_(0)K) is given by R_(theta)=R_(0)[1+a(theta-theta_(0))], where alpha= temperature coefficient of resistance. Now, if a Platinum resistance thermometer reads 0^(@)C when its resistance is 80Omega and 100^(@)C when its resistance is 90 Omega find the temperature at which its resistance is 86Omega.
Answer» Solution :Using the GIVEN relationship we have `(90-80)OMEGA=alpha(80 Omega)(100)`………..(i) `(86-80)Omega=alpha(80 Omega)(theta)`………..(ii) where `theta` is the desired TEMPERATURE. TAKING the RATIO of (i) and (ii) `10/6=100/(theta)impliestheta=60K`

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