Question
The resistance of pure platinum increases linearly with temperature (over a small range). We can use this property to measure temperature of different bodies, relative to a scale.
If we scale the thermometer such that a resistance of 80 Ω denotes 0oC and 90 Ω denotes 100oC, what temperature will a resistance of 86 Ω denote?
If we scale the thermometer such that a resistance of 80 Ω denotes 0oC and 90 Ω denotes 100oC, what temperature will a resistance of 86 Ω denote?
Answer: Option C
:
C
A "linear” relation between two quantities, say, Q1and Q2, means the following
Q1=mQ2+c
where mand care constants. This follows from the fact that for a straight line on a plane, yand xare related as,
y=mx+c
m being the slope and c the y-intercept respectively.
In our case, this should be true between the temperatureandthe resistance at that temperature,
Rθ=αθ+β
for some constantsα& β.
Let's find out what these constants are from the information given. We can write
80Ω=α×0oC+β
⇒β=80Ω
90Ω=α×100oC+β
⇒α=90−80100=0.1Ω/oC
Using these values, the temperature(θ)can be written as
86=0.1×θ+80
⇒θ=60oC
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:
C
A "linear” relation between two quantities, say, Q1and Q2, means the following
Q1=mQ2+c
where mand care constants. This follows from the fact that for a straight line on a plane, yand xare related as,
y=mx+c
m being the slope and c the y-intercept respectively.
In our case, this should be true between the temperatureandthe resistance at that temperature,
Rθ=αθ+β
for some constantsα& β.
Let's find out what these constants are from the information given. We can write
80Ω=α×0oC+β
⇒β=80Ω
90Ω=α×100oC+β
⇒α=90−80100=0.1Ω/oC
Using these values, the temperature(θ)can be written as
86=0.1×θ+80
⇒θ=60oC
Was this answer helpful ?
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