Answer
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Hint: In the above question the faulty thermometer has its lower fixed point as 5 and the higher fixed point is as 95. This thermometer is divided into 100 equal divisions. In case of a Celsius scale, the scale is divided into 100 equal parts with its lower fixed point as 0 degree Celsius and its upper fixed point as 100 degree Celsius. Hence we can use the relation between two temperature scales to determine the corresponding temperature on the faulty thermometer, when the temperature on the Celsius scale is 40.
Complete answer:
Both the temperature scales are divided into 100 equal divisions. Hence the relation between two above temperature scales is given by,
$\dfrac{C-LFP}{UFP-LFP}=\dfrac{F-LFP}{UFP-LFP}$, where C is the measurement on the Celsius scale, F is the corresponding measurement on the faulty thermometer scale, UFP stands for the upper fixed points and LFP stands for the lower fixed points on both the scales.
Substituting the measurement taken on the Celsius scale and the UFP and LFP of both the thermometers in the above equation we get,
$\begin{align}
& \dfrac{C-LFP}{UFP-LFP}=\dfrac{F-LFP}{UFP-LFP} \\
& \dfrac{40-0}{100-0}=\dfrac{F-5}{95-5} \\
& \dfrac{40\times 90}{100}=F-5 \\
& 36+5=F \\
& F={{41}^{\circ }}C \\
\end{align}$
Hence the corresponding reading on the faulty thermometer will be 41 degree Celsius.
Note:
The above question can also be solved in a much faster way. Both the scales are divided into 100 equal divisions each. Hence for every 1 degree rise in the Celsius scale there will be 0.9 degree rise on the faulty thermometer. Hence for a temperature of 40 degree Celsius on the Celsius scale, the rise in temperature on the faulty thermometer will be $40\times 0.9=36$. But the faulty thermometer is calibrated such that its initial point is 5. Hence the temperature value on the faulty thermometer will be 36+5=41.
Complete answer:
Both the temperature scales are divided into 100 equal divisions. Hence the relation between two above temperature scales is given by,
$\dfrac{C-LFP}{UFP-LFP}=\dfrac{F-LFP}{UFP-LFP}$, where C is the measurement on the Celsius scale, F is the corresponding measurement on the faulty thermometer scale, UFP stands for the upper fixed points and LFP stands for the lower fixed points on both the scales.
Substituting the measurement taken on the Celsius scale and the UFP and LFP of both the thermometers in the above equation we get,
$\begin{align}
& \dfrac{C-LFP}{UFP-LFP}=\dfrac{F-LFP}{UFP-LFP} \\
& \dfrac{40-0}{100-0}=\dfrac{F-5}{95-5} \\
& \dfrac{40\times 90}{100}=F-5 \\
& 36+5=F \\
& F={{41}^{\circ }}C \\
\end{align}$
Hence the corresponding reading on the faulty thermometer will be 41 degree Celsius.
Note:
The above question can also be solved in a much faster way. Both the scales are divided into 100 equal divisions each. Hence for every 1 degree rise in the Celsius scale there will be 0.9 degree rise on the faulty thermometer. Hence for a temperature of 40 degree Celsius on the Celsius scale, the rise in temperature on the faulty thermometer will be $40\times 0.9=36$. But the faulty thermometer is calibrated such that its initial point is 5. Hence the temperature value on the faulty thermometer will be 36+5=41.
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