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Question: The readings of the air thermometer at \({{0}^{o}}C\) and \({{100}^{o}}C\) are \(50cm\) and \(75cm\)...

The readings of the air thermometer at 0oC{{0}^{o}}C and 100oC{{100}^{o}}C are 50cm50cm and 75cm75cm of mercury column respectively. The temperature at which its reading is 85cm85cm of mercury column is
A.105oCA{{.105}^{o}}C.
B.110oCB{{.110}^{o}}C.
C.120oCC{{.120}^{o}}C
D.140oCD{{.140}^{o}}C

Explanation

Solution

We can apply the concept of thermal expansion to solve this particular problem. Thermal expansion is defined as the increase in length of the material due to increase in temperature. Mercury in thermometers is very sensitive to any change in temperature and used to measure temperature.
Formula used:
We are using relation, T=llol100lo×100oCT=\dfrac{l-{{l}_{o}}}{{{l}_{100}}-{{l}_{o}}}\times {{100}^{o}}Cto find the required temperature at the given length.

Complete answer:
From the problem given above we will write the different parameters first.
T=T= temperature at the given reading of length of mercury=?
l=l= given length of the mercury=85cm=85cm
lo={{l}_{o}}=length of mercury at 0oC=50cm{{0}^{o}}C=50cm
l100={{l}_{100}}=length of mercury at 100oC=75cm{{100}^{o}}C=75cm
Now using the formula, T=llol100lo×100oCT=\dfrac{l-{{l}_{o}}}{{{l}_{100}}-{{l}_{o}}}\times {{100}^{o}}C, we get
T=85507550×100oCT=\dfrac{85-50}{75-50}\times {{100}^{o}}C
T=3525×100T=\dfrac{35}{25}\times 100
Calculating further, we get,
T=35×4T=35\times 4
T=140oCT={{140}^{o}}C

So, the correct answer is “Option D”.

Additional Information:
Thermometer is the device which is used to measure temperature. Mercury thermometer is one of its kinds, which is generally used to measure the body temperature. Mercury is the metal which is used in this type of thermometer as the working fluid because of its property to expand sensitively on heating. Therefore application of thermal expansion is used in mercury thermometers to measure the temperature of the body. Nowadays, digital thermometer is preferred over traditional mercury thermometer because of its advanced technology.

Note:
We can also solve this problem using the relation l=lo(1+αΔt)l={{l}_{o}}(1+\alpha \Delta t). From this relation we can find α\alpha and then apply the same formula for the second temperature difference to find the length of mercury at the given temperature. Where l=l=new length, lo={{l}_{o}}=original length, α=\alpha =thermal coefficient and Δt=\Delta t=temperature difference. But this relation is a long method for this particular problem. Do all the calculations carefully.