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A faulty thermometer has LFP=10degree and UFP=110degree. What is actual temperature of an object if this thermometer reads 40degree celcius A faulty thermometer has LFP=10degree and UFP=110degree. What is actual temperature of an object if this thermometer reads 40degree celcius
Here faulty reading (x)=40Lower fixed point of faulty readind (L.F.P)=10Upper fixed point of faulty reading (U.F.P.)=110Fundamental interval (F.I.)=110- 10 = 100Correct reading(C)=?c/100*(x-L.F.P)/F.I.C/100=40/100 =0.4C=0.4*100=40 celsius scale
Dear student:usually the thermometer fixed points basically are:Lower point =0°cUpper point=100°cso in between there are 100 divisions.each one division corresponds to 1°cBut in the above mentioned case :Lower point=10°cupper point=110°cso total division=110-10=100thus, 1 degree= 1 division=100/100Now let us find the given temperature:measured temperature 40 °c=100x40/100=40°cHope it is clear.
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