Falling Behind in Studies? Join Our Performance Improvement Batch. Enroll For Free
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Live 1-1 coding classes to unleash the creator in your Child
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.
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -