IIT JEE , AIEEE Forum - Physics , Thermal Physics

sucess_seeker

A THERMOMETER HAS WRONG CALIBRATION,. IT READS THE MELTING POINT OF ICE -10 DEGREE. IT READS 60 DEGREE IN PLACE OF 50 DEGREE CALCULATE THE TEMPERATURE OF BOILING POINT OF WATER ON THIS SCALE. I KNOW THAT IF THERE IS SUCH A QUESTION THE FORMULA IS C/100=

2 -

1/N WHERE

1 IS THE LOWER FIXED POINT ON FAULTY THERMOMETER AND

2 READING ON THE FAULTY THERMOMETER, N= NO OF DIVISIONS BETWEEN UPPER AND LOWER POINTS
MORE I DON'T KNOW HOW TO WORK OUT PLEASE HELP.

edison

For 0 C, thermometer reads = -10 deg C

so here zero error = -10 deg C

for 50 C, it reads = 60 C

so for a difference of (50-0) C it shows a difference of [60- (-10)] = 70 C

so a difference of 50deg C is read as a difference of 70 deg C on thermometer scale

so 1 degree change should be read by thermometer as = 70/50 deg C

or calibration factor for 1 deg change = 1.4 deg C

so for boiling point of water which is at 100deg C it should read = 100 x 1.4 + zero error = 140 degree C + (-10) = 130 degree C

Asmita

Sir i think ur ans. is incorrect bcoz acc. 2 ur method

the reading shown by faulty thermometer when actual temp. is 0⁰C=0*1.4=0^oC

which is not so.

I think as u already said a actual diff. of 50⁰C=diff. of 70⁰C in the faulty thermometer & 100⁰C=50⁰C+50⁰C

so the ans. will b=60⁰C+70⁰C=130⁰C

Correct me if i m wrong .

001sri

asmita u r correct

001sri

other way -10+140=130
becoz 140 is diff btwn meltig n boiling points corresponding to 100.

edison

Incorporation of zero error will make the answer correct as here

zero error = -10 deg C

and adding this to the answer will give correct reading shown by the thermometer.

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