A thermometer has wrong calibration it recordes the melting point of ice `- 5^(@)C` If reads `55^(@)C` instead of `50^(@)C` Find the temperature of boiling point of water on the given scale

Answer

130

Explanation

Let
`theta_1` =Lower fixed point on faulty thermometer
`theta_2` = Reading on faulty thermometer
n= number of divisions between upper and lower fixed points
Now `(C)/(100)=(theta_2-theta_1)/(n)`
In the first case,
`theta_2=-10^@C, C=0^@C`
`therefore 0= (-10-theta_1)/(n)` or `theta_1=-10^@C`
In second case,
`theta_2=60^@C,C=50^@C`
`therefore (50)/(100)=(60-theta_1)/(n)` or `1/2=(60(-10))/(n)=(70)/(n)`
or n=140
As boiling point of water on Celsius scale is `100^@C`, so putting C = 100 in equation (i), we get
`(100)/(100)=(theta_2-(-10))/(140)` or `theta_2=130^@C`
`therefore` Boiling point of water on faulty thermometer = `130^@`C