A mercury therometer is transferred from melting ice to a hot liquid. The mercury rises 0.95 of the distance between lower and upper fixed points. What is the temperature of the liquied in `.^@C` and in `.^@F` ?

Explanation

Here, `I_t = 9.5, I_(100) = 1.0` say, then `I_0 = 0.`
Let `t_c` be the temperature of liquid on celsius scale. Then
`I_t = I_0 (1 + alpha t_c) = I_0 + I_0 alpha.t_c or I_t -I_0= I_0 alpha t_c`
and `I_(100) = I_0 (1 + alpha xx 100) = I_0 + I_0 alpha_(100)`
or `I_(100) - I_0 = I_0 alpha xx 100`
`:. (I_t - I_0)/(I_(100) -I_-0) = (t_c)/(100)`
`or t_c = (I_t - I-0)/(I_(100) - I_(10))xx 100 = (0.95 - 0)/(1.0 - 0)xx 100 = 95^@C`
If `t_F` be the temperature of liquied of Farenheight scale then `(t_c)/(100) = (t_F - 32)/(180)`
`t_F = (9)/(5) t_c + 32 = (9)/(5)xx95 + 32`
`=171 + 32 = 203^@F`