A thermometer is graduated in mm. It registers – 3mm when the bulb of thermometer is in pure melting ice and 22mm when the thermometer is in steam at a pressure of one atm. The temperature in °C when the thermometer registers 13mm is

To find the temperature in °C corresponding to a given reading on a custom scale, we can use the linear relationship between the custom scale readings and the Celsius scale.

Let the reading on the thermometer be $R$ in mm, and the corresponding temperature in Celsius be $T$ in °C.
We are given two reference points:
1. At the melting point of pure ice (0 °C), the thermometer registers -3 mm:
When $T=0$, $R=-3$ mm.
2. At the steam point (100 °C under a pressure of 1 atm), the thermometer registers 22 mm:
When $T=100$, $R=22$ mm.

The relationship between the Celsius scale and the thermometer scale can be written as:
$\frac{T-T_{\mathrm{ice}}}{T_{\mathrm{steam}}-T_{\mathrm{ice}}}=\frac{R-R_{\mathrm{ice}}}{R_{\mathrm{steam}}-R_{\mathrm{ice}}}$

Substituting the values:
$T_{\mathrm{ice}}=0$
$T_{\mathrm{steam}}=100$
$R_{\mathrm{ice}}=-3$
$R_{\mathrm{steam}}=22$

This gives:
$\frac{T-0}{100-0}=\frac{R-(-3)}{22-(-3)}$

Simplifying the equation:
$\frac{T}{100}=\frac{R+3}{25}$

Now, we want to find the temperature $T$ when the thermometer registers $R=13$ mm:
$\frac{T}{100}=\frac{13+3}{25}$

$\frac{T}{100}=\frac{16}{25}$

Solving for $T$:
$T=\frac{16\times 100}{25}$

Thus, the temperature in °C is $16\times 100/25$, which corresponds to Option 16x100/25.