A certain tiquid has a melting point of - 50 ° C and a boiling point of 150 ° C . A thermometer is designed with this liquid and its melting and boiling points are designated at 0 ° L and 100 ° L . The melting and boiling points of water on this scale are

Step-by-step solution

From principle of thermometry, T - T LFP T UFP - T LFP = a constant for every thermometric scale. Now, for any temperature L on a thermometer designed with given liquid and equivalent temperature C on centigrade scale, we have L - T LF T UFP - T LFP Liquid based scale  = C - T LFP T UFP - T LFP Centigrade Scale  ⇒    L - ( - 50 ) 150 - ( - 50 ) = C - 0 100 - 0 L + 50 150 + 50 = C 100 ⇒    L + 50 = 2 C Now at 0 ° L , centigrade scale reading will be 0 + 5