V. MASS. 91 



Example 1. 



T=the reservoir filled with mercury weighs at 19 Cels.= 1091-6 

 K = the empty reservoir - = 329*6 



Q = T li. Consequently mercury 762'0'grammes, 

 Consequently contents of the reservoir at 19 Cels. 



T-R 762-0 



= 13-59 (l-< 0-0018) = 13T5 ==56>236 Cub " Centim " 



If now 60 grammes (= P) gunpowder is placed in the reservoir, and the 

 remaining space filled with mercury, we obtain 

 611-6 grammes ( = T t ) 



Reservoir empty =329 -6 

 60 grammes powder = 60-0 



= 389-6 deducted, 



remains Q = 222.0 grammes weight of the mercury filling the intervening 

 space, occupying at 19 Cels. 



1 = 2^=16-444 CUD. centiin. 

 13'55 



Consequently volume of the 60 grammes powder= 56-236 16*444 



V= 39*792 cuh. centims. 

 p 



S y = specific weight of the 60 grammes gunpowder 



1.J07 



39-792 

 Example 2. 



If a powder prism weighs 42-0 grammes at 190 C. and weight of reservoir, 

 powder prism and mercury = 808-4 grammes. 



Reservoir empty 329-6 

 42 grammes powder 42-0 



371-6 

 consequently of 808-4 



371-6 deducted, 



= 436-8 grammes weight of mercury, 



or - - = 32-236 cuh. centim. occupied hy these 436'8 grammes. 

 13*55 



Consequently volume of the 42 grammes weighing powder prism =56-236 

 32-236 = 24-0 cub. centim. 



Thus, specific weight of the powder prism= _I_ = 1-75 



24-0 



General Formula. 

 S _P_ F P _ __ Q 



V I-I 1 T-R or -13-59(1-^0-0018), 



13-59(1- t 0-0018) 



g = P [13-59 (l-< 0-0018)] Since Q 1 = T' - R . - P 

 T-R-Qi . ' 



g _ P 13-59 (l-t 0-0018) 

 T-T' + P 



