﻿S66 
  Mr. 
  B.W. 
  Clack 
  on 
  the 
  

  

  enters 
  the 
  bulb 
  due 
  to 
  a 
  small 
  diminution 
  (dm) 
  in 
  the 
  mas 
  

   (m) 
  of 
  salt 
  in 
  the 
  bulb 
  

  

  =8x(l-5N>*m, 
  

  

  On 
  integrating, 
  the 
  mass 
  of 
  water 
  which 
  enters 
  the 
  bulb 
  

   due 
  to 
  a 
  given 
  diminution 
  

  

  = 
  | 
  Sx(l-6N) 
  

  

  </w. 
  

  

  Strictly 
  both 
  8 
  and 
  N 
  are 
  functions 
  of 
  m, 
  but 
  the 
  change 
  

   in 
  m 
  is 
  so 
  small 
  that 
  they 
  may 
  both 
  be 
  taken 
  as 
  constants; 
  

   thus 
  the 
  integral 
  is 
  equal 
  to 
  

  

  8 
  x 
  (1 
  — 
  /A 
  }(m2— 
  wii). 
  

  

  Now 
  in 
  one 
  second 
  ("if— 
  n%i)=c. 
  

  

  Hence 
  the 
  mas- 
  of 
  water 
  entering 
  the 
  bulb 
  per 
  second 
  

  

  = 
  8x(l-oN>. 
  

  

  Now 
  the 
  decrease, 
  i, 
  in 
  weigh! 
  per 
  second 
  of 
  the 
  bulb 
  

   = 
  the 
  weigh! 
  of 
  -alt 
  leaving 
  per 
  second 
  minus 
  the 
  weight 
  

   of 
  water^entering 
  per 
  second 
  ; 
  or 
  in 
  symbols 
  

  

  i. 
  e. 
  <■ 
  = 
  

  

  1-8(1-&N) 
  

  

  Again, 
  the 
  volume 
  of 
  water 
  entering 
  per 
  second, 
  c8, 
  is 
  equal 
  

   to 
  the 
  velocity 
  of 
  the 
  liquid 
  at 
  the 
  bottom 
  of 
  the 
  tube 
  

  

  - 
  r 
  ° 
  re 
  • 
  n 
  

  

  -f-eX 
  ( 
  fromi 
  

  

  ... 
  r 
  ^ 
  = 
  ^hi-bS). 
  

  

  2c 
  2 
  K 
  y 
  

  

  Again, 
  the 
  third 
  term 
  in 
  the 
  bracket 
  in 
  equation 
  iii. 
  is 
  

   found 
  to 
  be 
  negligibly 
  small 
  ; 
  hence 
  this 
  equation 
  is 
  written 
  

   for 
  practical 
  use 
  

  

  h= 
  — 
  *=* 
  \— 
  ( 
  . 
  . 
  (iv.) 
  

  

  