as the XXXXX content on July X is not given XXX us assume was dry that XXXX otherwise question XXXXXX be XXXXXX XXXXXXX the degree of XXXXXXXXXX
or XXXXX contains or XXXX XXXXX
= XXXXXX /m3
XXX volume of soil upto the XXXX zone XX equal to XX XXX x X.X = XX mX
mass XX XXX soil XXXXXX =ms
XXX available water on XXXX 15 XXX XX used XX XXXXX =
XXXXX XXXXXXX XXXX XX = XX.X% XXXXX XXXXXX use XXXXX XXXXXXX XXXXX XX %
MX = 0.079 x52500 = 4147.X kg
XXXXXX of water can be uptake = = X.15 mX
let us assume the XXXXX XXXXXXX on XXXX 1 was the XXXXXXX XXXXX, i.e. 22%
XXXX density XX soil XX XXXXX to rd= 1050 kg/mX
the volume of XXX XXXX is equal to 10 x 10x0.5 =XXX3
XX weight XX XXX XXXX ( XXXX solid + XXXXX XX + XX% w.c) = 1050x50=52500 kg
MS + XW = 52500
XXXXXX XX the soil now XX 15 July XXX XXXX XXXXX XXXXXXX 29.X -22 is equal XX X.9 % XS /XW = X.XXX MX XX XXXXX to 0.07 9x4303 X.79 = 3399.59
XXXXX of XXXX XXXXX XX
X]
same XXXXXXXX in XXXX XXXX as well
XX assuming the soil XXX XXX on 1st XXXX Ms = 50500 kg
XXXXXX XX water @ XX =
on 15 July XXXXX content XX equal XX XX.X %
XXXXXX of the XXXXX is equal XX XX /MS = X. 299
MW = 0.299 XXXXX = XXXXX.5 XX
XXXXXX XX XXXXX XXXXXXXX 24150- S XXXXX.5 = XXXX.5 XX
XXXXXX = XXXX.5 / XXXX = 8.45 m3
XX XXXX the amount of rainfall XXXXX, XXX XXXXXX XXXXXX be divided by XXXX of the field XX equal XX
assuming XXXXX content XX the XXXX XXXXXXXX XXXXX XX July 1
XX= XXXXX.XX kg
weight of XXXXX @ f c = XXXX X.29x 0.46= 1979 5.08 XX
weight XX XXXXX @ XXXX July = XXX XXX 2.2 XXX X.XXX= XXXXX . 8 kg
XXXXXX of water XXXXXX= 19795.XXXX - XXX XX.X= 6928.28 kg
volume is XXXXX XX= XXXX X.XX/ 1000= X.93m3
in terms of XXXXXXXX XXXXX= 6.XX/ 100= 0.0693M= X.XX XX XX