Irrigation water having the concentration of Na++, Ca++ and Mg++ as 20, 3 and 1 Milli equivalent per litre respectively will be classified as
A. Low sodium water
B. Medium sodium water
C. High sodium water
D. Very high sodium water
Answer: Option D
Solution (By Examveda Team)
Irrigation water is classified based on its sodium concentration relative to the concentrations of calcium and magnesium ions. The classification is determined by the sodium absorption ratio (SAR), which is calculated using the following formula:SAR = (Na++ / √(Ca++ + Mg++)) * 100
Given the concentrations provided:
Na++ = 20 meq/L
Ca++ = 3 meq/L
Mg++ = 1 meq/L
Plugging these values into the SAR formula:
SAR = (20 / √(3 + 1)) * 100
SAR ≈ (20 / √4) * 100
SAR ≈ (20 / 2) * 100
SAR ≈ 10 * 100
SAR ≈ 1000
According to the classification:
SAR < 10: Low sodium water
10 ≤ SAR < 18: Medium sodium water
18 ≤ SAR < 26: High sodium water
SAR ≥ 26: Very high sodium water
Since the calculated SAR is approximately 100, the irrigation water falls into the category of very high sodium water.
This Question Belongs to Civil Engineering >> Irrigation Engineering
Wrong answer
correct answer is: Medium
Na / ((Sqrt (Mg+Ca)/2))
= 20 / Sqrt((3+1)/2))
= 20/1.414 = 14.14
SAR < 10: Low sodium water
10 ≤ SAR < 18: Medium sodium water
18 ≤ SAR < 26: High sodium water
SAR ≥ 26: Very high sodium water
B. Medium sodium water
SAR=14.4(Medium sodium water)
How
SAR= Na+/Root of (Mg+2)+(Ca+2)/2 so
SAR=20/Root of 1+3/2
=20/Root2
=10×Root 2
=14.14 Which is in Medium Sodium Water
Correct answer is 14.14
So acc. to SAR value 10-18 value comes under medium sodium water
1_10. Tk low sodium water
10-18tk medium sodium water
18-26high sodium water
26< very high
SAR=Na++upon((ca++)+(Na++)/2) whole to the power half=14.14 so anwser b is correct
0
20/√((3+1)/2)
=14
The given ans is wrong plz check it will be 2304
Sol. 7% per 300m
For 1200m -28%will be increase
So 1.28*1800=2304m
1-10 = low sodium water
10-18 = medium sodium water
18-26 = high sodium water
>26 = very high sodium water