Q.1
ICAR AIEEA PG 2024 - Water Science and Technology (Q.53)
MCQ (Single Correct) +4 −1
A sample of moist soil with a wet mass of 1.0 kg and a volume of 0.64 liters (\(6.4 \times 10^{-4}\text{ m}^3\)) was dried in the oven and found to have a dry mass of 0.8 kg. Assuming the typical value of particle density (\(2650\text{ kg/m}^3\)) for a mineral soil, what will be the volumetric moisture content?
1.0 kg गीले द्रव्यमान और 0.64 लीटर (\(6.4 \times 10^{-4}\text{ m}^3\)) आयतन वाले नम मिट्टी के एक नमूने को ओवन में सुखाया गया और उसका सूखा द्रव्यमान 0.8 kg पाया गया। खनिज मिट्टी के लिए कण घनत्व का विशिष्ट मान (\(2650\text{ kg/m}^3\)) मानते हुए, आयतनी नमी की मात्रा (volumetric moisture content) क्या होगी?

Calculation of volumetric moisture content (

\(\theta\)): 1. Mass of water in the sample \(M_w = M_{\text{wet}} - M_{\text{dry}} = 1.0\text{ kg} - 0.8\text{ kg} = 0.2\text{ kg}\). 2. Taking density of water \(\rho_w = 1.0\text{ kg/L} = 1000\text{ kg/m}^3\): Volume of water \(V_w = \frac{M_w}{\rho_w} = \frac{0.2\text{ kg}}{1.0\text{ kg/L}} = 0.20\text{ liters}\). 3. Total soil volume \(V_t = 0.64\text{ liters}\). 4. Volumetric moisture content: \[\theta = \frac{V_w}{V_t} = \frac{0.20\text{ L}}{0.64\text{ L}} = \frac{20}{64} = 0.3125 = 31.25\%\]
Thus, the volumetric moisture content is 31.25%.

गणना:
1. जल का द्रव्यमान

\(M_w = 1.0\text{ kg} - 0.8\text{ kg} = 0.2\text{ kg}\) 2. जल का आयतन \(V_w = \frac{0.2\text{ kg}}{1.0\text{ kg/L}} = 0.20\) लीटर 3. कुल मृदा आयतन \(V_t = 0.64\) लीटर 4. आयतनी नमी प्रतिशत (\(\theta\)): \[\theta = \frac{V_w}{V_t} \times 100 = \frac{0.20}{0.64} \times 100 = 31.25\%\]
अतः सही उत्तर 31.25% है।
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