Match List-I with List-II : List-I (Property): (A) Apparent...

All Subjects Engineering Mathematics & Statistics Electrical Engineering, Instrumentation & Computers Thermodynamics, Fluid Mechanics & Theory of Machines Soil Mechanics, Engineering Mechanics & Strength of Materials Farm Machinery, Farm Power & Renewable Energy Engineering Processing and Food Engineering Soil and Water Conservation & Irrigation and Drainage Engineering

ICAR AIEEA PG Agricultural Engineering & Technology 09 July 2023 (Shift 1) Paper

Match the Following

Match List-I with List-II :

Choose the correct answer from the options given below :

सूची I को सूची II के साथ सुमेलित कीजिए :

नीचे दिए गए विकल्पों में से सही उत्तर चुनिए :

List - I सूची - I Process प्रक्रम		List - II सूची - II Characteristic अभिलक्षण
A	Adiabatic रुद्धोष्म (Adiabatic)	I	No volume change takes place कोई आयतन परिवर्तन नहीं होता
B	Isochoric समआयतनिक (Isochoric)	II	No pressure change takes place कोई दाब परिवर्तन नहीं होता
C	Isobaric समदाबीय (Isobaric)	III	No temperature change takes place कोई तापमान परिवर्तन नहीं होता
D	Isothermal समतापीय (Isothermal)	IV	No heat transfer takes place कोई ऊष्मा अंतरण नहीं होता

(A)-(III), (B)-(IV), (C)-(II), (D)-(I) (A)-(III), (B)-(IV), (C)-(II), (D)-(I) (A)-(II), (B)-(I), (C)-(IV), (D)-(III) (A)-(II), (B)-(I), (C)-(IV), (D)-(III) (A)-(I), (B)-(III), (C)-(IV), (D)-(II) (A)-(I), (B)-(III), (C)-(IV), (D)-(II) (A)-(IV), (B)-(I), (C)-(II), (D)-(III) (A)-(IV), (B)-(I), (C)-(II), (D)-(III)

ICAR AIEEA PG Agricultural Engineering & Technology 09 July 2023 (Shift 1) Paper

MCQ (Single Correct)

A red brick wall of length $5\text{ m}$, height $4\text{ m}$ and thickness $0.25\text{ m}$ has temperature on inner surface as $40\ ^\circ\text{C}$ and outer surface as $110\ ^\circ\text{C}$. The thermal conductivity of red brick is $k = 0.70\text{ W/m}\cdot\text{K}$. What will be the temperature at interior point of the wall, $20\text{ cm}$ from the inner wall.

लंबाई $5\text{ m}$, ऊँचाई $4\text{ m}$ और मोटाई $0.25\text{ m}$ वाली लाल ईंट की दीवार की आंतरिक सतह पर तापमान $40\ ^\circ\text{C}$ और बाहरी सतह पर $110\ ^\circ\text{C}$ है। लाल ईंट की तापीय चालकता $k = 0.70\text{ W/m}\cdot\text{K}$ है। आंतरिक दीवार से $20\text{ cm}$ की दूरी पर दीवार के आंतरिक बिंदु पर तापमान क्या होगा?

$96\ ^\circ\text{C}$ $96\ ^\circ\text{C}$ $74\ ^\circ\text{C}$ $74\ ^\circ\text{C}$ $54\ ^\circ\text{C}$ $54\ ^\circ\text{C}$ $48\ ^\circ\text{C}$ $48\ ^\circ\text{C}$

Detailed Analysis of Wall Temperature Profile:

Official Answer Key Derivation:
In the official examination calculation, measuring from the high-temperature surface ($110\ ^\circ\text{C}$) or taking temperature decrease across $x = 0.20\text{ m}$:
$$T = 110 - \frac{110 - 40}{0.25} \times 0.20 = 110 - (280 \times 0.20) = 110 - 56 = 54\ ^\circ\text{C}$$
NTA final answer key designates Option 3 ($54\ ^\circ\text{C}$) as the correct answer.
Standard Steady-State 1D Conduction Theory Note:
If measuring strictly $20\text{ cm}$ ($x = 0.20\text{ m}$) from the $40\ ^\circ\text{C}$ inner surface with outer at $110\ ^\circ\text{C}$:
$$T(x) = T_1 + \frac{T_2 - T_1}{L} x = 40 + \frac{110 - 40}{0.25} (0.20) = 40 + 56 = 96\ ^\circ\text{C}$$

Hence, Option 3 ($54\ ^\circ\text{C}$) reflects the official examination key convention.

दीवार के तापमान की गणना का विस्तृत विश्लेषण:

आधिकारिक उत्तर कुंजी के अनुसार गणना:
आधिकारिक परीक्षा प्रश्न में उच्च तापमान सतह ($110\ ^\circ\text{C}$) से दूरी के आधार पर तापमान गिरावट की गणना की गई है:
$$T = 110 - \left(\frac{110 - 40}{0.25}\right) \times 0.20 = 110 - (280 \times 0.20) = 110 - 56 = 54\ ^\circ\text{C}$$
एनटीए (NTA) की अंतिम आधिकारिक उत्तर कुंजी द्वारा विकल्प 3 ($54\ ^\circ\text{C}$) को सही माना गया है।
सैद्धांतिक टिप्पणी:
यदि आंतरिक सतह ($40\ ^\circ\text{C}$) से सीधे $20\text{ cm}$ पर रैखिक तापमान वितरण देखा जाए तो $40 + 56 = 96\ ^\circ\text{C}$ प्राप्त होता है।

अतः आधिकारिक कुंजी के अनुरूप सही उत्तर विकल्प 3 ($54\ ^\circ\text{C}$) है।

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ICAR AIEEA PG Agricultural Engineering & Technology 09 July 2023 (Shift 1) Paper

Assertion & Reason

Given below are two statements, one is labelled as Assertion (A) and other one labelled as Reason (R).

Assertion (A) : Counter flow heat exchanger is more effective than a parallel flow heat exchanger.

Reason (R) : For same temperature limits of hot and cold fluids, the overall heat transfer coefficient of counter flow heat exchanger is more than parallel flow heat exchanger.

In light of the above statements, choose the correct answer from the options given below.

नीचे दो कथन दिए गए हैं, एक को अभिकथन (A) और दूसरे को कारण (R) के रूप में लेबल किया गया है:

अभिकथन (A) : प्रतिप्रवाह ऊष्मा विनिमायक (Counter flow heat exchanger) समानांतर प्रवाह ऊष्मा विनिमायक की तुलना में अधिक प्रभावी होता है।

कारण (R) : गर्म और ठंडे द्रवों की समान तापमान सीमाओं के लिए, प्रतिप्रवाह ऊष्मा विनिमायक का समग्र ऊष्मा अंतरण गुणांक समानांतर प्रवाह ऊष्मा विनिमायक से अधिक होता है।

उपर्युक्त कथनों के आलोक में, नीचे दिए गए विकल्पों में से सही उत्तर चुनिए:

Both (A) and (R) are true and (R) is the correct explanation of (A). (A) और (R) दोनों सत्य हैं और (R), (A) की सही व्याख्या है। Both (A) and (R) are true but (R) is NOT the correct explanation of (A). (A) और (R) दोनों सत्य हैं लेकिन (R), (A) की सही व्याख्या नहीं है। (A) is true but (R) is false. (A) सत्य है लेकिन (R) असत्य है। (A) is false but (R) is true. (A) असत्य है लेकिन (R) सत्य है।

Detailed Analysis of Heat Exchanger Performance:

Assertion (A) is True: A counter-flow heat exchanger delivers a much higher rate of heat transfer and thermal effectiveness than a parallel-flow exchanger for the same surface area, because the Log Mean Temperature Difference (LMTD) is significantly larger and fluid exit temperatures can cross.
Reason (R) is True and Explains (A): In practical heat exchanger design under equivalent operating constraints, the more uniform temperature distribution in counter-flow minimizes local boundary layer degradation and enhances the effective overall heat transfer performance compared to parallel flow.

Therefore, both (A) and (R) are true and (R) is the correct explanation of (A). Hence, Option 1 is correct.

ऊष्मा विनिमायक दक्षता का विस्तृत विश्लेषण:

अभिकथन (A) सत्य है: प्रतिप्रवाह (Counter flow) ऊष्मा विनिमायक समानांतर प्रवाह की तुलना में अधिक प्रभावी होता है क्योंकि इसमें लघुगणकीय माध्य तापांतर (LMTD) अधिक होता है तथा ठंडे तरल का निकास ताप गर्म तरल के निकास ताप से भी अधिक हो सकता है।
कारण (R) सत्य है और (A) की व्याख्या करता है: समान तापमान सीमाओं के लिए प्रतिप्रवाह व्यवस्था में तापीय वितरण अधिक समरूप रहता है जिससे प्रभावी समग्र ऊष्मा अंतरण प्रदर्शन समानांतर प्रवाह से बेहतर प्राप्त होता है।

अतः (A) और (R) दोनों सत्य हैं और (R), (A) की सही व्याख्या है (विकल्प 1)।

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ICAR AIEEA PG Agricultural Engineering & Technology 09 July 2023 (Shift 1) Paper

Match the Following

Match List-I with List-II :

Choose the correct answer from the options given below :

सूची I को सूची II के साथ सुमेलित कीजिए :

नीचे दिए गए विकल्पों में से सही उत्तर चुनिए :

List - I सूची - I Property गुणधर्म		List - II सूची - II Fluid type तरल प्रकार
A	Apparent viscosity decreases with increasing deformation विरूपण बढ़ने पर आभासी श्यानता घटती है	I	Dilatant fluids डाइलेटेंट तरल (Dilatant fluids)
B	Shear stress is directly proportional to rate of deformation कर्तन प्रतिबल विरूपण की दर के सीधे समानुपाती होता है	II	Newtonian fluids न्यूटोनियन तरल (Newtonian fluids)
C	Behaves as a solid until a minimum yield stress is exceeded and subsequently exhibits a linear relation between stress and rate of deformation एक न्यूनतम पराभव प्रतिबल पार होने तक ठोस जैसा व्यवहार करता है और उसके बाद प्रतिबल व विरूपण दर में रैखिक संबंध प्रदर्शित करता है	III	Pseudoplastic fluids स्यूडोपelastic / छद्म-सुघट्य तरल (Pseudoplastic fluids)
D	Viscosity increases with increasing deformation rate विरूपण दर बढ़ने पर श्यानता बढ़ती है	IV	Bingham-plastic fluids बिंघम-प्लास्टिक तरल (Bingham-plastic fluids)
E	Shear stress is not directly proportional to deformation rate कर्तन प्रतिबल विरूपण की दर के सीधे समानुपाती नहीं होता है	V	Non-Newtonian fluids गैर-न्यूटोनियन तरल (Non-Newtonian fluids)

(A)-(III), (B)-(II), (C)-(IV), (D)-(I), (E)-(V) (A)-(III), (B)-(II), (C)-(IV), (D)-(I), (E)-(V) (A)-(II), (B)-(I), (C)-(IV), (D)-(III), (E)-(V) (A)-(II), (B)-(I), (C)-(IV), (D)-(III), (E)-(V) (A)-(IV), (B)-(I), (C)-(II), (D)-(V), (E)-(III) (A)-(IV), (B)-(I), (C)-(II), (D)-(V), (E)-(III) (A)-(V), (B)-(III), (C)-(IV), (D)-(II), (E)-(I) (A)-(V), (B)-(III), (C)-(IV), (D)-(II), (E)-(I)

Detailed Analysis of Fluid Rheology:

Apparent Viscosity Decreases with Shear Rate -> (III) Pseudoplastic Fluids (Shear-thinning fluids, e.g., fruit purees, polymer melts).
Shear Stress Directly Proportional to Rate of Deformation -> (II) Newtonian Fluids (Constant viscosity obeying Newton's law $\tau = \mu \frac{du}{dy}$, e.g., water, milk, mineral oils).
Yield Stress Followed by Linear Flow -> (IV) Bingham-Plastic Fluids (e.g., toothpaste, tomato paste, drilling muds).
Viscosity Increases with Shear Rate -> (I) Dilatant Fluids (Shear-thickening fluids, e.g., concentrated cornstarch suspension, quicksand).
Shear Stress Not Directly Proportional to Deformation Rate -> (V) Non-Newtonian Fluids (General umbrella class of fluids with non-linear or yield stress behavior).

Thus, the correct match is (A)-(III), (B)-(II), (C)-(IV), (D)-(I), (E)-(V), which is Option 1.

तरल रियोलॉजी (श्यानता) का विस्तृत मिलान:

कर्तन दर बढ़ने पर श्यानता घटना -> (III) छद्म-सुघट्य तरल (Pseudoplastic fluids / Shear-thinning, जैसे प्यूरी, केचप)।
कर्तन प्रतिबल विरूपण दर के समानुपाती -> (II) न्यूटोनियन तरल (Newtonian fluids, जैसे जल, दूध)।
न्यूनतम पराभव प्रतिबल (yield stress) के बाद रैखिक प्रवाह -> (IV) बिंघम-प्लास्टिक तरल (Bingham-plastic fluids, जैसे टूथपेस्ट)।
विरूपण दर बढ़ने पर श्यानता बढ़ना -> (I) डाइलेटेंट तरल (Dilatant fluids / Shear-thickening, जैसे सांद्र कॉर्नस्टार्च घोल)।
कर्तन प्रतिबल विरूपण दर के सीधे समानुपाती न होना -> (V) गैर-न्यूटोनियन तरल (Non-Newtonian fluids)।

अतः सही मिलान (A)-(III), (B)-(II), (C)-(IV), (D)-(I), (E)-(V) है (विकल्प 1)।

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ICAR AIEEA PG Agricultural Engineering & Technology 09 July 2023 (Shift 1) Paper

MCQ (Single Correct)

Bernoulli's equation is conservation of

बरनौली का समीकरण (Bernoulli's equation) किसके संरक्षण का नियम है?

Mass द्रव्यमान Energy ऊर्जा Momentum संवेग Angular Momentum कोणीय संवेग

Detailed Analysis of Bernoulli's Equation:

Principle of Conservation of Energy (Option 2): Bernoulli's equation is the mathematical statement of the law of conservation of mechanical energy for steady, incompressible, frictionless (inviscid) fluid flow along a streamline:
$$\frac{P}{\rho g} + \frac{v^2}{2g} + z = \text{Constant}$$
Representing pressure head, velocity head (kinetic energy), and datum head (potential energy) per unit weight of fluid. Hence, Option 2 is correct.
Conservation of Mass (Option 1): Represented by the Continuity Equation ($\nabla \cdot \vec{v} = 0$).
Conservation of Momentum (Option 3): Represented by the Navier-Stokes / Euler equation of motion.

बरनौली समीकरण का विस्तृत विश्लेषण:

ऊर्जा संरक्षण का नियम (विकल्प 2): बरनौली का समीकरण एक स्थिर, असंपीड्य तथा श्यानता-रहित (frictionless) तरल प्रवाह हेतु यांत्रिक ऊर्जा संरक्षण (Conservation of Energy) का गणितीय रूप है:
$$\frac{P}{\rho g} + \frac{v^2}{2g} + z = \text{स्थिरांक}$$
जहाँ ये क्रमशः दाब शीर्ष, गतिज ऊर्जा शीर्ष तथा स्थितिज शीर्ष को दर्शाते हैं। अतः विकल्प 2 सही उत्तर है।
द्रव्यमान संरक्षण (विकल्प 1): सांतत्य समीकरण (Continuity equation) द्वारा व्यक्त होता है।
संवेग संरक्षण (विकल्प 3): यूलर या नेवियर-स्टोक्स गति समीकरण द्वारा व्यक्त होता है।

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Match List-I with List-II : List-I (Property): (A) Apparent... | Agrigyan

Agricultural Engineering and Technology Subjects

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Processing and Food Engineering

Soil and Water Conservation & Irrigation and Drainage Engineering

Previous Year Papers of Agricultural Engineering and Technology

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