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GATE Storage Engineering, Material Handling & Packaging: practice questions

GATE — Agricultural Engineering (AG) · India · IITs and IISc on rotation (2026: IIT Guwahati)

Original practice questions written to the published syllabus · not past papers
10worked here, free
60written to this exam in this subject
11syllabus units

What this subject covers

Storing grain and produce safely, the loads it puts on bins and silos, and the conveyors and packages that move and protect it.

In the exams: GATE AG Section 6 (bin/silo pressure in 6 of 8 papers 2019–26); ICAR AICE subject 41 Units 6–7; NCEES PE (ANSI/ASAE S433 supplied); ASRB NET ASPE

What GATE asks from it

GATE — Agricultural Engineering (AG)

Section 6 (Storage Systems; Size Reduction and Material Handling; packaging material and machines). Bin and silo pressure has appeared in most recent papers: Janssen vertical and lateral pressure at a given depth, the limiting pressure, Rankine's k from the angle of internal friction, and the shallow-or-deep test. Also silo capacity from bulk density, bucket-elevator head-pulley speed, screw and belt conveyor capacity, and permeation or WVTR through a package film, often as numerical-answer questions with two or three linked steps.

Where the 60 questions fall

How our questions for GATE spread across the units of Storage Engineering, Material Handling & Packaging. Difficulty: 21 easy, 20 medium, 19 hard.

Janssen's and Airy's theories of bin pressure17
Barrier properties of packages7
Design of conveyors and elevators7
Aeration of stored grain6
Packaging materials: paper, plastics and laminates6
Causes of storage losses3
Bins and silos3
Fumigation3
Controlled and modified atmosphere storage3
Packaging machines3
Warehouse management2

10 practice questions, worked

Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.

1 · Janssen's and Airy's theories of bin pressure · hard

What is the maximum (limiting) lateral pressure that wheat of bulk density 780 kg/m³ can exert on the wall of a very deep 8 m diameter bin with a wall friction coefficient of 0.45?

  1. 34.0 kPa
  2. 3.47 kPa
  3. 68.0 kPa
  4. 6.89 kPa
Answer and working

A. 34.0 kPa

L_max = γR/μ′ with γ = 780 × 9.81 = 7.652 kN/m³ and R = D/4 = 2.0 m: 7.652 × 2.0/0.45 = 34.0 kPa. It does not depend on k. Using kg/m³ instead of weight density gives 3.47; using the radius D/2 instead of the hydraulic radius gives 68.0; multiplying by μ′ (6.89) is an algebra slip.

Further reading: Sahay and Singh, Unit Operations of Agricultural Processing

2 · Janssen's and Airy's theories of bin pressure · easy

For grain with φ = 30°, above what grain depth does a 4 m wide bin become a deep bin by the plane-of-rupture criterion?

  1. 2.31 m
  2. 4.00 m
  3. 3.46 m
  4. 6.93 m
Answer and working

D. 6.93 m

Critical depth = D·tan(45° + φ/2) = 4 × tan60° = 6.93 m. Using tan(45° − φ/2) gives 2.31 m; ignoring φ (tan45°) gives 4.00 m; using the radius instead of the width gives 3.46 m.

Further reading: Sahay and Singh, Unit Operations of Agricultural Processing

3 · Janssen's and Airy's theories of bin pressure · easy

By Rankine's theory, the resultant lateral force on the wall of a shallow bin with a level grain surface acts at

  1. Half the grain depth above the base
  2. Two thirds of the depth above the base
  3. One third of the grain depth above the base
  4. The grain surface
Answer and working

C. One third of the grain depth above the base

The pressure increases linearly from zero at the surface to γHk at the base, a triangular diagram whose centroid is H/3 above the base. H/2 would apply to a uniform pressure.

Further reading: Sahay and Singh, Unit Operations of Agricultural Processing

4 · Janssen's and Airy's theories of bin pressure · hard

In a 6 m diameter bin (μ′ = 0.4, k = 0.4), at roughly what depth does the Janssen pressure reach 95% of its limiting value?

  1. 9.4 m
  2. 56.2 m
  3. 28.1 m
  4. 11.2 m
Answer and working

C. 28.1 m

1 − e^−x = 0.95 when x = 3 (e^−3 = 0.050). So μ′kh/R = 3 → h = 3R/(μ′k) = 3 × 1.5/(0.4 × 0.4) = 28.1 m. At h = R/(μ′k) = 9.4 m the pressure is only 63% of the limit; using the bin radius (3 m) instead of the hydraulic radius R = D/4 = 1.5 m gives 56.2 m. 11.2 m omits k (3 × 1.5/0.4).

Further reading: Sahay and Singh, Unit Operations of Agricultural Processing

5 · Fumigation · medium

A 3 g tablet contains 56% aluminium phosphide (AlP, 57.95 g/mol). How much phosphine (PH3, 34.00 g/mol) can it release?

  1. 1.68 g
  2. 1.76 g
  3. 0.99 g
  4. 2.86 g
Answer and working

C. 0.99 g

AlP = 3 × 0.56 = 1.68 g = 0.0290 mol; 1 mol AlP gives 1 mol PH3, so PH3 = 0.0290 × 34.00 = 0.99 g (about 1 g). 1.68 g is the mass of AlP, not PH3; 1.76 g ignores the 56% content; inverting the molar mass ratio gives 2.86 g.

Further reading: FAO Manual of fumigation for insect control

6 · Fumigation · medium

A fumigation label concentration is given as 1 g/m³ of phosphine. What is this in ppm (v/v) at 25 °C?

  1. 659 ppm
  2. 719 ppm
  3. 1,000 ppm
  4. 1,391 ppm
Answer and working

B. 719 ppm

ppm = (mg/m³) × Vm/M = 1000 × 24.45/34.0 = 719 ppm. Using 22.41 L/mol gives 659; treating mg/m³ as ppm gives 1,000; inverting gives 1,391.

Further reading: FAO Manual of fumigation for insect control

7 · Controlled and modified atmosphere storage · hard

A 1 kg pack of produce respires 10 mL O2 kg⁻¹ h⁻¹ at storage temperature. The film has an oxygen transmission rate of 5000 cm³ m⁻² day⁻¹ atm⁻¹. What film area gives a steady-state internal O2 of 5% (outside 21%, total pressure 1 atm)?

  1. 0.048 m²
  2. 0.0125 m²
  3. 0.229 m²
  4. 0.300 m²
Answer and working

D. 0.300 m²

O2 consumed = 10 × 1 × 24 = 240 mL/day. At steady state this equals OTR × A × ΔpO2 = 5000 × A × (0.21 − 0.05) = 800A. A = 240/800 = 0.300 m². Ignoring the partial-pressure difference gives 0.048; using the hourly rate with a daily OTR gives 0.0125; using 0.21 as the driving force (0.229) assumes zero O2 inside.

Further reading: Kader (ed.), Postharvest Technology of Horticultural Crops

8 · Packaging materials: paper, plastics and laminates · medium

Arrange corrugated board flutes from tallest to shortest.

  1. E > B > C > A
  2. B > A > C > E
  3. A > C > B > E
  4. C > A > E > B
Answer and working

C. A > C > B > E

A flute is the tallest, then C, then B, then E (fine flute). Letters were assigned in the order of introduction, not size, which is why C sits between A and B.

Further reading: Robertson, Food Packaging: Principles and Practice

9 · Design of conveyors and elevators · hard

A bucket elevator has buckets of 2 L capacity, 80% filled, spaced 0.3 m apart on a belt running at 2.5 m/s, handling grain of bulk density 750 kg/m³. Capacity?

  1. 45.0 t/h
  2. 10.0 t/h
  3. 36.0 t/h
  4. 3.24 t/h
Answer and working

C. 36.0 t/h

Buckets per second = v/spacing = 2.5/0.3 = 8.333. Mass per bucket = 0.002 m³ × 0.8 × 750 = 1.20 kg. Mass flow = 1.20 × 8.333 = 10.00 kg/s = 36.0 t/h. Ignoring the fill fraction gives 45.0; reporting kg/s as t/h gives 10.0; multiplying by spacing instead of dividing gives 3.24.

Further reading: Sahay and Singh, Unit Operations of Agricultural Processing

10 · Design of conveyors and elevators · hard

A horizontal screw conveyor has a 200 mm flight, a 50 mm shaft, 200 mm pitch and runs at 120 rpm with a loading efficiency of 0.4. It conveys wheat of bulk density 750 kg/m³. Capacity?

  1. 12.72 t/h
  2. 13.57 t/h
  3. 31.81 t/h
  4. 763.4 t/h
Answer and working

A. 12.72 t/h

Swept volume per revolution = (π/4)(D² − d²)p = (π/4)(0.2² − 0.05²)(0.2) = 0.005890 m³. × 120 rpm × 0.4 = 0.28274 m³/min. Mass = 0.28274 × 750 × 60 = 12,723 kg/h = 12.72 t/h. Ignoring the shaft gives 13.57; ignoring loading efficiency gives 31.81; treating rpm as rev/s gives 763.4.

Further reading: Sahay and Singh, Unit Operations of Agricultural Processing

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Read the subject

Storage Engineering, Material Handling & PackagingThe full study manual, free GATEFormat, sections and how it has changed

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Storage Engineering, Material Handling & Packaging for other exams

ICAR AICE-JRF/SRF · IndiaICAR AIEEA-PG · IndiaASRB NET · India

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