Definition of Pressure

Pressure is the perpendicular force acting normally per unit area on a surface.

$P = \frac{F}{A}$

Its SI unit is the pascal (Pa), where 1 Pa = 1 N/m². Pressure increases when force increases or area decreases.

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Lesson Plan — ABC Model

Subject: Science (Physics)
Topic: Liquid Pressure
Grade: 10
Time: 45 minutes
Teaching Method: Activity-based, student-centred

A — Anticipatory Set (Introduction) — 5 minutes

Objectives:

• To define liquid pressure and state its unit.
• To explain the factors on which liquid pressure depends.
• To derive and apply the formula P = hρg.

Materials: Plastic bottle, water, nail, bucket, rubber tube, funnel, balloon, beaker, spring balance, stone, measuring cylinder.

Previous Knowledge Test:

1. What is pressure? What is its unit?
2. Why does a dam have a thicker base?
3. What happens when you press a balloon filled with water?

Motivation: The teacher pierces three holes at different heights in a plastic bottle filled with water. Water gushes out farthest from the lowest hole. Students are asked to explain why. This creates curiosity about liquid pressure.

B — Building the Concept (Development) — 35 minutes

Step 1: Definition and Formula (10 min)
The teacher explains that a liquid exerts pressure on the base and walls of its container due to its weight. The pressure at a depth h is:

$P = h\rho g$

where h = depth, ρ = density of liquid, g = acceleration due to gravity. Unit: pascal (Pa).

Step 2: Activity — Depth and Pressure (10 min)
Students use a plastic bottle with holes at 3 cm, 6 cm and 9 cm from the top. They observe the range of water jets. Conclusion: pressure increases with depth.

Step 3: Activity — Density and Pressure (8 min)
Equal volumes of water and kerosene are poured into two identical tubes connected to a manometer. The liquid with greater density shows greater pressure at the same depth. Conclusion: pressure increases with density.

Step 4: Activity — Pressure is Equal at the Same Depth (7 min)
A balloon filled with water is pricked at two points at the same level. Water oozes out with equal force in both directions. Conclusion: at a given depth, liquid pressure acts equally in all directions.

Step 5: Application (5 min)
Discussion on why dam walls are thicker at the bottom, why a diver feels pressure in the ears, and how a water tank is designed.

C — Conclusion and Evaluation — 5 minutes

Summary: Liquid pressure depends on depth, density of the liquid and acceleration due to gravity. It acts equally in all directions at the same depth.

Evaluation (Oral/Written):

1. Write the formula for liquid pressure and name its unit.
2. At what depth in water is the pressure 9800 Pa? (Take ρ = 1000 kg/m³, g = 9.8 m/s²)
3. Why is the wall of a dam made thicker at the bottom?

Homework: Solve numerical problems from the textbook on liquid pressure and draw a labelled diagram of a manometer.

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Strategies to Teach Archimedes' Principle

1. Activity Method (Demonstration):
A stone is weighed in air using a spring balance, then weighed again when fully immersed in water. The loss in weight is recorded. Water displaced is collected in a measuring cylinder and weighed. Students find that:

$\text{Loss in weight} = \text{Weight of displaced liquid}$

This leads to the statement of Archimedes' Principle.

2. Inquiry-Based Learning:
Students are asked to predict what happens to the weight of an object in water and then verify through experiment. They discover the principle themselves.

3. Use of Real-Life Examples:
Ships floating on water, submarines, hydrometers, and life jackets are discussed to connect the principle to daily life.

4. Mathematical Application:
Problems on upthrust, apparent weight and relative density are solved:

$\text{Upthrust} = W_{\text{air}} - W_{\text{water}}$

$\text{Relative density} = \frac{W_{\text{air}}}{W_{\text{air}} - W_{\text{water}}}$

5. Group Discussion and Presentation:
Students in groups perform the activity and present their findings, developing observation and communication skills.

6. Concept Mapping:
A chart linking density, upthrust, floating and sinking is prepared to consolidate understanding.

These strategies make the topic concrete, inquiry-driven and linked to real-life applications, ensuring lasting understanding.

Climate Change

Climate change is the long-term significant change in the average weather conditions of the Earth or of a particular region over a period of decades or more. It includes changes in temperature, rainfall pattern, wind pattern, humidity and the frequency of extreme events. The main causes are:

• Natural causes: volcanic eruption, changes in solar radiation, changes in the Earth's orbit.
• Human (anthropogenic) causes: burning of fossil fuels (coal, petrol, diesel), deforestation, industrialisation, use of chemical fertilisers, and emission of greenhouse gases such as carbon dioxide, methane and nitrous oxide.

These gases trap heat in the atmosphere (greenhouse effect), causing global warming, which leads to melting of glaciers, rise in sea level, floods, landslides, droughts and changes in cropping seasons. In Nepal, climate change is seen in the retreating of Himalayan glaciers, drying up of water sources, and more frequent landslides and flash floods.

Approaches Applicable to Teach "Natural Disaster" at the Secondary Level

Natural disaster is a practical, life-related and activity-based topic. So the following approaches are applicable:

1. Activity-based / Experiential approach: Students learn by doing — preparing a disaster map of the school, making an emergency kit, practising mock drills for earthquake and flood. This develops skill and readiness.

2. Discussion and brainstorming approach: Students discuss causes, effects and preventive measures of disasters in groups. It develops thinking and expression.

3. Problem-solving approach: A real or imaginary problem is given, e.g. "Your village is hit by a landslide — what will you do?" Students find solutions in steps. It develops decision-making ability.

4. Project work approach: Students collect information about a local disaster, prepare a report or chart and present it. It links school knowledge with community life.

5. Field visit / observation approach: Visiting a landslide-prone area, a river bank or a flood-affected settlement makes learning direct and lasting.

6. Demonstration approach: The teacher demonstrates first aid, rescue methods, or the use of fire extinguishers and sirens.

6. Integrated approach: Linking the topic with Science (causes), Social Studies (effects on society), Health (safety) and Mathematics (loss of life and property in numbers).

Justification with a Suitable Example

Suppose the teacher teaches "Earthquake" as a natural disaster in Class 9. The teacher first uses brainstorming to find out what students already know. Then a short video or picture of the 2015 Gorkha earthquake is shown. After that, students are divided into groups for discussion on causes and effects. Next, a mock drill is conducted — "Drop, Cover and Hold" — and students practise moving to an open place. Finally, each group prepares a project on "Earthquake safety rules for our school" and displays it on the notice board. In this way, the same topic is taught through several approaches, so knowledge, skill and attitude are all developed. This is why activity-based, discussion, problem-solving and project approaches are most suitable for this topic.

Two Higher Ability Level Objectives

According to Bloom's taxonomy, higher ability objectives fall in the analysis, synthesis and evaluation levels. Two such objectives are:

1. Analysis: Students will be able to analyse the causes and effects of a local natural disaster and distinguish between natural and human-made factors responsible for it.
2. Evaluation / Synthesis: Students will be able to evaluate the effectiveness of the disaster management measures adopted in their community and prepare a plan of action for disaster preparedness in their school or locality.

Preparation of Carbon Dioxide Gas in the Laboratory

1. Preparation Process

Chemicals used: Dilute hydrochloric acid (HCl) and marble pieces (calcium carbonate, CaCO₃).

Apparatus: Woulfe's bottle (or a conical flask fitted with a two-holed cork), thistle funnel, delivery tube, gas jar, trough, water.

Procedure:

1. Take some marble pieces in the Woulfe's bottle (or conical flask).
2. Fit the two-holed cork. Insert the thistle funnel through one hole so that its lower end dips into the acid, and the delivery tube through the other hole.
3. Pour dilute hydrochloric acid through the thistle funnel, drop by drop, over the marble pieces.
4. The reaction starts immediately and carbon dioxide gas is produced:

$\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2\uparrow$

5. The gas is collected in the gas jar by the downward displacement of air (since CO₂ is heavier than air) or by the upward displacement of water over a water trough.
6. When the gas jar is full, it is covered with a glass plate and kept inverted on the table.

Precautions:

• The lower end of the thistle funnel must dip into the acid, otherwise the gas will escape through the funnel.
• The delivery tube should not dip into the acid in the bottle.
• Marble pieces should not be powdered, otherwise the reaction becomes too fast.

2. Labelled Diagram

        Thistle funnel
             |
             v
      ______|______
     /             \
    |  Dilute HCl   |  <-- Woulfe's bottle
    |               |
    |  Marble pieces|
     \______ ______/
            |
            | Delivery tube
            |
            v
      _______________
     |               |
     |   Gas jar     |  <-- CO₂ collected by
     |   (CO₂ gas)   |      downward displacement
     |_______________|      of air

(In the answer sheet, draw the Woulfe's bottle with the thistle funnel dipping into the acid, the delivery tube leading into an inverted gas jar, and label: thistle funnel, dilute HCl, marble pieces, delivery tube, gas jar, carbon dioxide gas.)

3. Any Four Properties of Carbon Dioxide Gas

1. It is a colourless, odourless and tasteless gas.
2. It is heavier than air (about 1.5 times heavier) and highly soluble in water.
3. It is neither combustible nor a supporter of combustion; it extinguishes a burning candle.
4. It turns lime water milky due to the formation of insoluble calcium carbonate:

$\text{Ca(OH)}_2 + \text{CO}_2 \rightarrow \text{CaCO}_3\downarrow + \text{H}_2\text{O}$

4. Teaching the Topic Using the 5E Approach

Engage: Show a burning candle covered with a glass tumbler; the candle goes out. Ask students why. Also ask why lime water turns milky when we blow into it through a straw. This creates curiosity about the gas.

Explore: Provide marble pieces, dilute HCl, Woulfe's bottle, and gas jar. Let students set up the apparatus in groups and collect the gas. They test it with a burning candle and with lime water, and record observations.

Explain: Explain the chemical reaction, the equation, why the gas is collected by downward displacement of air, and the properties observed. Introduce the terms reactant, product, and the lime-water test.

Elaborate: Ask students to apply the knowledge — why CO₂ is used in fire extinguishers, why it is used in soft drinks, what happens when excess CO₂ is passed through lime water (milkiness disappears due to formation of soluble calcium bicarbonate). Give a numerical or a comparison table of O₂ and CO₂.

Evaluate: Assess through oral questions (name the chemicals used, write the equation), a short written test, and checking the labelled diagram drawn by students. Also evaluate their laboratory observation notes.

Adaptation is the process by which a plant or animal develops special features (structural, behavioural or physiological) over many generations that help it survive and reproduce in its particular habitat. For example, a cactus has spines instead of leaves and a thick waxy stem to reduce water loss in the desert.

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Interactive activity: "Match the Creature to its Home" (Habitat Detective Game)

Time: 35–40 minutes | Class size: any | Materials: picture cards of 6 plants/animals, 6 habitat cards (desert, pond, snow region, forest, ocean, grassland), chart paper, markers, sticky notes.

Step 1 – Set up (5 min). Place the six habitat cards at different corners of the room. Give each group of 4–5 students one organism card (camel, cactus, fish, lotus, polar bear, pine tree).

Step 2 – Think–Pair–Share (8 min). Each group discusses: What special body parts does our organism have? How do these parts help it live in its habitat? They write 2–3 features on sticky notes.

Step 3 – Move and justify (12 min). Groups walk to the habitat corner they think matches their organism. One member explains the reason, e.g. "The camel has a hump to store fat and long eyelashes to keep out sand — so it belongs in the desert."

Step 4 – Peer challenge (8 min). Other groups may challenge the placement by giving a counter-reason. The teacher moderates and corrects misconceptions.

Step 5 – Consolidation (7 min). On chart paper the class builds a table: Organism → Habitat → Adaptive feature → Advantage. The teacher links this to the three types of adaptation (structural, behavioural, physiological).

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Sample checklist for assessment

Name: ...................... Roll No. ...... Date ......

Scoring: 1 mark per "Yes" (total 7). Interpretation: 6–7 = excellent, 4–5 = satisfactory, below 4 = needs remedial support.

Critical analysis of present Secondary Level Science Curriculum (Grades 9–10)

Recommendations:

1. Reduce content load and revise the curriculum as competency-based, integrating local science–technology and emerging areas with more lab, field and project work.
2. Strengthen practical/performance-based assessment using standard rubrics, and invest in teacher training, laboratory equipment and ICT facilities.

Carnot's Theorem from the Second Law of Thermodynamics

Carnot's theorem states:

1. No engine operating between two given thermal reservoirs (temperatures $T_1$ and $T_2$) can be more efficient than a reversible Carnot engine working between the same two reservoirs.
2. All reversible engines operating between the same two reservoirs have the same efficiency, regardless of the working substance.

Demonstration using the second law

Part 1 — No engine is more efficient than a Carnot engine.

Suppose an engine $E$ (any engine) works between a hot reservoir at $T_1$ and a cold reservoir at $T_2$, and suppose its efficiency $\eta_E$ is greater than that of a reversible Carnot engine $C$ working between the same reservoirs, i.e. $\eta_E > \eta_C$.

Let both engines extract the same heat $Q_1$ from the hot reservoir. Then:

• Work done by $E$: $W_E = \eta_E Q_1$
• Work done by $C$: $W_C = \eta_C Q_1$

Since $\eta_E > \eta_C$, we have $W_E > W_C$.

Now reverse the Carnot engine $C$ and drive it as a refrigerator using part of the work $W_E$ from engine $E$. The Carnot refrigerator takes heat $Q_2$ from the cold reservoir and rejects $Q_1$ to the hot reservoir. Since $W_C < W_E$, the surplus work $W_E - W_C$ remains available as useful output.

Net result: The hot reservoir is unchanged (it gives $Q_1$ to $E$ and receives $Q_1$ from $C$). Engine $E$ rejects $Q_1 - W_E$ to the cold reservoir, while the Carnot refrigerator takes $Q_1 - W_C$ from it, so the cold reservoir loses a net heat $(Q_1 - W_C) - (Q_1 - W_E) = W_E - W_C$. This heat is converted entirely into work $W_E - W_C$, with no other change. This violates the Kelvin–Planck statement of the second law. Hence $\eta_E > \eta_C$ is impossible. So $\eta_E \le \eta_C$.

Part 2 — All reversible engines have equal efficiency.

Let $C_1$ and $C_2$ be two reversible engines between the same reservoirs. By Part 1, $\eta_{C_1} \le \eta_{C_2}$ and $\eta_{C_2} \le \eta_{C_1}$. Therefore $\eta_{C_1} = \eta_{C_2}$.

Thus Carnot's theorem is established. The efficiency depends only on $T_1$ and $T_2$:

$\eta = 1 - \frac{T_2}{T_1}$

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Comparison of the Four Thermodynamic Processes

Key points

• Isothermal: Slow process in a conducting cylinder; temperature stays constant, so internal energy does not change and all heat supplied appears as work.
• Adiabatic: Fast process in an insulated cylinder; no heat enters or leaves, so work is done at the cost of internal energy, causing a temperature change.
• Isobaric: Heating at constant pressure (e.g., boiling water in an open vessel); heat supplied raises both internal energy and does expansion work.
• Isochoric: Heating at constant volume (e.g., in a rigid closed vessel); no work is done, so all heat supplied increases internal energy only.

a) Interference and diffraction

Proof: Refractive index of a medium is equal to tangent of the angle of polarization

Let a ray of unpolarized light be incident from air on a transparent medium at the polarizing angle $i_p$. At this angle, the reflected ray is completely plane polarized. According to Brewster’s law, at the polarizing angle the reflected ray and the refracted ray are perpendicular to each other.

Let the angle of refraction be $r$. Then

[ i_p + r = 90^\circ ]

so,

[ r = 90^\circ - i_p ]

By Snell’s law,

[ n = \frac{\sin i_p}{\sin r} ]

Substituting $r = 90^\circ - i_p$,

[ n = \frac{\sin i_p}{\sin(90^\circ - i_p)} ]

[ n = \frac{\sin i_p}{\cos i_p} ]

[ \boxed{n = \tan i_p} ]

Thus, the refractive index of a transparent medium is equal to the tangent of the angle of polarization.

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b) Gauss theorem

Gauss theorem: The total electric flux through a closed surface is equal to (\frac{1}{\varepsilon_0}) times the net electric charge enclosed by the surface.

[ \oint \vec{E}\cdot d\vec{A} = \frac{q_{\text{enclosed}}}{\varepsilon_0} ]

Electric field due to an infinite plane sheet of charge

Let an infinite plane sheet have uniform surface charge density (\sigma). By symmetry, the electric field is perpendicular to the sheet and has the same magnitude at equal distances on both sides.

Consider a cylindrical Gaussian surface of cross-sectional area (A), passing through the sheet and extending equal distances on both sides.

• Flux through the curved surface is zero, because (\vec{E}) is parallel to the curved surface.
• Flux through the two flat faces is

[ \Phi = EA + EA = 2EA ]

Charge enclosed by the Gaussian surface is

[ q_{\text{enclosed}} = \sigma A ]

Using Gauss theorem,

[ 2EA = \frac{\sigma A}{\varepsilon_0} ]

[ \boxed{E = \frac{\sigma}{2\varepsilon_0}} ]

In vector form,

[ \vec{E} = \frac{\sigma}{2\varepsilon_0}\hat{n} ]

where (\hat{n}) is a unit vector normal to the sheet, directed away from it for positive charge.

Note: For a charged conducting sheet, the field near its surface is
[ E = \frac{\sigma}{\varepsilon_0} ] because the charge resides only on the outer surfaces.

(a) Why gold is preferred over iron for making jewellery

Both gold and iron are metals, but gold is preferred for making jewellery for the following reasons:

1. Least reactivity (noble metal): Gold is a noble metal. It does not react with air, moisture, oxygen, CO₂ or sulphur even after a long time. Hence, it neither rusts nor corrodes and keeps its shine for years. Iron, however, reacts with moist air and forms rust (hydrated iron(III) oxide), which destroys its shine and strength.
2. Permanent lustre and colour: Gold has a bright, lasting yellow lustre that does not fade. Iron quickly loses its shine due to corrosion.
3. Highest malleability and ductility: Gold is the most malleable and ductile metal. It can be beaten into extremely thin leaves and drawn into very fine wires, so it can be shaped into delicate and intricate designs. Iron is hard and less malleable, so fine ornamental work is difficult.
4. Softness and comfort: Pure gold is soft and is alloyed with copper or silver to get the required hardness for jewellery. Iron is hard and heavy, so ornaments made of it are uncomfortable to wear.
5. Rarity and value: Gold is a rare metal, so it is costly and is regarded as a symbol of wealth. Iron is abundant and cheap, so it is not used for costly ornaments.

Because of its non-corrosive nature, lasting lustre, high malleability and rarity, gold is preferred over iron for making jewellery.

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(b) Extraction of iron from its ore

The main ore of iron is haematite (Fe₂O₃). Extraction is carried out in a blast furnace.

Steps involved:

1. Concentration of ore: The ore is crushed and then concentrated by gravity separation or magnetic separation to remove gangue (sand, clay, etc.).
2. Calcination/Roasting: The concentrated ore is heated strongly in the presence of air to remove moisture, carbon dioxide and sulphur impurities. $\text{Fe}_2\text{O}_3\cdot x\text{H}_2\text{O} \xrightarrow{\text{heat}} \text{Fe}_2\text{O}_3 + x\text{H}_2\text{O}$
3. Smelting in blast furnace: The roasted ore is mixed with coke (carbon) and limestone (CaCO₃) in the ratio 8 : 4 : 1 and fed into the blast furnace from the top through the charging hopper. Hot air is blown through the bottom.
   • Reduction: Coke burns to form CO, which reduces iron oxide to molten iron. $\text{C} + \text{O}_2 \rightarrow \text{CO}_2$ $\text{CO}_2 + \text{C} \rightarrow 2\text{CO}$ $\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2$
   • Removal of impurities: Limestone decomposes to CaO, which combines with silica (SiO₂) to form slag (calcium silicate). $\text{CaCO}_3 \xrightarrow{\text{heat}} \text{CaO} + \text{CO}_2$ $\text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3 \text{ (slag)}$
4. Separation of slag: Molten iron being heavier settles at the bottom, while molten slag floats on top. They are tapped out separately through the iron outlet and slag outlet respectively.
5. Conversion to steel: The molten iron (pig iron) is purified in a Bessemer converter to obtain steel.

Labelled diagram of blast furnace:

              Charging hopper
             (ore + coke + limestone)
                     |
                     v
        ┌─────────────────────────────┐
        │                             │ → Waste gas outlet
        │      Blast Furnace          │
        │                             │
        │                             │
        │                             │
        │                             │
        │  Molten slag ───────────────┼ → Slag outlet
        │  Molten iron ───────────────┼ → Iron outlet
        └─────────────────────────────┘
                     ↑
              Hot air blast inlet
                 (bottom)

Labelled parts: Charging hopper (top), waste gas outlet (top), blast furnace body, hot air blast inlet (bottom), slag outlet (upper side), molten iron outlet (bottom).

In this way, iron is extracted from haematite ore by reduction with carbon monoxide in a blast furnace.

(a) Difference between Porifera and Coelenterata

Classification of Coelenterata (Cnidaria)

Coelenterata is divided into three classes:

1. Hydrozoa — e.g., Hydra, Obelia, Physalia
2. Scyphozoa — e.g., Aurelia (jellyfish)
3. Anthozoa — e.g., Adamsia (sea anemone), Corallium (coral)

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(b) Ecosystem

An ecosystem is a self-regulating, interacting unit in which the biotic community (producers, consumers and decomposers) interacts with the abiotic components (air, water, soil, sunlight, temperature) and with one another, resulting in a flow of energy and a cycling of nutrients. Example: a pond, a forest.

Why areas around national parks in Nepal are declared as buffer zones

1. To reduce human–wildlife conflict — animals straying out of the park destroy crops and attack people; the buffer zone provides a transition area that lessens such conflict.
2. To supply forest products to local people — people living around the park depend on fuelwood, fodder, timber and grazing land; the buffer zone meets these needs and thereby reduces pressure on the core park area.
3. To conserve biodiversity — the buffer zone acts as additional habitat and a corridor for wildlife, supporting the park's conservation goal.
4. To share park revenue with local people — 30–50% of the park's income is returned to buffer zone communities, which encourages them to support conservation.
5. To promote local participation in conservation — Buffer Zone User Committees involve local people in managing and protecting natural resources, making conservation sustainable.

Tissue Culture

Tissue culture is the technique of growing plant cells, tissues, or organs in an artificial nutrient medium under sterile, controlled conditions in vitro (in glass), separate from the whole plant. It is based on the principle of totipotency — the ability of a single plant cell to divide and regenerate into a complete plant.

Steps:

1. Selection of explant — a small piece of plant tissue (shoot tip, leaf, stem, anther, etc.) is chosen from a healthy, disease-free mother plant.
2. Sterilization — the explant and all materials are made free of microbes.
3. Inoculation — the sterile explant is placed on the nutrient medium in a laminar flow cabinet.
4. Incubation — cultures are kept at 25 ± 2 °C with light and humidity control.
5. Callus formation / proliferation — cells divide to form an undifferentiated mass (callus) or direct shoots.
6. Organogenesis / embryogenesis — shoots and roots differentiate under the influence of auxins and cytokinins.
7. Hardening (acclimatization) — plantlets are transferred to soil gradually.

Nutrient medium (usually MS medium) contains macro- and micronutrients, vitamins, amino acids, sugar (sucrose), growth regulators (auxin, cytokinin) and a gelling agent (agar).

Advantages: rapid multiplication, production of disease-free plants, propagation of sterile or rare plants, year-round production, and genetic improvement.

Sterilization Techniques in Plant Tissue Culture

Contamination by bacteria and fungi is the greatest problem in tissue culture, so every step must be sterile. The main techniques are:

1. Sterilization of the working area (aseptic room / laminar flow cabinet)

• The inoculation room is cleaned with detergents and disinfected with 70% ethanol or 2% sodium hypochlorite.
• The laminar flow cabinet is wiped with 70% alcohol and its air is sterilized by UV light (switched off during work) and by HEPA-filtered air.

2. Sterilization of glassware and instruments

• Glassware, scalpels, forceps, needles and petri dishes are wrapped and sterilized in an autoclave at 121 °C and 15 psi for 15–20 minutes.
• Metal instruments are also flame-sterilized by dipping in 95% alcohol and flaming, or by a glass-bead sterilizer (250 °C).
• Dry heat sterilization in a hot-air oven (160–180 °C for 2–3 hours) is used for glassware.

3. Sterilization of the nutrient medium

• Media are sterilized by autoclave at 121 °C, 15 psi for 15–20 minutes.
• Heat-labile substances (some vitamins, hormones, antibiotics) are sterilized separately by filtration through a 0.22 µm membrane filter and added after cooling.

4. Sterilization of the explant (surface sterilization)

• The explant is washed in running tap water, then treated with a detergent (Tween-20) to remove dust and microbes.
• It is then dipped in a disinfectant solution:
  • Sodium hypochlorite (0.5–2%) for 5–20 minutes, or
  • Mercuric chloride (0.1%) for 2–10 minutes, or
  • Ethanol (70%) for 30 seconds to 1 minute.
• Finally it is rinsed 3–4 times with sterile distilled water to remove the disinfectant.

5. Sterilization of the transfer / culture room

• The culture room is kept clean, filtered air is used, and the room is fumigated or sprayed with disinfectants.
• Cultures are handled only with sterile instruments and in the laminar flow cabinet.

6. Antibiotic treatment

• Antibiotics (e.g., streptomycin) are sometimes added to the medium to suppress bacterial growth.

Thus, by combining autoclaving, filtration, flame sterilization, UV radiation and chemical disinfectants, a completely aseptic environment is maintained for successful tissue culture.