Madhya Pradesh Board Class 9 Science — NCERT Solutions

Formula used: $T(°C) = T(K) - 273$

(a) 293 K to Celsius:
$$T(°C) = 293 - 273 = 20°C$$

(b) 470 K to Celsius:
$$T(°C) = 470 - 273 = 197°C$$

Formula used: $T(K) = T(°C) + 273$

(a) 25°C to Kelvin:
$$T(K) = 25 + 273 = 298 \text{ K}$$

(b) 373°C to Kelvin:
$$T(K) = 373 + 273 = 646 \text{ K}$$

(a) Sodium chloride from its solution in water:
Technique — Evaporation (or Crystallisation).
Water (solvent) evaporates on heating, leaving behind solid sodium chloride.

(b) Ammonium chloride from a mixture of sodium chloride and ammonium chloride:
Technique — Sublimation.
Ammonium chloride sublimes (converts directly from solid to vapour) on heating, while sodium chloride does not. The vapours are collected and condensed to get pure ammonium chloride.

(c) Small pieces of metal from engine oil:
Technique — Filtration (using a fine mesh/filter) or using a magnet if the metal is magnetic (e.g., iron).
The metal pieces are insoluble in oil and can be filtered out.

(d) Different pigments from an extract of flower petals:
Technique — Chromatography.
Different pigments travel at different speeds on the chromatography paper due to their different solubilities, separating them into distinct bands.

(e) Butter from curd:
Technique — Centrifugation.
Curd is churned at high speed; the denser butter separates out due to centrifugal force.

(f) Oil from water:
Technique — Separating funnel (Separatory funnel).
Oil and water are immiscible liquids with different densities. They form two distinct layers in the separating funnel; the lower layer (water) is drained out first, then oil is collected.

(g) Tea leaves from tea:
Technique — Filtration.
Tea is poured through a strainer/filter; tea leaves (residue) are retained and the liquid tea (filtrate) passes through.

(h) Iron pins from sand:
Technique — Magnetic separation.
A magnet is moved over the mixture; iron pins (being magnetic) are attracted to the magnet and separated from sand.

(i) Wheat grains from husk:
Technique — Winnowing.
The mixture is allowed to fall from a height in a gentle breeze; the lighter husk is blown away while the heavier wheat grains fall straight down.

(j) Fine mud particles suspended in water:
Technique — Centrifugation or Sedimentation followed by Decantation.
On centrifugation (or on allowing to stand), the heavier mud particles settle at the bottom; the clear water is then carefully poured off (decanted).

Steps for Making Tea:

Step 1: Take water in a pan and heat it. Water acts as the solvent.

Step 2: Add tea leaves to the boiling water. Tea leaves are partially soluble in hot water — the colour and flavour compounds dissolve in water, while the leaf material itself is insoluble.

Step 3: Add sugar (a solute) to the mixture. Sugar readily dissolves in hot water to form a solution.

Step 4: Add milk (another solute/component) and allow the mixture to boil for a short time, forming a uniform solution of tea.

Step 5: Pour the tea through a strainer (filtration). The liquid tea that passes through is the filtrate — this is our tea. The undissolved tea leaves left behind on the strainer are the residue.

Result: The final tea is a solution in which water is the solvent and the dissolved substances (sugar, colour, flavour compounds from tea leaves) are the solutes.

Given:

• Total mass of compound = 0.24 g
• Mass of boron = 0.096 g
• Mass of oxygen = 0.144 g

Formula used:
$$\text{Percentage by weight} = \frac{\text{Mass of element}}{\text{Total mass of compound}} \times 100$$

Step 1: Percentage of Boron
$$\% \text{ Boron} = \frac{0.096}{0.24} \times 100 = 40\%$$

Step 2: Percentage of Oxygen
$$\% \text{ Oxygen} = \frac{0.144}{0.24} \times 100 = 60\%$$

Verification: $40\% + 60\% = 100\%$ ✓

Answer: The compound contains 40% Boron and 60% Oxygen by weight.

Given (first case):

• Mass of carbon = 3.0 g
• Mass of oxygen = 8.00 g
• Mass of CO₂ produced = 11.00 g

Observation from first case:
Carbon and oxygen combine in the ratio:
$$\text{C : O} = 3.0 : 8.0 = 3 : 8$$
This means 3 g of carbon requires exactly 8 g of oxygen to produce 11 g of CO₂.

Second case:

• Mass of carbon = 3.00 g
• Mass of oxygen available = 50.00 g

Step 1: Find oxygen required to burn 3.00 g of carbon completely.
From the fixed ratio, oxygen required $= 8.00$ g.

Step 2: Since 50.00 g of oxygen is available but only 8.00 g is needed, oxygen is in excess. Carbon is the limiting reactant.

Step 3: Mass of CO₂ formed
$$= 3.00 + 8.00 = 11.00 \text{ g}$$

Answer: 11.00 g of carbon dioxide will be formed.

Law governing this: This is governed by the Law of Constant Proportions (Law of Definite Proportions), which states that in a chemical compound, elements are always present in definite proportions by mass, regardless of the source or method of preparation.

The properties of the three sub-atomic particles are compared below:

Key points:

• Electrons are negatively charged and revolve around the nucleus in fixed shells.
• Protons are positively charged and are present in the nucleus.
• Neutrons are neutral (no charge) and are also present in the nucleus.
• The mass of an electron is negligible compared to protons and neutrons.

Given: We need to compare plant cells and animal cells.

Differences between Plant Cells and Animal Cells:

Conclusion: Plant cells have a cell wall, chloroplasts, and a large central vacuole which are absent in animal cells, while animal cells have centrioles and lysosomes which are generally absent in plant cells.

Definition: A tissue is a group of cells that are similar in structure, origin, and function, and work together to perform a specific function in the body.

Example: Muscle tissue is made up of muscle cells that work together to bring about movement.

Given:

• Diameter of circular track = 200 m, so radius $r = 100$ m
• Time for one round = 40 s
• Total time = 2 min 20 s = 140 s

Step 1: Find the number of rounds completed.
$$\text{Number of rounds} = \frac{\text{Total time}}{\text{Time per round}} = \frac{140}{40} = 3.5 \text{ rounds}$$

Step 2: Find the distance covered.

Circumference of the track (distance per round):
$$C = \pi d = \pi \times 200 = 628 \text{ m (approx.)}$$

$$\text{Distance covered} = 3.5 \times 628 = 2200 \text{ m}$$

Step 3: Find the displacement.

After 3.5 rounds, the athlete is at the diametrically opposite end of the starting point.

$$\text{Displacement} = \text{diameter} = 200 \text{ m}$$

Answer:

• Distance covered = 2200 m
• Displacement = 200 m

Given/Concept: Net external unbalanced force = 0. By Newton's First Law of Motion, an object continues in its state of rest or of uniform motion in a straight line unless acted upon by an unbalanced external force.

Answer: Yes, it is absolutely possible for the object to be travelling with a non-zero velocity even when the net external unbalanced force is zero.

Conditions:

• The object must be moving with a constant (uniform) velocity — i.e., no change in speed.
• The direction of motion must remain unchanged (straight line).
• There should be no acceleration (since $F = ma$, if $F = 0$ then $a = 0$).

For example, a hockey puck sliding on a frictionless ice surface moves with constant velocity because no net unbalanced force acts on it.

Given: Distance between two objects is reduced to half, i.e., new distance $r' = \dfrac{r}{2}$.

Formula used (Newton's Law of Gravitation):
$$F = G\frac{m_1 m_2}{r^2}$$

Working:

Original force:
$$F = G\frac{m_1 m_2}{r^2}$$

New force when distance is halved:
$$F' = G\frac{m_1 m_2}{\left(\dfrac{r}{2}\right)^2} = G\frac{m_1 m_2}{\dfrac{r^2}{4}} = 4 \times G\frac{m_1 m_2}{r^2} = 4F$$

Conclusion: When the distance between two objects is reduced to half, the gravitational force between them becomes four times the original force.

Concept: Work is done when a force acts on an object AND the object is displaced in the direction of (or having a component along) the applied force. W = F × d × cos θ.

1. Suma is swimming in a pond.

Suma applies force on water with her arms and legs, and her body moves forward (displacement occurs in the direction of applied force). Work is done.

2. A donkey is carrying a load on its back.

The donkey exerts an upward force (normal reaction) to support the load, but the displacement of the load is horizontal. The angle between force (vertical) and displacement (horizontal) is 90°, so W = F × d × cos 90° = 0. Work is NOT done by the donkey on the load (in the scientific sense).

3. A wind-mill is lifting water from a well.

The windmill applies force on water and the water is displaced upward (in the direction of force). Work is done.

4. A green plant is carrying out photosynthesis.

Photosynthesis is a biochemical process. There is no mechanical force causing displacement of an object. Work is NOT done (in the mechanical sense).

5. An engine is pulling a train.

The engine applies force on the train and the train moves in the direction of the force. Work is done.

6. Food grains are getting dried in the sun.

Drying is a physical/chemical process involving heat energy. No mechanical force causes displacement of the grains. Work is NOT done (in the mechanical sense).

7. A sailboat is moving due to wind energy.

Wind exerts force on the sail and the boat is displaced in the direction of the force. Work is done.

Sound:
Sound is a form of energy that produces the sensation of hearing in our ears. It is a mechanical wave that requires a material medium (solid, liquid, or gas) for its propagation.

How sound is produced:
Sound is produced by the vibration of objects. When an object vibrates, it causes the particles of the surrounding medium to vibrate. These vibrations travel through the medium in the form of a wave, which we perceive as sound.

Example: When we pluck a stretched rubber band, it vibrates and produces sound. When a drum is beaten, its membrane vibrates and produces sound. The vibrating object acts as the source of sound.

Given: We need to explain one method of crop production that ensures high yield.

Method: HYV (High Yielding Variety) Seeds combined with Crop Rotation

One important method is the use of High Yielding Variety (HYV) seeds along with proper agronomic practices.

Explanation:

Step 1 – Selection of HYV Seeds:
Seeds of high-yielding varieties are developed through plant breeding. These varieties are selected for:

• Higher yield per hectare
• Resistance to biotic stresses (pests, diseases)
• Resistance to abiotic stresses (drought, salinity, waterlogging)
• Shorter maturity duration

Step 2 – Proper Nutrient Management:
HYV seeds respond well to fertilizers and manures. Adequate supply of macronutrients (N, P, K) and micronutrients ensures the crop reaches its genetic yield potential.

Step 3 – Irrigation:
Timely and adequate irrigation ensures that the crop does not suffer water stress at critical growth stages.

Step 4 – Pest and Disease Control:
Use of pesticides, weedicides, and biological control agents protects the crop from losses.

Result: The combination of HYV seeds with proper nutrient supply, irrigation, and protection measures ensures significantly higher yield compared to traditional varieties.

Conclusion: Use of HYV seeds is one of the most effective methods to ensure high yield in crop production.