CBSE Class 9 Science Experiments Worksheet

Experiment 9

Objective

To study the third law of motion using two spring balances.

Materials Required

• Two identical spring balances
• Weight box
• Thread
• A frictionless pulley
• A pan of known mass

Theory

• Newton's third law of motion states that whenever one object applies a force on another, the second object exerts an equal and opposing force on the first.
• Forces in nature always occur in matched pairs; an isolated single force cannot exist.
• These two forces are equal in magnitude and point in opposite directions.
• The mutual forces between interacting objects are termed action and reaction.
• Action and reaction forces never act on the same object; they act simultaneously on two distinct bodies.

Procedure

• Determine the measuring range and least count of both spring balances.
• Confirm that the two balances have identical calibrations and spring stiffness.
• Hang both balances vertically without any load to confirm that the indicators point exactly to zero.
• Connect balance A and balance B together by hooking their ends. Anchor balance B securely to a fixed vertical support.
• Attach a strong thread to the free ring of balance A, pass the thread over a smooth pulley clamped to the table edge, and tie it to a scale pan.
• Place a known mass M (for example, 150 g) in the pan. The combined suspended mass becomes (M + m).
• Read and record the values indicated on both spring balances once they come to rest.
• Repeat the measurement with at least four different loads placed in the pan and record the readings in the observation table.

Observation and Calculations

(i) Range of each spring balance = 0 - 5 N (or 0 - 500 g).
(ii) Least count of each spring balance = 0.1 N (or 10 g).
(iii) Acceleration due to gravity (g) = 9.8 m s^-2.
(iv) Mass of pan (m) = ......... g = ......... kg.
(v) Weight of empty pan (w = m × g) = ......... N.

Precautions

• Verify that the pointer on each spring balance reads zero when no load is attached.
• Record scale readings only after the oscillations damp out and the indicators are fully stationary.
• Use balances that possess identical construction, range, and spring constant.
• Ensure the connecting string does not stretch during the experiment and the pulley rotates freely without friction.

Viva-Voce

Question 1. Do action and reaction act on the same body or on different bodies ?
Answer: Action and reaction always exert themselves on two distinct objects, never on a single body.
In simple words: When one object pushes a second object, the second object pushes back on the first object.

Exam Tip: Never say action and reaction cancel each other out; because they act on separate bodies, their net effect is not zero on either individual body.

Question 2. State Newton's third law of motion.
Answer: Newton's third law indicates that for every exerted force, an opposing force of identical strength arises simultaneously on the interacting partner.
In simple words: Every action creates an equal push in the opposite direction on another body.

Exam Tip: Always include three key aspects when defining this law: equal in magnitude, opposite in direction, and acting on different objects.

Question 3. Define action and reaction forces.
Answer: The force applied by the initial entity on the receiving partner is termed action, whereas the responsive push exerted simultaneously by the receiving entity back onto the initial one is known as reaction.
In simple words: The first push is called action, and the push coming right back is called reaction.

Exam Tip: Either force in an interaction can technically be labeled as action; the corresponding mutual force automatically becomes the reaction.

Question 4. Is action and reaction act on same body ?
Answer: No, these paired forces operate on separate interacting objects.
In simple words: Action and reaction never apply to the exact same thing.

Exam Tip: A direct single-word "No" followed by a concise reason is sufficient for full marks in oral and short-answer questions.

Question 5. Do action and reaction forces act simultaneously ?
Answer: Yes, both forces occur at the exact same instant without any time delay.
In simple words: Action and reaction happen together at the very same moment.

Exam Tip: Clarify that action is not the "cause" that later produces a reaction; both are created at the exact same instant.

Question 6. Is action and reaction forces are equal ?
Answer: Yes, their magnitudes are always completely identical.
In simple words: The size of the forward push always matches the backward push.

Exam Tip: Mention that equality refers to numerical strength (magnitude), while directions remain strictly opposite.

Question 7. Is action and reaction forces are balanced ?
Answer: No, they do not neutralize each other under normal circumstances because they act on separate bodies. They can only be regarded as balanced if both interacting bodies are analyzed together as one unified composite system.
In simple words: They do not cancel out because they affect two different objects, unless you view both objects as one single unit.

Exam Tip: Balanced forces must act on the same body; since action and reaction act on different bodies, they cannot balance each other.

Question 8. Give examples in the support of the third law of motion.
Answer: Common illustrations include:
(i) The backward kick or recoil felt when firing a bullet from a firearm.
(ii) A person stepping forward onto a riverbank causing the boat beneath them to drift backward.
(iii) Exhaust gases expelled downward from a rocket propelling the craft upward.
In simple words: Firing a gun pushes the bullet forward and kicks the gun backward. Stepping off a boat pushes the boat away.

Exam Tip: For each example, explicitly name which force acts as the action and which acts as the reaction.

Question 9. Which scientist gave the famous three laws of motion ?
Answer: Sir Isaac Newton formulated the three fundamental laws governing body movements.
In simple words: Isaac Newton discovered and wrote the three laws of motion.

Exam Tip: Newton published these principles in his milestone 1687 work titled Philosophiae Naturalis Principia Mathematica.

Question 10. Define one newton force.
Answer: A single newton represents the amount of force needed to impart an acceleration of 1 m/s² to a mass measuring exactly 1 kilogram.
In simple words: One newton is the push needed to speed up a 1-kilogram weight by 1 meter per second every second.

Exam Tip: Use the formula \( F = ma \) to verify definitions: \( 1\text{ N} = 1\text{ kg} \times 1\text{ m/s}^2 \).

Question 11. Define the term force.
Answer: Force denotes any external effort in the guise of a push or pull capable of altering an object's resting state, linear speed, or travel orientation.
In simple words: A force is a push or pull that can move an object, stop it, or change its direction.

Exam Tip: Mention that force can also deform an object by changing its shape or dimensions.

Question 12. Is force a vector quantity ?
Answer: Yes, force is fundamentally a vector since full description requires stating both its magnitude and its directional path.
In simple words: Yes, because every force has an amount as well as a specific direction.

Exam Tip: Stating the SI unit (newton) along with direction confirms your understanding of vector representation.

Question 13. What do you mean by the term interaction in physics ?
Answer: An interaction refers to a mutual process wherein two bodies exert influence on one another, modifying their respective states of motion through applied forces.
In simple words: An interaction means two objects acting upon each other by exerting mutual forces

Exam Tip: Emphasize that forces never exist alone; an interaction requires at least two participating bodies.

Multiple Choice Questions

Question 1. The action and reaction forces referred to in Newton's third law.
(a) Must act on the same object
(b) May act on the different objects
(c) Must act on different objects
(d) Need not be equal in magnitude but must have the same direction
Answer: (c) Must act on different objects
Newton's third law mandates that mutual interaction forces must exert themselves on two distinct bodies simultaneously.
In simple words: Action pushes one object while reaction pushes back on the other object.

Exam Tip: Remembering that action and reaction act on separate bodies explains why they do not neutralize each other.

Question 2. Identify A in the given Fig. 9.2.
(a) Action
(b) Reaction
(c) Motion
(d) Acceleration
Answer: (a) Action
In the diagram, arrow A portrays the backward and downward force exerted by the horse's hooves against the ground surface, representing action.
In simple words: Arrow A shows the horse pushing backward against the dirt, which is the action force.

Exam Tip: The backward push on the ground constitutes the action; the forward push from the ground onto the feet represents the reaction.

Question 3. Out of these which one is not an example of Newton's third law of motion :
(a) Motion of an aeroplane
(b) Walking of man
(c) Flight of jet
(d) Catching the ball by a cricketer
Answer: (d) Catching the ball by a cricketer
Pulling one's hands backward while receiving a fast-moving cricket ball demonstrates impulse and momentum variation per Newton's second law of motion, not the third law.
In simple words: Cushioning a ball catch illustrates the second law of motion by reducing impact force over extra time.
Exam Tip: Questions involving prolonging stopping time to lessen impact force relate to Newton's second law (\( F = \Delta p / \Delta t \)).

Question 4. If action and reaction were to act on the same body :
(a) the resultant would be zero
(b) the body would not move at all
(c) both (a) and (b) are correct
(d) neither (a) nor (b) is correct
Answer: (c) both (a) and (b) are correct
Were two forces of identical magnitude and opposite orientation applied to one solitary object, their net sum would vanish, keeping the object permanently stationary.
In simple words: Equal opposite forces applied to the same item cancel out completely, preventing any acceleration.

Exam Tip: Balanced forces act on one body and yield zero net force; action-reaction pairs act on two different bodies.

Question 5. When we kick a stone, we get hurt. Due to which one of the following properties of the stone does it happen ?
(a) Inertia
(b) Velocity
(c) Reaction
(d) Momentum
Answer: (c) Reaction
Striking a stone subjects it to an impact; per Newton's third law, the stone strikes our foot simultaneously with an identical opposing force, causing pain.
In simple words: The foot hurts because the stone delivers an equal reaction force right back onto our toes.

Exam Tip: The sensation of pain experienced during kicking or pushing a rigid barrier stems directly from the reaction force.

Question 6. A man is standing in a boat in still water. If he tries to walk towards the shore, the boat will :
(a) move away from the shore
(b) remain stationary
(c) sink
(d) move towards the shore
Answer: (a) move away from the shore
To step forward toward the dry land, the person must push backward on the boat hull; this action propels the floating craft away from the bank.
In simple words: Walking forward pushes the boat backward into the deeper water.

Exam Tip: The backward displacement of the boat directly illustrates the reaction to the passenger's forward propulsive step.

Question 7. A fisherman is stranded in a lake because the motor of his motor-boat has failed. What should he do come to the shore ? He should start :
(a) throwing the fish he has collected, away from the shore
(b) throwing the fish he has collected, towards the shore
(c) walking in his boat towards the shore
(d) crying for help
Answer: (a) throwing the fish he has collected, away from the shore
Hurling the catch in the direction pointing away from the land exerts a forward reaction push onto the fisherman and his vessel, driving them toward the bank.
In simple words: Tossing objects away from the coast pushes the boat toward the shore by reaction force.

Exam Tip: Applying conservation of momentum or Newton's third law shows that backward momentum of thrown mass yields forward boat motion.

Question 8. During a planned manoeuvre in a space flight, a free-floating astronaut A pushes another free-floating astronaut B, the mass of A being greater than that of B. Then, the magnitude of the force exerted by the astronaut A on astronaut B is
(a) equal to zero
(b) equal to the force exerted by B on A
(c) greater than the force exerted by B on A
(d) less than the force exerted by B on A
Answer: (b) equal to the force exerted by B on A
Irrespective of discrepancies in their individual masses, the mutual contact forces experienced by the two astronauts remain strictly identical in magnitude per Newton's third law.
In simple words: Even if one person is heavier, both astronauts feel the exact same strength of push against each other.

Exam Tip: Unequal masses do not change the equality of action and reaction forces; they only produce unequal accelerations.

Question 9. In the above question during the push :
(a) the acceleration of A is greater than that of B
(b) the acceleration of A is less than that of B
(c) neither is acclerated
(d) their accelerations are equal in magnitude but opposite in direction
Answer: (b) the acceleration of A is less than that of B
Since \( a = F/m \) and both individuals experience matching force magnitudes, the more massive astronaut A acquires a smaller rate of acceleration than astronaut B.
In simple words: The heavier astronaut speeds up more slowly than the lighter astronaut under the same push.

Exam Tip: Use the inverse relationship between acceleration and mass (\( a \propto 1/m \)) when forces are identical.

Question 10. It is difficult to walk on ice because of :
(a) more inertia
(b) more friction
(c) absence of friction
(d) absence of inertia
Answer: (c) absence of friction
Slick ice provides negligible frictional resistance, preventing the feet from gripping and exerting a backward action push; in turn, no forward reaction force is generated.
In simple words: Without friction, our feet cannot push backward against the ground, so we cannot move forward.

Exam Tip: Friction provides the essential grip required to produce the backward action force needed for walking.

Question 11. Two balls at the same temperature collide. What is conserved ?
(a) Momentum
(b) Velocity
(c) Temperature
(d) Kinetic energy
Answer: (a) Momentum
In all physical impacts occurring in an isolated system without external forces, total linear momentum is strictly conserved.
In simple words: Total momentum always stays constant during any collision.

Exam Tip: Total momentum is conserved in all collisions, whereas kinetic energy is only conserved in perfectly elastic collisions.

Question 12. A rocket or a jet engien works on the principle of :
(a) conservation of energy
(b) conservation of momentum
(c) conservation of mass
(d) Newton's second law of motion
Answer: (b) conservation of momentum
Jet turbines and rocket nozzles operate by discharging exhaust gases rearward at high velocity, imparting forward momentum to the craft per momentum conservation and Newton's third law.
In simple words: High-speed gas shooting backward propels the rocket forward to conserve total momentum.
Exam Tip: Conservation of momentum and Newton's third law of motion are fundamentally equivalent descriptions of rocket propulsion.

Question 13. Force of action and reaction on object are :
(a) equal
(b) opposite
(c) balanced
(d) equal and opposite
Answer: (d) equal and opposite
Newton's third law establishes that interacting entities generate forces having equal numerical strength and contrary orientations.
In simple words: Action and reaction forces have identical sizes but point in opposite directions.

Exam Tip: Always select the choice specifying both properties ("equal and opposite") rather than incomplete answers mentioning only one.

Question 14. Consider two spring balances hooked as shown in Fig. 9.3. We pull them in opposite directions. If the reading shown by A is 1.5 N, the reading shown by B will be
(a) 1.5 N
(b) 2.5 N
(c) 3.0 N
(d) zero
Answer: (a) 1.5 N
Because balance A exerts a pull of 1.5 N on balance B, balance B simultaneously pulls balance A with a reaction force of identical magnitude (1.5 N).
In simple words: The tension along connected balances is identical, so balance B reads exactly 1.5 N like balance A.
Exam Tip: Connected spring balances experience equal internal tension and will always display identical values.

Question 15. A loded gun has a bullet inside it. When the gun's trigger is pressed, the power inside cartrige, it is the example of a
(a) First law of motion
(b) Second law of motion
(c) Third law of motion
(d) Zeroth law of thermodynamics
Answer: (c) Third law of motion
Expanding gunpowder forces the projectile outward at high velocity (action), simultaneously thrusting the weapon backward against the shooter's shoulder (reaction).
In simple words: The bullet flying forward and the gun kicking backward is a direct example of Newton's third law.

Exam Tip: Recoil of a firearm is one of the classic textbook examples used to demonstrate Newton's third law of motion.

Experiment 10

Objective

To determine the percentage of water imbibed by raisins.

Materials Required

• Weight box
• Physical balance
• Raisins (5-10 g) with intact stalk
• 250 ml beaker
• Water
• Petridish
• Filter paper

Theory

• Osmosis is the spontaneous passage of solvent molecules across a selectively permeable membrane toward a more concentrated solution.
• The minimum mechanical pressure needed to stop this inward osmotic migration is known as osmotic pressure.
• Commercial raisins represent dehydrated grape berries whose cell membranes remain intact.
• Submerging dried raisins in pure water causes them to swell through the inward movement of water molecules known as endosmosis.
• A selectively permeable membrane restricts the movement of dissolved solutes while facilitating the transit of water molecules from a hypotonic region to a hypertonic environment.
• Endosmosis denotes inward water influx causing cellular swelling, whereas exosmosis describes outward water egress resulting in shrinkage.

Procedure

• Select healthy dry raisins possessing unbroken stalks and record their initial mass using a balance.
• Submerge the weighed raisins in a beaker filled with approximately 100 ml of distilled water.
• Allow the sample to soak undisturbed for roughly 24 hours (or overnight).
• Remove the turgid raisins and gently blot their surface with filter paper sheets to eliminate clinging water droplets.
• Weigh the hydrated raisins again and calculate the percentage increase in mass.

Observation

1. Mass of dry raisins taken (x₁) = ......... g.
2. Mass of soaked, swollen raisins (x₂) = ......... g.

Calculations

1. Mass of water absorbed by raisins = (x₂ - x₁) g = ......... g
2. Percentage of water absorbed = \( \frac{x_2 - x_1}{x_1} \times 100 \) = ......... %

Inference

The percentage of water absorbed by the raisins through endosmosis is ......... %.

Precautions

• Select healthy raisins with dry, undamaged skins and intact stalks to prevent solute leakage.
• Ensure all raisins remain fully covered beneath the water surface throughout the soaking interval.
• Allow sufficient immersion time (overnight or 24 hours) for maximum absorption.
• Before taking the second weight, carefully roll the raisins over absorbent paper to remove external liquid without crushing them.

Note for the Teacher

• Adsorption describes the physical surface accumulation of liquid layers on solids.
• Dry chickpeas or gram seeds can be substituted effectively for raisins in this experiment.
• The initial rapid intake of moisture by dry colloidal plant walls is termed imbibition.

Viva-Voce

Question 1. What is osmosis ?
Answer: Osmosis represents the spontaneous transit of water molecules from a diluted (hypotonic) medium into a more concentrated (hypertonic) solution across a semipermeable boundary.
In simple words: Osmosis is water moving from a watery area into a denser sugary area through a fine filter skin.

Exam Tip: Mentioning the presence of a "semipermeable membrane" is mandatory to differentiate osmosis from general diffusion.

Question 2. What is endosmosis ?
Answer: Endosmosis identifies the inward osmotic migration of water molecules into a cell or tissue across its semipermeable membrane when placed in a hypotonic medium.
In simple words: Endosmosis is water entering and swelling up a cell from the outside.

Exam Tip: Remember the prefix: "Endo" signifies entering/inward transit, leading to increased cell volume and turgor.

Question 3. How does water enter into the raisins ?
Answer: Water enters the raisin cells primarily via endosmosis because the external water bath is hypotonic compared to the concentrated internal sugars.
In simple words: Water enters raisins by endosmosis because the sugary inside draws water through the skin.

Exam Tip: State that the sugar concentration inside the raisin exceeds that of pure water outside, driving endosmosis.

Question 4. What is isotonic solution ?
Answer: An isotonic solution possesses an osmotic concentration perfectly matching the internal solute concentration of the immersed cell, resulting in zero net water movement.
In simple words: An isotonic liquid has the same strength as the cell fluid, so the cell neither swells nor shrinks.

Exam Tip: Specify that cell volume remains completely unchanged in an isotonic solution because water influx equals water efflux.

Question 5. What is hypotonic solution ?
Answer: A hypotonic solution features a lower solute concentration and higher water potential than the internal cell sap, encouraging water to move inside.
In simple words: A hypotonic liquid is more watery than the cell inside, causing water to rush into the cell.

Exam Tip: Pure distilled water represents the ideal hypotonic medium used to demonstrate endosmosis.

Question 6. Define hypertonic solution.
Answer: A hypertonic solution exhibits a higher solute concentration and lower water potential than the internal cell contents, drawing water out of the cell.
In simple words: A hypertonic liquid is very concentrated, which draws moisture out of the cell.

Exam Tip: A concentrated brine or sugar syrup serves as a standard hypertonic medium in laboratory experiments.

Question 7. What will happen if grapes dipped into concentrated solution of sugar for sometime ? Why does it happen ?
Answer: The grape berries undergo shrinkage. Because the surrounding sugar syrup is hypertonic relative to the juice inside the grape, water exits through the outer skin via exosmosis.
In simple words: Grapes shrink in thick sugar syrup because their internal water gets sucked out by exosmosis.

Exam Tip: Identify the outer skin of the grape as the semipermeable barrier and cite exosmosis as the operative mechanism.

Question 8. What is exosmosis ?
Answer: Exosmosis defines the outward movement of water across a semipermeable membrane from a cell into a hypertonic surrounding environment.
In simple words: Exosmosis is water exiting a cell, causing it to deflate or wrinkle.

Exam Tip: Associate "Exo" with exit/outward flow, which leads to cell flaccidity and plasmolysis.

Question 9. What is plasmolysis ?
Answer: Plasmolysis denotes the shrinkage and withdrawal of the living protoplast away from the rigid plant cell wall following excessive water loss in a hypertonic medium.
In simple words: Plasmolysis happens when a plant cell loses so much water that its inner contents pull away from the wall.

Exam Tip: Emphasize that plasmolysis occurs specifically in walled cells such as plants, bacteria, and fungi.

Question 10. Name two important aspects in which plants use osmosis process.
Answer: Major roles include:
(i) Uptake of ground moisture by root hair cells from the surrounding soil.
(ii) Intercellular distribution of water and solutes throughout vegetative tissues to maintain turgor pressure.
In simple words: Plants use osmosis to drink water from the soil and to keep their stems and leaves upright and firm.

Exam Tip: Mentioning "absorption by root hairs" and "maintenance of cell turgidity" guarantees full marks.

Question 11. Define imbibition.
Answer: Imbibition is a distinctive type of absorption where solid or colloidal substances take up liquids without forming a true solution, independent of any semipermeable membrane.
In simple words: Imbibition is a dry solid soaking up liquid and swelling, like dry seeds or wooden doors in rain.

Exam Tip: Clarify that imbibition does not require a semipermeable membrane, whereas osmosis strictly does.

Question 12. Turgidity represents a condition in the cell when it is fully stretched. Why does a cell not burst after attaining turgidity ?
Answer: As internal hydrostatic pressure (turgor pressure) rises, the rigid plant cell wall exerts an equal inward mechanical resistance (wall pressure) that halts further water inflow.
In simple words: The tough outer plant cell wall pushes back against the water, stopping the cell from bursting.

Exam Tip: Key terms examiners look for: "turgor pressure" balanced by "wall pressure".

Question 13. How is osmosis different from diffusion ?
Answer: In osmosis, only solvent molecules move across a mandatory semipermeable barrier from lower to higher solute concentration. In contrast, diffusion involves the unrestricted dispersion of any particle type (solute or solvent) directly down its concentration gradient without needing a membrane.
In simple words: Osmosis requires a semipermeable skin for water to pass through, while diffusion lets particles spread freely anywhere.

Exam Tip: Provide two distinct contrast points: the presence/absence of a semipermeable membrane, and the types of diffusing particles.

Multiple Choice Questions

Question 1. Osmosis is the :
(a) Process of selective transmission of a solvent in preference to another
(b) Process of selective transmission of a solvent (water) in preference to the solute through semipermeable membrane
(c) Both of the options
(d) None of the options
Answer: (b) Process of selective transmission of a solvent (water) in preference to the solute through semipermeable membrane
Osmosis strictly describes the selective permeation of solvent molecules through a semipermeable barrier while solute molecules are barred.
In simple words: Osmosis is the movement of water through a selective membrane that blocks the dissolved particles.

Exam Tip: Look for the phrase "through semipermeable membrane" to identify the correct formal definition of osmosis.

Question 2. Water enters into the raisins by the process of :
(a) Osmosis
(b) Endosmosis
(c) Exosmosis
(d) None of the options
Answer: (b) Endosmosis
Because water flows inward into the concentrated raisin cells, the specific biological term is endosmosis.
In simple words: When water moves into a dried fruit to make it swell, it is called endosmosis.

Exam Tip: When both "Osmosis" and "Endosmosis" appear among options, choose "Endosmosis" as the more precise biological description.

Question 3. Endosmosis is defined as :
(a) The inflow of water by the cell
(b) The inflow of water through a permeable membrane
(c) The inflow of water through a semipermeable membrane
(d) None of the options
Answer: (c) The inflow of water through a semipermeable membrane
Endosmosis demands both an inward trajectory of water and transit across a semipermeable membrane.
In simple words: Endosmosis means water flowing into a cell through a semipermeable filter wall.

Exam Tip: A permeable membrane allows both solute and solvent through; only a semipermeable membrane facilitates osmosis.

Question 4. What is exosmosis ?
(a) Outflow of water from the cell
(b) Outflow of water from the cell through a permeable membrane
(c) Outflow of water from the cell through a semipermeable membrane
(d) None of the options
Answer: (c) Outflow of water from the cell through a semipermeable membrane
Exosmosis describes the outward passage of water across a semipermeable membrane toward an exterior hypertonic medium.
In simple words: Exosmosis is water leaving the cell through its selective boundary layer.

Exam Tip: Pair "Exo" with exit/outflow through a semipermeable membrane to avoid confusion with basic seepage.

Question 5. Percentage of water absorbed by raisins is calculated by a formula :
(a) \( \frac{\text{Increase in wt. of raisins}}{\text{Weight of water taken}} \times 100 \)
(b) \( \frac{\text{Increase in wt. of raisins}}{\text{Weight of raisins taken}} \times 100 \)
(c) \( \frac{\text{Increase in wt. of raisins}}{\text{Weight of water absorbed}} \times 100 \)
(d) None of the options
Answer: (b) \( \frac{\text{Increase in wt. of raisins}}{\text{Weight of raisins taken}} \times 100 \)
The percentage gain in mass compares the absorbed water mass directly against the baseline initial dry mass of the raisins.
In simple words: Divide the gained weight by the starting dry weight of the raisins, then multiply by 100.

Exam Tip: The denominator in percentage calculations is always the initial dry weight of the raisins (\( x_1 \)), never the volume of water.

Question 6. Which is not correct regarding the absorption of water by raisins ?
(a) Weighing must be accurate
(b) Petridish must have sufficient water
(c) Raisins must be without stalk
(d) Raisins must be wiped off gently before weighing
Answer: (c) Raisins must be without stalk
Raisins must have intact stalks attached; missing stalks tear the skin and cause sugar solute leakage instead of pure endosmosis.
In simple words: Raisins must keep their stalks; missing stalks tear the fruit skin and ruin the experiment.

Exam Tip: Intact stalks are an essential precaution in this practical to preserve the integrity of the cell membranes.

Question 7. In determining the percentage of water absorbed by raisins, which is necessary :
(a) Raisins should be clean and completely immersed into water
(b) Raisins should be soaked in water for sufficient time (overnight)
(c) Before weighing, gently dry the raisins with the help of filter paper
(d) All of the options
Answer: (d) All of the options
All these procedural precautions are required to guarantee uniform hydration and prevent extra mass from clinging surface moisture.
In simple words: Every listed step is needed to ensure the experiment gives accurate results.

Exam Tip: Always blot off surface moisture gently; rubbing aggressively can rupture the turgid skin and reduce sample mass.

Question 8. What will happen if grapes dipped into concentrated solution of sugar for sometime ?
(a) The grapes swell
(b) The grapes shrink
(c) No change in grapes
(d) Swell in the beginning then shrink
Answer: (b) The grapes shrink
Because the external syrup is hypertonic compared to the grape interior, water leaves the fruit cells via exosmosis, causing shrinkage.
In simple words: Dense syrup pulls water out of the grapes, making them wrinkle and shrink.

Exam Tip: A hypertonic environment always draws water out of living plant cells via exosmosis.

Question 9. A student dissolved 1 g of sugar in 100 ml of distilled water in beaker A. She dissolved 100 g of sugar in 100 ml of distilled water in beaker B. Then she dropped a few raisins of equal weight in each beaker. After overnight she found the raisins in A swollen and those in B shrunken. The inference drawn is that :
(a) Sugar concentration of raisins is lower than that of solution A and higher than that of solution B
(b) Sugar concentration of raisins is higher than that of solution A and lower than that of solution B
(c) In B the cell membrane of raisins was damaged resulting in leaching
(d) In A the permeability to water of the cell membrane of raisins was enhanced
Answer: (b) Sugar concentration of raisins is higher than that of solution A and lower than that of solution B
Swelling in beaker A indicates solution A is hypotonic (raisin sap is more concentrated). Shrinkage in beaker B indicates solution B is hypertonic (raisin sap is less concentrated).
In simple words: The raisins are sweeter than weak liquid A so they take in water, but less sweet than syrup B so they lose water.

Exam Tip: Water always moves from lower solute concentration (hypotonic) to higher solute concentration (hypertonic).

Question 10. What are the important aspects in which plants use osmosis process ?
(a) Plant root hairs absorb water from the soil by osmosis
(b) Movement of water/solution from cell to cell in the body of plants takes place by osmosis
(c) Both of the options
(d) None of the options
Answer: (c) Both of the options
Plants employ osmosis both to draw ground water into root hairs and to conduct cell-to-cell water distribution throughout plant tissues.
In simple words: Plants rely on osmosis both for absorbing soil water and for transporting moisture between cells.

Exam Tip: Root absorption and internal fluid transport are the two primary biological applications of osmosis in botany.

Question 11. The grapes shrink when dipped into conc. sugar solution because :
(a) Solution in grapes is dilute than sugar solution outside
(b) Solution in grapes in conc. than sugar solution outside
(c) Concentration of solution in grapes and sugar solution is same
(d) Concentration of solution in grapes and sugar is different
Answer: (a) Solution in grapes is dilute than sugar solution outside
The sap inside the grape has a lower solute concentration (more dilute) than the exterior syrup, creating a gradient that drives water outwards.
In simple words: The liquid inside the grape is more watery than the thick syrup outside, so water flows out.

Exam Tip: Exosmosis always occurs when the internal cell sap is more dilute than the external solution.

Question 12. The grapes shrink when dipped in concentrated sugar solution, then peeling of the grapes act as :
(a) Semipermeable membrane
(b) Permeable membrane
(c) Non-permeable
(d) None of the options
Answer: (a) Semipermeable membrane
The outer skin of the grape functions as a natural selectively permeable boundary that permits solvent transit while blocking sugar molecules.
In simple words: The grape skin acts like a semipermeable filter that lets water pass but keeps sugar locked out.

Exam Tip: The living epidermal layer of fruit acts as a natural semipermeable membrane.

Question 13. A student dissolved 1 g of sugar in 100 ml of distilled water in a beaker A. He dissolved 10 g of sugar in 1000 ml of distilled water in beaker B. Then he dropped a few raisins, in each. After overnight he found the raisins :
(a) swollen in A and shrunken in B
(b) shrunken in A and swollen in B
(c) swollen in both
(d) shrunken in both
Answer: (c) swollen in both
Both beakers have an identical concentration of 1% sugar (1 g/100 ml = 10 g/1000 ml). Because both solutions are highly dilute compared to the raisin sap, endosmosis occurs in both vessels.
In simple words: Both beakers have the exact same weak sugar strength (1%), so raisins swell in both.

Exam Tip: Check the concentration ratio (\( \text{mass}/\text{volume} \)) rather than absolute mass when comparing solutions.

Question 14. The pressure which is just sufficient to stop the flow of solvent (water) from weaker solution to concentrated solution through semipermeable membrane is termed as :
(a) Atmospheric pressure
(b) Osmotic pressure
(c) Vapour pressure
(d) Exopressure
Answer: (b) Osmotic pressure
Osmotic pressure is defined as the exact hydrostatic pressure required to prevent the inward passage of pure solvent across a semipermeable membrane.
In simple words: Osmotic pressure is the counter-push needed to halt the incoming water flow through the membrane.

Exam Tip: Higher solute concentration produces higher osmotic pressure.

Question 15. Which balance will you use to weigh the raisins ?
(a) Chemical balance
(b) Physical balance
(c) Spring balance
(d) Electronic balance
Answer: (b) Physical balance
A physical balance is the standard school laboratory equipment specified for measuring small solid masses in practical biology exams.
In simple words: A physical balance is standard for measuring the weight of raisins in the laboratory.

Exam Tip: Spring balances measure force/weight in newtons, whereas a physical balance measures mass in grams.

Question 16. A student soaked 10 g of raisins in 100 ml of distilled water in two beakers A and B each. She maintained beaker A at 30°C and beaker B at 60°C. After two hours, the percentage of water absorbed will be :
(a) the same in both A and B
(b) more in A than in B
(c) more in B than in A
(d) exactly twice as much in B as in A
Answer: (c) more in B than in A
Elevated temperature increases the kinetic energy of water molecules and boosts membrane permeability, accelerating the rate of endosmosis in beaker B.
In simple words: Warmer water speeds up absorption, so raisins soak up more water at 60°C than at 30°C.

Exam Tip: Diffusion and osmosis rates increase with rising temperature due to enhanced kinetic energy.

Question 17. If the wt. of raisins taken 7 gm and weight of swollen raisins x gm then the amount of water absorbed by raisins will be :
(a) (7 - x) gm
(b) (x - 7) gm
(c) (2x - 7) gm
(d) None of the options
Answer: (b) (x - 7) gm
The absorbed water mass equals the final swollen mass minus the initial dry mass, giving \( (x - 7)\text{ g} \).
In simple words: Subtract the starting 7 grams from the final swollen weight x to find the water absorbed.

Exam Tip: Water absorbed is always final mass minus initial mass: \( x_2 - x_1 \).

Question 18. In the determination of percentage of water absorbed by raisins, for 5 gm raisins the amount of water required is nearly
(a) 5 ml
(b) 100 ml
(c) 10 ml
(d) any amount can be taken
Answer: (b) 100 ml
A standard volume of roughly 100 ml ensures that the raisins remain completely submerged throughout their swelling period.
In simple words: Around 100 ml of water gives plenty of liquid so the raisins stay fully submerged.

Exam Tip: Using too little water (like 5 ml or 10 ml) risks the raisins absorbing all the water before reaching full turgidity.

Question 19. The grapes shrink when dipped in concentrated sugar solution for sometime. It is because of :
(a) Water from sugar solution flows into grapes and grapes shrink
(b) Now flow of water from sugar solution into grapes
(c) Water from grapes flows into concentrated sugar solution and grapes shrink
(d) None of the options
Answer: (c) Water from grapes flows into concentrated sugar solution and grapes shrink
Water exits the grape cells via exosmosis into the exterior hypertonic sugar syrup, resulting in grape shrinkage.
In simple words: Water moves out of the grapes into the strong syrup, making them shrivel.

Exam Tip: In exosmosis, net water movement is directed from the interior of the cell to the outside.

Question 20. Student A, B and C were given five raisins each of equal weight. The raisins were soaked in distilled water at room temperature. A removed the raisins after 10 minutes, B after overnight and C after 60 minutes. If P_A, P_B and P_C denote percentage absorption of water obtained by students A, B and C respectively, then :
(a) P_A > P_B > P_C
(b) P_A < P_B < P_C
(c) P_A < P_B > P_C
(d) P_A = P_B = P_C
Answer: (c) P_A < P_B > P_C
Absorption increases with prolonged soaking time until saturation. Thus, after 10 minutes absorption is minimal, after 60 minutes it is intermediate, and overnight it reaches maximum: \( P_A < P_C < P_B \), which means \( P_B \) is greater than both \( P_A \) and \( P_C \).
In simple words: Raisins soak up more water the longer they sit, so the overnight sample absorbs the most.

Exam Tip: Water absorption by endosmosis is time-dependent until maximum cellular turgidity is reached.

Question 21. What are the two types of osmosis ?
(a) Osmosis and reosmosis
(b) Osmosis and endosmosis
(c) Osmosis and exosmosis
(d) Endosmosis and exosmosis
Answer: (d) Endosmosis and exosmosis
Osmosis is classified based on water flow direction into endosmosis (inward movement) and exosmosis (outward movement).
In simple words: The two types of osmosis are water moving inward (endosmosis) and water moving outward (exosmosis).

Exam Tip: Endosmosis leads to turgidity; exosmosis leads to flaccidity and plasmolysis.

Question 22. In the determination of % of water absorbed by raisins, raisins should be soaked in water for :
(a) 2-5 hr
(b) 24 hr
(c) 5-10 hr
(d) 1/2 hr
Answer: (b) 24 hr
Soaking raisins for approximately 24 hours (or overnight) ensures complete endosmosis and full cellular saturation.
In simple words: Soaking for 24 hours ensures the raisins absorb as much water as possible.

Exam Tip: Overnight or 24 hours is the standard duration prescribed in the CBSE Class 9 science lab manual.

Question 23. To demonstrate the absorption of water raisins should be immersed into water :
(a) Completely
(b) Partially
(c) Both of the options
(d) None of the options
Answer: (a) Completely
Complete immersion provides uniform exposure over the entire surface area of each raisin for even endosmosis.
In simple words: The raisins must be totally covered in water so every side absorbs moisture evenly.

Exam Tip: Partial immersion leaves upper cells unhydrated, causing inconsistent experimental data.

Question 24. To determine the % of water absorbed the raisins, before weighing are dried by :
(a) clothes
(b) filter paper
(c) air
(d) sunlight
Answer: (b) filter paper
Gentle blotting with filter paper strips away external surface droplets without puncturing the delicate outer skin.
In simple words: Filter paper dabs off outside water drops cleanly without damaging the swollen raisins.

Exam Tip: Never dry swollen raisins in sunlight or hot air; that would reverse the process via evaporation and exosmosis.