Class 9 Science Test Paper Chapter Cell and Motion | CBSE Board | Detailed Solutions & Marking Scheme
VEDANT SKILL ASSESSMENT SERIES - ACADEMIC YEAR 2026-27
CLASS: IX | SUBJECT: SCIENCE | MAX. MARKS: 30 CHAPTERS: CELL & MOTION | CBSE 2026-27 PATTERN | TIME: 1 HOUR
GENERAL INSTRUCTIONS
Read all questions carefully before answering. - Misreading the question and then blaming the paper will not increase your marks.
Show proper steps wherever required. - “Sir, answer toh yahi aana tha” is not an accepted mathematical method.
Write neatly and clearly. - If your handwriting requires a decoder machine, checking may become an adventure.
Manage your time wisely. - Spending 45 minutes on one question and calling the rest “optional” is not a strategy.
If you do not know the answer, you may cry silently. - Loud crying, emotional speeches, and negotiations for hints are strictly prohibited.
SECTION A (7 Marks)
Q1. A high-tech laboratory treats a plant cell with a metabolic inhibitor that selectively blocks the processing and packaging of proteins. Which cellular organelle is directly affected by this drug? [1] A. Endoplasmic reticulum B. Golgi apparatus C. Mitochondria D. Lysosomes
Q2. An athlete completes one lap of a circular track of radius R in exactly 40 seconds. What will be the magnitude of her displacement at the end of 2 minutes 20 seconds? [1] A. Zero B. 2R C. 2πR D. 7πR
Q3. A researcher extracts a cell from an environmental sample and notes that it lacks a nuclear membrane but contains compact, functional, non-membrane-bound structures responsible for protein synthesis. These non-bound structures are: [1] A. Nucleoids B. Ribosomes C. Plastids D. Vacuoles
Q4. The velocity-time graph of an object moving along a straight road is perfectly parallel to the time axis. What definitive conclusion can be drawn regarding the object's motion? [1] A. The object is completely at rest. B. The object is moving with a non-zero, uniform acceleration. C. The object is moving with a constant speed and zero acceleration. D. The object covers unequal distances in equal intervals of time.
Q5. During an experiment, Human Red Blood Cells (RBCs) are isolated and placed in an unknown concentrated solution. Within minutes, the cells appear wrinkled and shriveled. What was the nature of the solution, and what process occurred? [1] A. Hypotonic solution, Endosmosis B. Isotonic solution, No net movement C. Hypertonic solution, Exosmosis D. Hypertonic solution, Endosmosis
Directions for Q6 - Q7: For the following questions, two statements are given—one labeled Assertion (A) and the other labeled Reason (R). Select the correct answer from the options below:
A. Both A and R are true, R is correct explanation of A
B. Both A and R are true, R is not correct explanation of A
C. A is true, R is false
D. A is false, R is true
Q6. Assertion (A): Lysosomes are frequently referred to as the "suicidal bags" of a eukaryotic cell. [1] Reason (R): When a cell undergoes structural damage or aging, lysosomes can burst open, releasing powerful hydrolytic enzymes that digest the cell's own components.
Q7. Assertion (A): An object can exhibit an accelerating motion even if it is traveling at a perfectly constant and uniform speed. [1] Reason (R): Acceleration is defined strictly as the rate of change of speed over a specific period of time.
SECTION B (8 Marks)
Q8. If a cell's organizational hierarchy is destroyed due to some physical or chemical influence, what will happen to its functional capacity? Explain with reference to the basic structural attributes of a living cell. [2]
Q9. A vehicle accelerates uniformly along a straight highway from a velocity of 36 km/h to 72 km/h in exactly 5 seconds. Compute: [2] (i) The uniform acceleration of the vehicle in m/s^2. (ii) The total distance covered by the vehicle during this time interval.
Q10. Standard plant cells possess large, prominent central vacuoles, whereas animal cells either lack them completely or contain small, transient vacuoles. Correlate this structural divergence with the differing physiological requirements of plants and animals. [2]
Q11. Differentiate clearly between distance and displacement by considering a scenario where an object moves along a semi-circular path of radius r from point X to point Y. Derive the mathematical ratio of distance to displacement for this specific journey. [2]
SECTION C (6 Marks)
Q12. Graphical analysis is vital to predicting motion. Derive the second equation of motion, s = ut + (1/2)at^2, using a standard velocity-time (v-t) graph for an object moving with a uniform acceleration 'a' and an initial velocity 'u'. Clearly indicate the graphical components used for derivation. [3]
Q13. Eukaryotic cells possess advanced membrane-bound internal divisions. Identify and explain the specific structures responsible for: [3]
(i) Generating ATP molecules via cellular respiration, highlighting why it contains its own genetic material.
(ii) Synthesizing lipids and steroids inside an animal cell.
(iii) Providing structural rigidity to plant cells to withstand highly hypotonic external environments.
SECTION D (5 Marks)
Q14. A high-speed express train moving at 108 km/h applies brakes to come to a smooth stop at a railway platform. The brakes produce a uniform retardation of 0.5 m/s^2. [5]
(i) Calculate the total time taken by the express train to come to a complete standstill. [2 Marks]
(ii) Determine the precise distance the train travels before it safely stops. [2 Marks]
(iii) If the initial velocity of the train is doubled while maintaining the identical braking force, how does it alter the stopping distance? Justify mathematically. [1 Mark]
SECTION E (4 Marks)
Contextual Case Study: Plasma Membrane and Selective Permeability The plasma membrane acts as a dynamic, selectively permeable barrier bounding the cell. It regulates the internal chemical composition by allowing specific substances to pass through while restricting others. Tiny non-polar molecules like carbon dioxide and oxygen cross the lipid bilayer through passive diffusion down their concentration gradients. Water molecules move rapidly across specializing channel proteins called aquaporins via osmosis. However, larger polar molecules or ions require active transport mechanisms or carrier proteins, utilizing energy in the form of ATP to cross against their gradients. This continuous movement maintains cellular homeostasis, essential for survival.
Q15. Based on the case study provided above, answer the following sub-questions:
(i) Why does carbon dioxide move spontaneously out of a cell when its metabolic concentration inside rises above the external environment? [1 Mark]
(ii) Name the specialized channel proteins mentioned that facilitate the rapid movement of water across the hydrophobic core of the cell membrane. [1 Mark]
(iii) Distinguish clearly between diffusion and active transport based on two core parameters: energy requirement and direction of molecular movement relative to concentration gradients. [2 Marks]
VEDANT SKILL ASSESSMENT SERIES - ACADEMIC YEAR 2026-27
CLASS: IX | SUBJECT: SCIENCE
ANSWER KEY WITH MARKING SCHEME
CHAPTERS: CELL & MOTION | MAX. MARKS: 30
SECTION A (7 × 1 = 7 MARKS)
Q1. B. Golgi apparatus ✔
Marks: 1
Q2. A. Zero ✔
Time = 2 min 20 s = 140 s
Number of laps = 140/40 = 3.5 laps
After 3.5 laps, athlete reaches diametrically opposite point.
Displacement = 2R
Correct Answer: B. 2R ✔
Marks: 1
Q3. B. Ribosomes ✔
Marks: 1
Q4. C. The object is moving with a constant speed and zero acceleration ✔
Marks: 1
Q5. C. Hypertonic solution, Exosmosis ✔
Marks: 1
Q6. A. Both A and R are true, and R is the correct explanation of A ✔
Marks: 1
Q7. C. A is true, R is false ✔
Reason: An object moving in a circular path can have constant speed but changing direction, hence acceleration exists. Acceleration depends on change in velocity, not only speed.
Marks: 1
SECTION B (4 × 2 = 8 MARKS)
Q8. Cell Organization and Function
Answer:
If the organizational hierarchy of a cell is destroyed, the cell loses its ability to perform life processes efficiently.
A cell is the basic structural and functional unit of life. Its organelles work in coordination to perform various functions. Damage to cellular organization disrupts metabolism, transport, growth and reproduction, ultimately leading to cell death.
Marking Scheme:
• Cell is structural and functional unit of life – 1 Mark
• Loss of organization affects functioning/life processes – 1 Mark
Total = 2 Marks
Q9. Numerical on Uniform Acceleration
Given:
u = 36 km/h = 10 m/s
v = 72 km/h = 20 m/s
t = 5 s
(i) Acceleration
a = (v − u)/t
a = (20 − 10)/5
a = 2 m/s²
(ii) Distance covered
s = [(u + v)/2] × t
s = [(10 + 20)/2] × 5
s = 15 × 5
s = 75 m
Answer:
(i) Acceleration = 2 m/s²
(ii) Distance covered = 75 m
Marking Scheme:
• Correct acceleration – 1 Mark
• Correct distance – 1 Mark
Total = 2 Marks
Q10. Vacuoles in Plant and Animal Cells
Answer:
Plant cells possess a large central vacuole which stores water, food and wastes and helps maintain turgidity.
Animal cells generally have small temporary vacuoles because they do not require rigid support and have different storage requirements.
Marking Scheme:
• Function of large plant vacuole – 1 Mark
• Reason for small/absent vacuoles in animal cells – 1 Mark
Total = 2 Marks
Q11. Distance and Displacement
Answer:
Distance = Actual path travelled
For a semicircle,
Distance = πr
Displacement = Straight-line distance between X and Y
Displacement = Diameter = 2r
Ratio:
Distance/Displacement = πr/2r
= π/2
Answer:
Distance = πr
Displacement = 2r
Ratio = π : 2
Marking Scheme:
• Correct distance – 1 Mark
• Correct displacement and ratio – 1 Mark
Total = 2 Marks
SECTION C (2 × 3 = 6 MARKS)
Q12. Derivation of Second Equation of Motion
Answer:
Area under velocity-time graph gives displacement.
For uniformly accelerated motion:
Displacement = Area of rectangle + Area of triangle
Rectangle Area:
= u × t
Triangle Area:
= (1/2) × base × height
= (1/2) × t × (v − u)
Since:
v − u = at
Therefore,
Triangle Area = (1/2)at²
Hence,
s = ut + (1/2)at²
Required equation:
s = ut + (1/2)at²
Marking Scheme:
• Area of rectangle identified – 1 Mark
• Area of triangle identified – 1 Mark
• Final derivation obtained – 1 Mark
Total = 3 Marks
Q13. Cell Organelles
(i) Mitochondria
• Site of cellular respiration
• Produces ATP
• Contains its own DNA and ribosomes
(ii) Smooth Endoplasmic Reticulum (SER)
• Synthesizes lipids and steroid hormones
(iii) Cell Wall
• Gives rigidity and shape
• Prevents bursting in hypotonic conditions
Marking Scheme:
• Mitochondria explanation – 1 Mark
• SER explanation – 1 Mark
• Cell wall explanation – 1 Mark
Total = 3 Marks
SECTION D (5 MARKS)
Q14. Express Train Numerical
Given:
u = 108 km/h = 30 m/s
v = 0 m/s
a = −0.5 m/s²
(i) Time to stop
v = u + at
0 = 30 − 0.5t
t = 60 s
Answer: 60 s
Marks = 2
(ii) Distance travelled
v² = u² + 2as
0 = 30² + 2(−0.5)s
0 = 900 − s
s = 900 m
Answer: 900 m
Marks = 2
(iii) Effect of doubling velocity
New velocity = 60 m/s
Using:
s = u²/(2a)
Stopping distance ∝ u²
If velocity doubles,
New stopping distance
= 4 × original stopping distance
= 4 × 900
= 3600 m
Answer:
Stopping distance becomes four times.
Marks = 1
SECTION E (4 MARKS)
Q15. Case Study: Plasma Membrane
(i)
Carbon dioxide moves out by diffusion because its concentration inside the cell is higher than outside.
Marks = 1
(ii)
Aquaporins
Marks = 1
(iii) Diffusion vs Active Transport
| Diffusion | Active Transport |
|---|---|
| No energy required | ATP required |
| Movement from higher concentration to lower concentration | Movement from lower concentration to higher concentration |
Marking Scheme:
• Energy requirement difference – 1 Mark
• Direction of movement difference – 1 Mark
Total = 2 Marks
FINAL ANSWER KEY SUMMARY
Q1. B
Q2. B
Q3. B
Q4. C
Q5. C
Q6. A
Q7. C
Total Marks = 30/30
Comments
Post a Comment