This is a fresh, 100% original Periodic Assessment 1 practice paper for Class 9 Science, written for the 2026-27 session and the new NCERT Exploration textbook. Every question is followed by a fully worked answer in the answer key at the end of the page, and both the question paper and the answer key can be downloaded as PDFs.
Periodic Assessment pattern — school pattern se thoda alag ho sakta hai.
Before You Start
- Time: 40 minutes
- Maximum marks: 25
- Chapter coverage: NCERT Exploration Chapters 1 to 4 — Entering the World of Secondary Science, Cell – The Building Block of Life, Tissues in Action, and Describing Motion Around Us.
- Every question carries a chapter tag such as [Ch 3]. If your school has covered fewer chapters, simply skip the questions tagged beyond your syllabus and adjust your total accordingly.
- All questions are compulsory. There is no overall choice.
- Show all steps in numerical answers. Marks are given for correct working, not just the final number.
Section A – Objective Type (5 × 1 = 5 marks)
Q1. [Ch 1] A student writes the length of a leaf as 8.4 cm. Written in the SI base unit of length, the same measurement is:
(a) 0.084 m (b) 0.84 m (c) 8.4 m (d) 84 m
Q2. [Ch 2] Which of these structures is found in a plant cell but not in an animal cell?
(a) Ribosome (b) Mitochondrion (c) Plastid (d) Golgi apparatus
Q3. [Ch 3] A gardener notices that a young plant is growing taller every week. The tissue directly responsible for this increase in height is:
(a) Lateral meristem (b) Apical meristem (c) Sclerenchyma (d) Parenchyma
Q4. [Ch 4] An athlete runs exactly one full round of a circular track of length 400 m and stops at the starting point. The distance covered and the magnitude of displacement are respectively:
(a) 400 m and 400 m (b) 400 m and 0 (c) 0 and 400 m (d) 0 and 0
Q5. [Ch 2] Which one of the following is a prokaryotic cell?
(a) Onion peel cell (b) Human cheek cell (c) Bacterial cell (d) Amoeba
Section B – Very Short Answer (4 × 2 = 8 marks)
Q6. [Ch 2] State any two differences between the cell wall and the plasma membrane.
Q7. [Ch 3] Name the tissue that (a) joins a muscle to a bone, and (b) forms the outermost protective covering of the skin. Give one feature of each.
Q8. [Ch 4] A cyclist covers 300 m in 40 s and then covers the next 200 m in 60 s along a straight road. Calculate her average speed for the whole journey.
Q9. [Ch 1] A student measures the time taken for 20 complete oscillations of a simple pendulum and records 32.0 s. Find the time period of the pendulum. Why does she time 20 oscillations instead of just one?
Section C – Short Answer (3 × 3 = 9 marks)
Q10. [Ch 2] State three differences between a plant cell and an animal cell.
Q11. [Ch 4] A car starts from rest and accelerates uniformly at 2 m/s2 for 8 s along a straight road. Calculate (a) its velocity at the end of 8 s, (b) the distance it travels in this time, and (c) its average velocity over the 8 s.
Q12. [Ch 3] Name the three types of simple permanent tissue found in plants and give one function of each.
Section D – Case-Based Question (1 × 3 = 3 marks)
Q13. [Ch 4 and Ch 1] Read the case and answer the questions that follow.
Ria places a wind-up toy car on a long straight table and marks the table every metre. She releases the car and uses a stopwatch to note how far it has gone at fixed moments. Her table of readings is given below.
Time (s): 0, 2, 4, 6
Distance from start (m): 0, 5, 10, 15
(a) Is the motion of the toy car uniform or non-uniform? Justify your answer using the readings.
(b) Calculate the speed of the toy car.
(c) Assuming the car keeps moving in the same way, how far from the start will it be at t = 10 s?
Answer Key
Show the Full Answer Key
Section A – Objective Type (5 × 1 = 5 marks)
Q1. (a) 0.084 m
1 cm = 0.01 m, so 8.4 cm = 8.4 × 0.01 m = 0.084 m. The SI base unit of length is the metre, so the value must be divided by 100, not multiplied.
Q2. (c) Plastid
Ribosomes, mitochondria and the Golgi apparatus are present in both plant and animal cells. Plastids (including chloroplasts) are found only in plant cells. The cell wall and a large central vacuole are the other two plant-only features.
Q3. (b) Apical meristem
Apical meristem sits at the tips of roots and shoots and adds length to the plant. Lateral meristem adds girth (thickness). Sclerenchyma and parenchyma are permanent tissues and have stopped dividing.
Q4. (b) 400 m and 0
Distance is the total length of the path actually covered, which is one full round = 400 m. Displacement is the shortest straight line from the starting point to the finishing point. Since the athlete finishes exactly where she started, the displacement is zero.
Q5. (c) Bacterial cell
A prokaryotic cell has no nucleus bound by a membrane and no membrane-bound organelles. Bacteria are prokaryotic. Onion peel cells, human cheek cells and Amoeba all have a true nucleus, so they are eukaryotic.
Section B – Very Short Answer (4 × 2 = 8 marks)
Q6. Any two of the following (1 mark each):
1. The cell wall is present only in plant cells, fungi and bacteria; the plasma membrane is present in every living cell.
2. The cell wall is made mainly of cellulose in plants; the plasma membrane is made of lipids and proteins.
3. The cell wall is rigid and gives shape and mechanical support; the plasma membrane is thin and flexible.
4. The cell wall is fully permeable; the plasma membrane is selectively permeable and controls what enters and leaves the cell.
Q7. (a) Tendon, a type of dense connective tissue. Feature: it is made of tightly packed collagen fibres, so it has great strength with very limited flexibility. (1 mark)
(b) Epithelial tissue, specifically stratified squamous epithelium in the skin. Feature: its cells are arranged in tightly packed layers with almost no material between them, which stops germs and water loss. (1 mark)
Q8. Total distance = 300 m + 200 m = 500 m (½ mark)
Total time = 40 s + 60 s = 100 s (½ mark)
Average speed = total distance ÷ total time = 500 ÷ 100 = 5 m/s (1 mark)
Common error to avoid: averaging the two separate speeds (7.5 m/s and about 3.33 m/s) gives about 5.42 m/s, which is wrong. Average speed must always be total distance divided by total time.
Q9. Time period = total time ÷ number of oscillations = 32.0 ÷ 20 = 1.6 s (1 mark)
Timing 20 oscillations spreads the small human reaction-time error of starting and stopping the stopwatch across 20 swings instead of one. When the total is divided by 20, that error is also divided by 20, so the value of the time period is far more reliable. (1 mark)
Section C – Short Answer (3 × 3 = 9 marks)
Q10. Any three of the following (1 mark each):
1. A plant cell has a rigid cell wall outside the plasma membrane; an animal cell has only the plasma membrane.
2. A plant cell contains plastids, including chloroplasts for photosynthesis; an animal cell has no plastids.
3. A plant cell usually has one large central vacuole that fills most of the cell; an animal cell has small vacuoles or none at all.
4. Centrioles are normally absent in plant cells but present in animal cells.
5. A plant cell tends to have a fixed rectangular or polygonal shape; an animal cell has a flexible, often rounded shape.
Q11. Given: u = 0, a = 2 m/s2, t = 8 s.
(a) v = u + at = 0 + (2 × 8) = 16 m/s (1 mark)
(b) s = ut + ½at2 = 0 + ½ × 2 × 8 × 8 = 64 m (1 mark)
(c) Average velocity = total displacement ÷ total time = 64 ÷ 8 = 8 m/s (1 mark)
Check: for uniform acceleration the average velocity is also (u + v) ÷ 2 = (0 + 16) ÷ 2 = 8 m/s, which matches. A second check using v2 = u2 + 2as gives 162 = 0 + 2 × 2 × 64, that is 256 = 256.
Q12. 1. Parenchyma – thin-walled living cells that store food and water; when they contain chloroplasts (chlorenchyma) they carry out photosynthesis. (1 mark)
2. Collenchyma – cells thickened at the corners that give flexible mechanical support, which is why a young stem or a leaf stalk can bend in the wind without breaking. (1 mark)
3. Sclerenchyma – dead cells with thick lignified walls that give hardness and rigidity, as in the husk of a coconut or the shell of a nut. (1 mark)
Section D – Case-Based Question (1 × 3 = 3 marks)
Q13. (a) The motion is uniform. In each 2 s interval the car covers the same distance of 5 m (0 to 5, 5 to 10, and 10 to 15). Equal distances in equal intervals of time means uniform motion. (1 mark)
(b) Speed = distance ÷ time = 5 m ÷ 2 s = 2.5 m/s. Using the whole run gives the same value: 15 m ÷ 6 s = 2.5 m/s. (1 mark)
(c) At constant speed, distance = speed × time = 2.5 × 10 = 25 m from the start. (1 mark)
Apna Score Kaise Padhein
- 21–25: Chapters 1–4 are solid. Move on to Chapter 5 and keep revising motion numericals.
- 15–20: Good base. Re-read the two topics where you lost the most marks and re-attempt only those questions.
- Below 15: Go back to the chapter notes first, then attempt this paper again after two days. Repetition, not speed, fixes this.
More chapter tests and unit test papers are collected on the Unit Test Practice hub.
Kaizen note: One paper a day, honestly attempted and honestly checked, beats ten papers skimmed the night before the exam. Improve by a little, every day.