FIRST MID TERM TEST - 2024
Standard - IX | SCIENCE
Solved Paper & Answer Key
Time: 1.30 hrs
Marks: 50
I. Choose the correct answers:- (Solved)
- 1 metric ton is equal to b) 10 quintalsExplanation: 1 metric ton = 1000 kg. 1 quintal = 100 kg. Therefore, 1000 kg / 100 kg = 10 quintals.
- Which of the following is not a device to measure mass? a) Spring balanceExplanation: A spring balance measures weight (force due to gravity, W=mg), not mass. Beam, physical, and digital balances measure mass.
- The centrifugal force is b) the force of reaction of centripetal forceExplanation: In an inertial frame, centripetal force is the real force acting inwards. The outward-acting centrifugal force is a pseudo or virtual force, often described as the reaction to the centripetal force.
- When we mix a drop of ink in water, we get a c) homogenous mixtureExplanation: Ink particles dissolve and disperse uniformly throughout the water, forming a solution, which is a homogenous mixture.
- Change in number of neutrons in an atom changes it to b) an isotopeExplanation: Isotopes are atoms of the same element (same number of protons) but with different numbers of neutrons.
- Identify the animal having four chambered heart: c) birdand d) batExplanation: Both birds (Aves) and bats (Mammalia) have four-chambered hearts. Lizards and snakes (Reptilia, except crocodilians) have three-chambered hearts. Both c) and d) are correct.
- Air sacs and pneumatic bones are seen in c) birdExplanation: These are adaptations for flight, characteristic of the class Aves (birds). Pneumatic bones are hollow and filled with air, reducing weight.
- Poikilothermic organisms are c) fish, frog, lizard, snakeExplanation: Poikilothermic (cold-blooded) animals cannot regulate their internal body temperature. Fish, amphibians (frog), and reptiles (lizard, snake) are poikilothermic. Man, cow, and crow are homeothermic (warm-blooded).
- The fibre consists of b) sclerenchymaExplanation: Plant fibres (like jute, flax) are primarily composed of sclerenchyma cells, which provide mechanical strength and support.
- Smooth muscles occur in d) all the aboveExplanation: Smooth muscles are involuntary muscles found in the walls of hollow internal organs, including the uterus, arteries, and veins.
II. Answer in one or two sentences: (Solved)
- Define least count of any device.
The smallest measurement that can be accurately made by a measuring instrument is called its least count.
- i) Speed is a ______ quantity whereas
ii) Velocity is a ______ quantity.i) Speed is a scalar quantity whereas
ii) Velocity is a vector quantity. - Match the following:
Correct Matches:
- 1) Sclereids (Sclerenchyma) → Sclerenchyma
- 2) Chloroplast → Chorenchyma (Parenchyma containing chloroplasts)
- 3) Simple tissue → Collenchyma (An example of simple tissue)
- 4) Companion cells → Phloem (Part of the phloem complex tissue)
- Write the electronic configuration of K and Cl.
Potassium (K): Atomic Number = 19. Electronic Configuration is 2, 8, 8, 1.
Chlorine (Cl): Atomic Number = 17. Electronic Configuration is 2, 8, 7. - Name the apparatus that you used to separate two: i) miscible liquids ii) immiscible liquids
i) Miscible liquids: Fractional distillation apparatus (based on different boiling points).
ii) Immiscible liquids: Separating funnel (based on different densities). - What is nematocyst?
A nematocyst (or cnidocyte) is a specialized explosive cell found in animals of the phylum Cnidaria (e.g., jellyfish, corals). It contains a coiled, barbed filament used for capturing prey and for defense.
- Say true/false:
i) Larva of amphibian is tadpole. → True.
ii) Canal System is seen in Coelenterata. → False. (It is a characteristic of Phylum Porifera/sponges). - Define Sublimation.
Sublimation is the process where a substance transitions directly from a solid state to a gaseous state, without passing through the intermediate liquid phase. Example: Camphor, dry ice (solid CO₂).
- What are Complex tissues?
Complex tissues are made of more than one type of cell that work together as a unit to perform a common function. In plants, the primary examples are Xylem (for water conduction) and Phloem (for food transport).
- A racing car has a uniform acceleration of $4 \text{ ms}^{-2}$. What distance does it cover in 10s after the start?
Given: Initial velocity (u) = 0 m/s, Acceleration (a) = 4 m/s², Time (t) = 10 s.
Formula: $s = ut + \frac{1}{2}at^2$
Calculation: $s = (0)(10) + \frac{1}{2}(4)(10)^2$
$s = 0 + 2 \times 100$
$s = 200$ meters.
Answer: The car covers a distance of 200 m.
III. Answer in brief. (Solved)
- Differentiate between flatworm and round worms.
Feature Flatworms (Platyhelminthes) Roundworms (Nematoda) Body Shape Dorso-ventrally flattened Cylindrical, circular in cross-section Body Cavity Acoelomate (no body cavity) Pseudocoelomate (false body cavity) Digestive System Incomplete (single opening) Complete (mouth and anus) Example Tapeworm, Planaria Ascaris (Roundworm), Wuchereria (Filarial worm) - Write four rules for writing symbols of units in the S.I. System.
- 1. Symbols for units are written in lowercase letters (e.g., m for metre, s for second).
- 2. Symbols for units named after scientists are written with a capital initial letter (e.g., N for Newton, W for Watt).
- 3. Symbols are never followed by a full stop (unless at the end of a sentence).
- 4. Symbols are not expressed in plural form (e.g., write 25 kg, not 25 kgs).
- i) What is Brownian movement? ii) Write about Tyndall effect.
i) Brownian Movement: It is the random, zig-zag motion of microscopic particles suspended in a fluid (liquid or gas). This movement is caused by the continuous and uneven bombardment of the suspended particles by the molecules of the surrounding medium.
ii) Tyndall Effect: It is the phenomenon of scattering of a beam of light by the particles of a colloid or a very fine suspension. This scattering makes the path of the light beam visible. True solutions do not exhibit the Tyndall effect because their particles are too small to scatter light. - In which stage of mitosis do the chromosomes align in an equatorial plate? How?
Stage: Chromosomes align at the equatorial plate during Metaphase.
How: During metaphase, the spindle fibres, which originate from the opposite poles (centrosomes) of the cell, attach to the centromere of each chromosome. The equal pulling force exerted by the spindle fibres from both poles causes the chromosomes to move and line up precisely along the cell's center, forming a structure called the metaphase or equatorial plate.
IV. Answer in detail: (Solved)
- Explain a method to find the thickness of a hollow tea cup.
Aim: To measure the thickness of the wall of a hollow teacup.
Apparatus Required: Vernier caliper.
Procedure:- Find Least Count (LC) and Zero Error: First, determine the least count of the vernier caliper using the formula LC = 1 MSD - 1 VSD. Check for any zero error by closing the jaws and noting if the zero of the main scale coincides with the zero of the vernier scale.
- Measure Internal Diameter ($d_1$): Place the internal measuring jaws of the caliper inside the teacup. Expand the jaws until they gently touch the inner opposite walls of the cup. Record the Main Scale Reading (MSR) and the Vernier Scale Coincidence (VSC). Calculate the internal diameter using the formula: $d_1 = \text{MSR} + (\text{VSC} \times \text{LC}) - \text{Zero Correction}$.
- Measure External Diameter ($d_2$): Place the teacup between the external measuring jaws of the caliper. Close the jaws until they gently touch the outer opposite walls of the cup. Record the MSR and VSC. Calculate the external diameter using the formula: $d_2 = \text{MSR} + (\text{VSC} \times \text{LC}) - \text{Zero Correction}$.
- Calculate Thickness (t): The thickness of the teacup wall is half the difference between its external and internal diameters.
Thickness, $t = \frac{\text{External Diameter} - \text{Internal Diameter}}{2} = \frac{d_2 - d_1}{2}$
The final result is expressed in cm or mm. -
a) Draw the structure of Oxygen and Sulphur atoms. Write its electronic configuration.
Oxygen (O):b) Match the following:
- Atomic Number: 8
- Electronic Configuration: 2, 6
- Structure (Bohr Model): A nucleus with 8 protons. The first shell (K-shell) has 2 electrons. The second and outermost shell (L-shell) has 6 electrons.
- Atomic Number: 16
- Electronic Configuration: 2, 8, 6
- Structure (Bohr Model): A nucleus with 16 protons. The first shell (K-shell) has 2 electrons, the second shell (L-shell) has 8 electrons, and the third and outermost shell (M-shell) has 6 electrons.
The matching in the question paper is incorrect. The correct pairs are:- i) Dalton → First atomic theory
- ii) Chadwick → Discovery of neutrons
- iii) Rutherford → Discovery of nucleus
- iv) J.J. Thomson → Plum pudding model
- Write about Phylum Arthropoda.
Arthropoda (from Greek: arthros - "joint", podos - "foot") is the largest phylum in the animal kingdom, including insects, spiders, crabs, and centipedes.
Key Characteristics:- Exoskeleton: They have a hard, protective outer body covering made of chitin, which they must shed periodically (molting or ecdysis) to grow.
- Segmented Body: The body is bilaterally symmetrical and typically segmented into three regions: head, thorax, and abdomen. In some, the head and thorax are fused to form a cephalothorax.
- Jointed Appendages: They possess paired, jointed limbs that are modified for various functions like walking, feeding, sensing, and defense.
- Circulatory System: They have an open circulatory system. Blood, called hemolymph, is pumped by a heart into the body cavity (hemocoel), where it bathes the tissues directly.
- Respiration: Diverse respiratory organs exist, including gills in aquatic forms (e.g., prawns), book lungs in arachnids (e.g., spiders), and a network of air tubes called tracheae in insects.
- Examples: Cockroach (Insecta), Spider (Arachnida), Crab (Crustacea), Centipede (Myriapoda).
- Distinguish between mitosis and meiosis.
Basis of Difference Mitosis Meiosis Purpose Growth, repair of tissues, asexual reproduction. Formation of gametes (sperm and eggs) for sexual reproduction. Occurs in Somatic (body) cells. Germ (reproductive) cells. Number of Divisions One nuclear division. Two successive nuclear divisions (Meiosis I and Meiosis II). Number of Daughter Cells Two diploid (2n) cells. Four haploid (n) cells. Chromosome Number Remains the same as the parent cell (2n → 2n). Is halved in the daughter cells (2n → n). Genetic Variation Daughter cells are genetically identical to the parent. Daughter cells are genetically different due to crossing over. Crossing Over Does not occur. Occurs during Prophase I, leading to genetic recombination.