HBSE Class 9 Science Question Paper 2023 Answer Key

Haryana Board (HBSE) Class 9 Science Question Paper 2023 Answer Key. Haryana Board Class 9th Science Solved Question Paper 2023 PDF Download. HBSE Board Solved Question Paper Class 9 Science 2023. Haryana Board Class 9th Science Question Paper 2023 Pdf Download with Answer. HBSE 9th Science Solved Question Paper 2023. HBSE 9th Class Science Solved Question Paper 2023.

HBSE Class 9 Science Question Paper 2023 Answer Key


Q1. The Change in velocity per unit time is called :
(A) Velocity
(B) Acceleration
(C) Speed
(D) Distance
Ans – (B) Acceleration

Q2. What is the momentum of an object of Mass m, moving with a velocity v ?
(A) mv
(B) mv²
(C) ½mv²
(D) (mg)²
Ans – (A) mv

Q3.The time taken by the wave for one complete oscillation of the density of the medium is called the ………….
Ans – time period

Q4. Define average speed. Write the formula of average speed.
Ans – Average speed is defined as the total distance travelled by the body in total time. It is a scalar quantity. It’s SI unit is m/s.
Average Speed = Total Distance / Total Time

Q5. Why is it advised to tie any luggage kept on the roof of a bus with a rope ?
Ans – When the bus stops suddenly, the luggage on the roof top will fall forward due to inertia of motion.

Q6. Distinguish between loudness and intensity of sound.
Ans : Loudness – Loudness is the measure of the response of the ear to the sound. Loudness is measured in decibels. Loudness is dependent on the sensitivity of the human ears.
Intensity of Sound – Intensity is the sound power per unit area. Intensity is measured in Watt per meter square. Intensity is independent of the sensitivity of the human ears.

Q7.(i) An object weight 10 N when measured on the surface of the earth. What would be its weight when measured on the surface of the moon ?
Ans : Weight of object on moon = 1/6 × (weight of object on the earth) = 1/6 × 10 = 1.67 N

(ii) What are the importance of universal law of Gravitation ?
Ans – It explains the motion of planets around the sun. It helps in determining the trajectory of astronomical bodies and to predict their motion. It also explains rainfall, the motion of water in the oceans, rivers etc.

Q8.(i) What is Potential Energy ? Write its formula.
Ans – The energy of the body due to its position or its compressed state is called potential energy. Its SI unit is Joule.
Potential Energy (P.E.) = mgh

(ii) Find the energy in kWh consumed in 10 hours by four devices of power 500 W each.
Ans : Energy = 4 × 10 × 500 = 20000 Wh = 20 kWh


(i) Write the Law of conservation of energy.
Ans – The Law of Conservation of Energy states that energy neither be created nor be destroyed. It may be transformed from one form to another.

(ii) An object of Mass 40 kg is raised to a height of 5 m above the ground. What is its potential energy? If the object is allowed to fall, find its kinetic energy when it is half-way down. (g = 10 m/sec²)
Ans : Potential Energy = mgh = 40 × 10 × 5 = 2000 Joule
Kinetic Energy = 2000/2 = 1000 J

Q9. S.I. unit of density is :
(A) Newton
(B) Kilogram per cubic metre
(C) Cubic metre
(D) Pascal
Ans – (B) Kilogram per cubic metre or kg/m³ or kg m-³

Q10. Which one of the following is a correct electronic configuration of sodium ?
(A) 2, 8
(B) 2, 1, 8
(C) 2, 8, 1
(D) 8, 2, 1
Ans – (C) 2, 8, 1

Q11. How does the water kept in an earthen pot become cool during summer ?
Ans – The walls of an earthen pot have a high number of microscopic pores, and some water molecules regularly seep through these pores to the outside. This water evaporates indefinitely, absorbing the latent heat needed for vaporisation from the remaining water. As a result, the remaining water loses heat and cools.

Q12. Name the three sub-atomic particles of an atom.
Ans – Protons, Neutrons and Electrons

Q13. Define mass number. Write the mass number of carbon.
Ans – The total number of protons and neutrons present in the nucleus of an atom is called mass number.
Mass Number = no. of protons + no. neutrons
Mass number of carbon = 12

Q14. Write four differences between mixtures and compounds.
Ans : Compounds – (i) Compounds are pure substances.
(ii) They are made up of two or more elements combined chemically.
(iii) The constituents of a compound are present in a fixed ratio.
(iv) Compounds have fixed properties. For example, a particular compound will have fixed temperatures at which it melts and boils.
(v) A compound can have properties different from its constituents, as a new substance is formed when the constituents are chemically combined.
(vi) The constituents of a compound can be separated only by chemical methods.

Mixtures – (i) Mixtures are impure substances.
(ii) They are made up of two or more substances mixed physically.
(iii) The constituents of a mixture are present in varying ratios.
(iv) Mixtures do not have fixed properties. Their properties depend on the nature of their compo­nents and the ratios in which they are combined.
(v) In mixtures, no new substance is formed. The properties of a mixture are the same as the properties of its constituents.
(vi) The constituents of a mixture can be separated easily by physical methods.

Q15.(i) Convert into mole :
(a) 12 gm of oxygen gas
Ans : Molar mass of oxygen = 32 gm
32 gm of oxygen gas = 1 mole
12 gm of oxygen gas = 1/32 × 12 = 12/32 = 0.375 moles

(b) 22 gm of carbon dioxide
Ans : Molar mass of carbon dioxide = 44 gm
44 gm of carbon dioxide = 1 mole
22 gm of carbon dioxide = 1/44 × 22 = 22/44 = 0.5 moles

(ii) Write the formulae of the following :
(a) Sodium oxide
Ans : Na2O

(b) Aluminium chloride
Ans : AlCl3


(i) Calculate the molar mass of the following substances :
(a) Hydrochloric acid (HCl)
Ans : Molar mass = 1×1 + 1×35.5 = 36.5 g

(b) Nitric acid (HNO3)
Ans : Molar mass = 1×1 + 1×14 + 3×16 = 1 + 14 + 48 = 63 g

(ii) Why is it not possible to see an atom with naked eyes?
Ans – because they are very small in size.

Q16. Cell membrane is
(A) Permeable
(B) impermeable
(C) Selectively permeable
(D) None of these
Ans – (C) Selectively permeable

Q17. How many heart chambers crocodiles have ?
(A) 1
(B) 2
(C) 3
(D) 4
Ans – (D) 4

Q18. Which disease are air-borne ?
(A) Pneumonia
(B) Common-cold
(C) Tuberculosis
(D) All of these
Ans – (D) All of these

Q19. Which of the following is a green house gas ?
(A) Carbon dioxide
(B) Nitrogen
(C) Oxygen
(D) All of these
Ans – (A) Carbon dioxide

Q20.What is the percentage of Nitrogen gas in our atmosphere ?
Ans : 78%

Q21. What are the forms of oxygen found in the atmosphere ?
(A) O2
(B) O3
(C) O5
(D) Both (A) and (B)
Ans – (D) Both (A) and (B)

Q22. Macro nutrients are :
(A) Nitrogen
(B) Phosphorus
(C) Sulphur
(D) All of these
Ans – (D) All of these

Q23. Where do the lipids and proteins constituting the cell membrane get synthesized ?
Ans – The synthesis of proteins and lipids takes place in the Endoplasmic Reticulum.
It is of two types – Smooth Endoplasmic Reticulum (SER) and Rough Endoplasmic Reticulum (RER).

Q24. What are infectious diseases? Give one example.
Ans – Infectious diseases are those diseases that can be easily transmitted from one person to another. They can be caused by bacteria, viruses, fungi. e.g. common cold, influenza.

Q25. How animal husbandry practices are beneficial for farmers ?
Ans – Animal husbandry practices result in increased offspring survival, better growth, better production, better reproduction, and higher profits. Yields in good quality cattle. Better quality of milk production. Use in agriculture for carting, irrigation and tilling.

Q26.(i) How do Poriferan animals differ from Coelenterate animals ?
Ans : Poriferan Animals – Poriferans are asymmetric and non-motile organisms, usually found attached to rocks. Poriferan Animals have pores or holes throughout their body.

Coelenterate Animals – Coelenterates are radially symmetrical and motile organisms, that either live in colonies or as solitary organisms. Coelenterate Animals have a coelenteron, a body cavity with only one opening.

(ii) What are the major divisions in the plantae ?
Ans – Thallophyta, Bryophyta, Pteridophyta, Gymnosperms and Angiosperms.

Q27.(i) Differentiate between Parenchyma, Collenchyma and Sclerenchyma on the basis of their cell wall.
Ans : Parenchyma – Primary cell wall is present. Thin cell wall, primarily composed of cellulose is present.

Collenchyma – Primary cell wall is present. Cell wall has localised thickening at the corners because of the deposition of cellulose.

Sclerenchyma – Secondary cell wall is present. Extremely thick cell wall is present because of the presence of lignin.

(ii) What is the role of epidermis in plants ?
Ans – (i) The plant epidermis is a protective tissue that covers the entire surface of the plant.
(ii) The epidermis protects the plant from infection and water loss.
(iii) It regulates the gas exchange in plant cells.
(iv) The epidermis regulates the secretion of metabolic substances.


(i) Differentiate between striated, unstriated and cardic muscles on the basis of their structure and site in the body.
Ans : Striated Muscles (Skeletal Muscles) – Cells are cylindrical. Cells are not branched. These muscles are present in body structures like hands, legs and tongue.

Unstriated Muscles (Smooth Muscles) – Cells are elongated. Cells are not branched. They are present on the linings of the alimentary canal and the blood vessels.

Cardiac Muscles – Cells are cylindrical. Cells are branched. These muscles are present on the walls of the heart.

(ii) What are the functions of the stomata ?
Ans – Stomata helps in exchange of gases in plants (carbon dioxide and oxygen). It helps in evaporation of water in plants during the process of transpiration. Stomata closes or opens it pores to keep the moisture content balanced depending on weather conditions. It creates upward pull for absorption of water from the roots of plant.


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