Question 1

Question 1A student took Sodium Sulphate solution in a test tube and added Barium Chloride solution to it. He observed that an insoluble substance has formed. The colour and molecular formula of the insoluble substance is:Grey, Ba2SO4Yellow, Ba(SO4)2White, BaSO4Pink, BaSO4

Accepted Answer

White, BaSO4Reason — When sodium sulphate is reacted with barium chloride, a white ppt. of barium sulphate is produced.Na2SO4 (aq) + BaCl2 (aq) ⟶ BaSO4 (s) + 2NaCl (aq)

Question 2

Question 9Study the following table and choose the correct option:S.No.SaltParent AcidParent BaseNature of Salt(a)Sodium ChlorideHClNaOHBasic(b)Sodium CarbonateH2CO3NaOHNeutral(c)Sodium SulphateH2SO4NaOHAcidic(d)Sodium AcetateCH3COOHNaOHBasic

Accepted Answer

Sodium AcetateCH3COOHNaOHBasicReason — Sodium acetate made from acetic acid and sodium hydroxide is basic in nature.

Question 3

Question 11Consider the following statements in connection with the functions of the blood vessels marked A and B in the diagram of a human heart as shown.(i) Blood vessel A — It carries carbon dioxide rich blood from the lungs.(ii) Blood vessel B — It carries oxygen rich blood from the lungs.(iii) Blood vessel B — Left atrium relaxes as it receives blood from this blood vessel.(iv) Blood vessel A — Right atrium has thick muscular wall as it has to pump blood to this blood vessel.The correct statements are(i) and (ii) only(ii) and (iii) only(ii), (iii) and (iv)(i), (ii) and (iii)

Accepted Answer

(i), (ii) and (iii)Reason— Blood vessel A is Pulmonary artery that takes deoxygenated blood to lungs and Blood vessel B is Pulmonary vein that takes oxygenated blood from lungs to left atrium.

Question 4

Question 18A student wants to obtain a magnified image of an object AB as on a screen. Which one of the following arrangements shows the correct position of AB for him/her to be successful?1.2.3.4.

Accepted Answer

Reason — When the object is kept between F and 2F, enlarged, real and inverted image is obtained on the screen.

Question 5

Question 20A ray of light starting from the air passes through medium A of refractive index 1.50, enters medium B of refractive index 1.33 and finally enters medium C of refractive index 2.42. If this ray emerges out in the air from C, then for which of the following pairs of media the bending of light is least?air-AA-BB-CC-air

Accepted Answer

A-BReason — The bending of light is least when the refractive index change is least. When light is travelling from medium A to medium B, the change in refractive index is from 1.50 to 1.33 which is the least in this case. Hence, the bending of light is also the least for A-B.

Question 6

Question 26Consider the pH value of the following acidic samples :S.No.SamplepH Value1.Lemon Juice2.22.Gastric Juice1.23.Vinegar3.764.Dil. Acetic acid3.0The decreasing order of their H+ ion concentration is3 > 4 > 1 > 22 > 1 > 3 > 42 > 1 > 4 > 33 > 4 > 2 > 1

Accepted Answer

2 > 1 > 4 > 3Reason — The number of H+ ion concentration determines the acidic strength of the sample. The pH and hydrogen ion concentration are inversely related i.e., as the value of pH increases, the H+ ion concentration decreases hence decreasing order of their H+ ion concentration is 2 > 1 > 4 > 3.

Question 7

Question 27Study the experimental set-up shown in given figure and choose the correct option from the following:PQChange observed in calcium hydroxide solution1.K2CO3Cl2 gasNo change2.KHCO3CO2 gasNo change3.KHCO3H2 gasTurns milky4.K2CO3CO2 gasTurns milky

Accepted Answer

K2CO3CO2 gasTurns milkyReason — The reaction between potassium carbonate and HCl gas produces carbon dioxide gas.K2CO3 + 2HCl (dil.) ⟶ 2KCl + H2O + CO2Calcium hydroxide reacts with carbon dioxide gas and forms calcium carbonate, which turns the solution milky.Ca(OH)2 + CO2 ⟶ CaCO3 + H2O

Question 8

Question 29The pair(s) which will show displacement reaction is/are(i) NaCl solution and copper metal(ii) AgNO3 solution and copper metal(iii) Al2(SO4)3 solution and magnesium metal(iv) ZnSO4 solution and iron metal(ii) only(ii) and (iii)(iii) and (iv)(i) and (ii)

Accepted Answer

(ii) and (iii)Reason — AgNO3 solution and copper metal will give displacement reaction where copper being more reactive than silver, displaces it from silver nitrate . Similarly, Al2(SO4)3 solution and magnesium metal will give displacement reaction where magnesium being more reactive than aluminium, displaces it from aluminium sulphate.

Question 9

Question 30Which of the following salts do not have the water of crystallisation?(i) Bleaching Powder(ii) Plaster of Paris(iii) Washing soda(iv) Baking soda(ii) and (iv)(i) and (iii)(ii) and (iii)(i) and (iv)

Accepted Answer

(i) and (iv)Reason — Bleaching Powder [CaOCl2] and baking soda [NaHCO3] do not have water of crystallization whereas plaster of Paris [CaSO4.12\dfrac{1}{2}21​ H2O] and washing soda [Na2CO3.10H2O] have.

Question 10

Question 31Assertion (A): Sodium hydrogen carbonate is used as an ingredient in antacids.Reason (R): NaHCO3 is a mild non-corrosive basic salt.Both A and R are true, and R is the correct explanation of A.Both A and R are true, and R is not the correct explanation of A.A is true but R is false.A is false but R is true.

Accepted Answer

Both A and R are true and R is the correct explanation of A.Reason — Sodium hydrogen carbonate [NaHCO3] is used as an ingredient in antacids, as it is a mild non-corrosive basic salt and it neutralizes the excess acid produced in our stomach.

Question 11

Question 34Assertion (A): Sun appears reddish at the time of Sunrise and Sunset.Reason (R): Distance travelled by sunlight in the atmosphere is lesser during sunrise and sunset as compared to noon.Both A and R are true, and R is the correct explanation of A.Both A and R are true, and R is not the correct explanation of A.A is true but R is false.A is false but R is true.

Accepted Answer

A is true and R is false.Reason — During the sunrise and sunset, the sun is near the horizon and most of the blue light and shorter wavelengths are scattered away by the particles. Therefore, the light that reaches our eyes is of longer wavelength. This gives rise to the reddish appearance of the sun.During sunrise and sunset, sun is at a farther distance while at noon sun is directly overhead. Hence, distance travelled by sunlight in the atmosphere is more during sunrise and sunset as compared to noon.

Question 12

Question 37Respiratory structures of two different animals — a fish and a human being are as shown.Observe (a) and (b) and select one characteristic that holds true for both of them.Both are placed internally in the body of animal.Both have thin and moist surface for gaseous exchange.Both are poorly supplied with blood vessels to conserve energy.In both the blood returns to the heart after being oxygenated.

Accepted Answer

Both have thin and moist surface for gaseous exchange.Reason — Gills as well as alveoli have moist surface. Also, fishes have only two chambers to their hearts, and the blood is pumped to the gills, is oxygenated there, and passes directly to the rest of the body.

Question 13

Question 38Observe the diagram of an activity given below. What does it help to conclude, when the person exhales into the test-tube?Percentage of carbon dioxide is more in inhaled air.Fermentation occurs in the presence of oxygen.Percentage of carbon dioxide is more in exhaled air.Fermentation occurs in presence of carbon dioxide

Accepted Answer

Percentage of carbon dioxide is more in exhaled air.Reason — Exhaling through the straw turns the lime water milky showing that percentage of carbon dioxide is more in exhaled air.

Question 14

Question 39If a lens can converge the sun rays at a point 20 cm away from its optical centre, the power of this lens is –+2D–2D+5D–5D

Accepted Answer

+5DReason — Given,Convex lens as converging rays.Focal length (f) = 20 cmP = 100f(cm)\dfrac{100}{	ext{f(cm)}}f(cm)100​Substituting we get,P = 10020\dfrac{100}{20}20100​ = +5D [positive sign as convex lens.]

Question 15

Question 40The radius of curvature of a converging mirror is 30 cm. At what distance from the mirror should an object be placed so as to obtain a virtual image?Infinity30 cmBetween 15 cm and 30 cmBetween 0 cm and 15 cm

Accepted Answer

Between 0 cm and 15 cmReason — Given, concave mirror as it is a converging mirror.R = 30 cmWe know, R = 2fSubstituting we get, 30 = 2 x ff = 302\dfrac{30}{2}230​ = 15 cmIn a concave mirror, virtual image is obtained when the object is placed between focus and optical centre hence, object should be placed between 0 cm and 15 cm.

Question 16

Question 41The length of small intestine in a deer is more as compared to the length of small intestine of a tiger. The reason for this is :Mode of intake of foodType of food consumedPresence or absence of villi in intestinePresence or absence of digestive enzymes

Accepted Answer

Type of food consumedReason — The length of the small intestine differs in various animals depending on the food they eat. Herbivores eating grass need a longer small intestine to allow the cellulose to be digested. Meat is easier to digest, hence carnivores like tigers have a shorter small intestine.

Question 17

Question 43A converging lens forms a three-times magnified image of an object, which can be taken on a screen. If the focal length of the lens is 30 cm, then the distance of the object from the lens is–55 cm–50 cm–45 cm–40 cm

Accepted Answer

–40 cmReason — Given, convex lens as converging lensf = 30 cmFrom formula,m = vu\dfrac{	ext{v}}{	ext{u}}uv​-3 = vu\dfrac{	ext{v}}{	ext{u}}uv​ [-ve sign as real and inverted image is formed.]v = -3uUsing formula,1f\dfrac{1}{	ext{f}}f1​ = 1v\dfrac{1}{	ext{v}}v1​ - 1u\dfrac{1}{	ext{u}}u1​We get,130\dfrac{1}{	ext{30}}301​ = 1-3u\dfrac{1}{	ext{-3u}}-3u1​ - 1u\dfrac{1}{	ext{u}}u1​130\dfrac{1}{	ext{30}}301​ = −1−33u\dfrac{-1-3}{	ext{3u}}3u−1−3​ = −43u\dfrac{-4}{	ext{3u}}3u−4​3u = -4 x 30u = −1203\dfrac{-120}{3}3−120​ = -40 cmHence, distance of the object from the lens = -40 cm.

Question 18

Question 45In the diagram shown below n1, n2 and n3 are refractive indices of the media 1, 2 and 3 respectively. Which one of the following is true in this case?n1 = n2n1 > n2n2 > n3n3 > n1

Accepted Answer

n3 > n1Reason — According to the figure, the light ray bends towards the normal when it passes from medium 1 to medium 2 and then it bends further when it passes from medium 2 to 3. We know, ray of lights bends towards the normal when it passes from rarer to denser medium hence medium 3 is denser than medium 1 i.e., n3 > n1.

Question 19

Question 46The refractive index of medium A is 1.5 and that of medium B is 1.33. If the speed of light in the air is 3 × 108 m/s, what is the speed of light in medium A and B respectively?(a) 2 × 108 m/s and 1.33 × 108 m/s(b) 1.33 × 108 m/s and 2 × 108 m/s(c) 2.25 × 108 m/s and 2 × 108 m/s(d) 2 × 108 m/s and 2.25 × 108 m/s

Accepted Answer

2 × 108 m/s and 2.25 × 108 m/sReason — We know,refractive index of medium = speed of light in airspeed of light in medium\dfrac{	ext{speed of light in air}}{	ext{speed of light in medium}}speed of light in mediumspeed of light in air​Substituting we get,For medium A,1.5 = 3×108speed of light in medium A\dfrac{3 	imes 10^8}{	ext{speed of light in medium A}}speed of light in medium A3×108​speed of light in medium A = 3×1081.5\dfrac{3 	imes 10^8}{1.5}1.53×108​Hence, speed of light in medium A = 2 × 108 m/sFor medium B,1.33 = 3×108speed of light in medium B\dfrac{3 	imes 10^8}{	ext{speed of light in medium B}}speed of light in medium B3×108​speed of light in medium B = 3×1081.33\dfrac{3 	imes 10^8}{1.33}1.333×108​Hence, speed of light in medium B =2.25 × 108 m/s

Question 20

Question 47An object of height 4 cm is kept at a distance of 30 cm from the pole of a diverging mirror. If the focal length of the mirror is 10 cm, the height of the image formed is+3.0 cm+2.5 cm+1.0 cm+0.75 cm

Accepted Answer

+1.0 cmReason — Given, convex mirror as diverging mirror.focal length (f) = 10 cmu = -30 cmho = 4 cmv = ?Using formula,1f\dfrac{1}{	ext{f}}f1​ = 1v\dfrac{1}{	ext{v}}v1​ + 1u\dfrac{1}{	ext{u}}u1​We get,110\dfrac{1}{	ext{10}}101​ = 1v\dfrac{1}{	ext{v}}v1​ - 130\dfrac{1}{	ext{30}}301​1v\dfrac{1}{	ext{v}}v1​ = 110\dfrac{1}{	ext{10}}101​ + 130\dfrac{1}{	ext{30}}301​1v\dfrac{1}{	ext{v}}v1​ = 3+130\dfrac{3+1}{	ext{30}}303+1​ = 430\dfrac{4}{	ext{30}}304​1v\dfrac{1}{	ext{v}}v1​ = 215\dfrac{2}{	ext{15}}152​v = 152\dfrac{15}{	ext{2}}215​ = 7.5m = - vu\dfrac{	ext{v}}{	ext{u}}uv​ = hiho\dfrac{	ext{h}_i}{	ext{h}_o}ho​hi​​Substituting we get,-7.5−30\dfrac{7.5}{-30}−307.5​ = hi4\dfrac{	ext{h}_i}{4}4hi​​height of the image = 7.5×430\dfrac{7.5 	imes 4}{30}307.5×4​ = +1 cmHence, height of the image = +1 cm

Solved 2022 Term 1 Question Paper CBSE Class 10 Science

Section A

Section B

Section C