The formation of magnesium oxide is correctly shown in option:
(A) MgMgMg ⟶ OOO ⟶ Mg2+Mg^{2+}Mg2+ [O2−][O^{2-}] [O2−](B) MgMgMg ⟶ OOO ⟶ Mg2+Mg^{2+}Mg2+ [O2−][O^{2-}] [O2−](C) Mg2Mg_2Mg2 ⟶ O2O_2O2 ⟶ Mg22+Mg^{2+}_2Mg22+ [O2−]2[O^{2-}]_2 [O2−]2(D) 2Mg2Mg2Mg ⟶ O2O_2O2 ⟶ [Mg22+][Mg^{2+}_2] [Mg22+] [O22−][O^{2-}_2] [O22−]
Answer:
(B) MgMgMg ⟶ OOO ⟶ Mg2+Mg^{2+}Mg2+ [O2−][O^{2-}] [O2−]
Explanation:
Magnesium (Mg) is a metal that loses two electrons to form Mg²⁺.
Oxygen (O) is a non-metal that gains two electrons to form O²⁻.
The resulting ions Mg²⁺ and O²⁻ attract each other to form an ionic bond, creating magnesium oxide (MgO).
2. The metals obtained from their molten chlorides by the process of electrolytic reduction are:
Answer: (B) Calcium and magnesium
3. In one formula unit of salt 'X', seven molecules of water of crystallization are present. The salt 'X' is:
4. Reaction between two elements A and B forms a compound C. A loses electrons, and B gains electrons. Which one of the following properties will not be shown by compound C?
Answer: (B) It is highly soluble in water.
5. Electrolysis of water is a decomposition reaction. The mass ratio (MH_HH : MO_OO) of hydrogen and oxygen gases liberated at the electrodes during the electrolysis of water is:
6. Consider the following reactions:(i) Dilute hydrochloric acid reacts with sodium hydroxide.(ii) Magnesium oxide reacts with dilute hydrochloric acid.(iii) Carbon dioxide reacts with sodium hydroxide.
It is found that in each case:Answer: (A) Salt and water are formed.
Question 7:
The products formed when Aluminium and Magnesium are burnt in the presence of air respectively are:
(A) Al₂O₃ and MgO₂(B) Al₂O₄ and MgO(C) Al₂O₃ and MgO(D) Al₂O₃ and MgO₂
Answer:
Explanation:
Question 8:
The plant hormone whose concentration stimulates the cells to grow longer on the side of the shoot which is away from light is:
(A) Cytokinins(B) Gibberellins(C) Adrenaline(D) Auxins
Answer:
Explanation:
Question 9:
Secretion of less saliva in the mouth will affect the conversion of:
(A) Proteins into amino acids(B) Fats into fatty acids and glycerol(C) Starch into simple sugars(D) Sugars into alcohol
Answer:
(C) Starch into simple sugars
Explanation:
Saliva contains amylase, an enzyme that helps in breaking down starch into maltose (a simple sugar).
If saliva secretion is reduced, starch digestion is affected.
Question 10:
If pea plants with round and green seeds (RRyy) are crossed with pea plants having wrinkled and yellow seeds (rrYY), the seeds developed by the plants of F1 generation will be:
(A) 50% round and green(B) 75% wrinkled and green(C) 100% round yellow(D) 75% wrinkled and yellow
Answer:
Explanation:
RRyy (Round, Green) × rrYY (Wrinkled, Yellow) results in RrYy offspring in F1 generation.
Round (R) is dominant over Wrinkled (r).
Yellow (Y) is dominant over Green (y).
So, all F1 plants will have Round Yellow (RrYy) seeds.
Question 11:
The correct/true statement(s) for a bisexual flower is/are:
(i) They possess both stamen and pistil.(ii) They possess either stamen or pistil.(iii) They exhibit either self-pollination or cross-pollination.(iv) They cannot produce fruits on their own.
(A) Only (i)(B) Only (iv)(C) (i) and (iii)(D) (i) and (iv)
Answer:
Explanation:
Bisexual flowers have both stamen and pistil (Statement (i) is correct).
They can undergo self-pollination or cross-pollination (Statement (iii) is correct).
Statement (ii) is incorrect because bisexual flowers contain both reproductive organs.
Statement (iv) is incorrect because bisexual flowers can self-pollinate and produce fruits.
Question 12:
The breakdown of glucose has taken the following pathway:
Glucose → (a) → Pyruvate + Energy → (b) → Lactic acid + Energy
The sites 'a' and 'b' respectively are:
(A) Mitochondria and Oxygen-deficient muscle cells(B) Cytoplasm and Oxygen-rich muscle cells(C) Cytoplasm and Yeast cells(D) Cytoplasm and Oxygen-deficient muscle cells
Answer:
(D) Cytoplasm and Oxygen-deficient muscle cells
Explanation:
(a) Cytoplasm: The first step occurs in the cytoplasm, where glucose undergoes glycolysis to form pyruvate.
(b) Oxygen-deficient muscle cells: In the absence of oxygen, pyruvate undergoes anaerobic respiration in muscle cells, leading to the production of lactic acid.
Question 13:
The white light entering a glass prism splits into its constituent colours. It is observed that:
(A) Red light deviates the most.(B) Violet light deviates the least.(C) Yellow light deviates more than the blue light.(D) Green light deviates more than the orange light.
Answer:
(D) Green light deviates more than the orange light.
Explanation:
Dispersion of light occurs because different wavelengths refract differently.
Violet light deviates the most, and red light deviates the least because violet has a shorter wavelength and undergoes more bending.
Green light is refracted more than orange light as its wavelength is shorter than orange.
Question 14:
Mirror 'X' is used to concentrate sunlight in a solar furnace, and Mirror 'Y' is fitted on the side of the vehicle to see the traffic behind the driver.
Which of the following statements are true for the two mirrors?
(i) The image formed by mirror 'X' is real, diminished, and at its focus.(ii) The image formed by mirror 'Y' is virtual, diminished, and erect.(iii) The image formed by mirror 'X' is virtual, diminished, and erect.(iv) The image formed by mirror 'Y' is real, diminished, and at its focus.
(A) (i) and (ii)(B) (ii) and (iii)(C) (iii) and (iv)(D) (ii) and (iv)
Answer:
Explanation:
Mirror X (Concave Mirror) is used in solar furnaces to concentrate sunlight at its focus, forming a real and diminished image.
Mirror Y (Convex Mirror) is used in vehicles as a rear-view mirror, which always forms a virtual, erect, and diminished image.
Question 15:
The percentage of solar energy that is not converted into food energy by the leaves of green plants in a terrestrial ecosystem is about:
(A) 1%(B) 10%(C) 90%(D) 99%
Answer:
Explanation:
Only about 1% of solar energy is actually converted into chemical energy (food) by plants through photosynthesis.
The remaining 99% is reflected, absorbed as heat, or lost in other ways.
Question 16:
Which of the following groups do not constitute a food chain?
(i) Wolf, rabbit, grass, lion(ii) Plankton, man, grasshopper, fish(iii) Hawk, grass, snake, grasshopper, frog(iv) Grass, snake, wolf, tiger
(A) (i) and (iv)(B) (i) and (iii)(C) (ii) and (iii)(D) (ii) and (iv)
Answer:
Explanation:
A food chain must follow a linear energy flow: producer → primary consumer → secondary consumer → tertiary consumer.
Option (ii) is incorrect because man and grasshopper do not fit in a logical sequence.
Option (iv) is incorrect because wolves and tigers are both apex predators and do not follow a logical food chain sequence.
Question 17:
Assertion (A): A human child bears all the basic features of human beings.Reason (R): It looks exactly like its parents, showing very little variation.
(A) Both A and R are true, and R is the correct explanation of A.(B) Both A and R are true, but R is not the correct explanation of A.(C) A is true, but R is false.(D) A is false, but R is true.
Answer:
(C) A is true, but R is false.
Explanation:
Question 18:
Assertion (A): The amount of ozone in the atmosphere began to drop sharply in the 1980s.Reason (R): The oxygen atoms combine with molecular oxygen to form ozone.
(A) Both A and R are true, and R is the correct explanation of A.(B) Both A and R are true, but R is not the correct explanation of A.(C) A is true, but R is false.(D) A is false, but R is true.
Answer:
(B) Both A and R are true, but R is not the correct explanation of A.
Explanation:
The ozone layer depletion was due to CFCs (Chlorofluorocarbons), not just the natural process of ozone formation.
Ozone formation (O₂ + O → O₃) is a continuous process, but its depletion in the 1980s was due to human activities.
Question 19:
Assertion (A): Decomposition reactions are generally endothermic reactions.Reason (R): Decomposition of organic matter into compost is an exothermic process.
(A) Both A and R are true, and R is the correct explanation of A.(B) Both A and R are true, but R is not the correct explanation of A.(C) A is true, but R is false.(D) A is false, but R is true.
Answer:
(C) A is true, but R is false.
Explanation:
Most decomposition reactions require heat (endothermic reactions).
However, organic decomposition (composting) releases heat (exothermic process), contradicting the reason given.
Question 20:
Assertion (A): No two magnetic field lines are found to cross each other.Reason (R): The compass needle cannot point towards two directions at the point of intersection of two magnetic field lines.
(A) Both A and R are true, and R is the correct explanation of A.(B) Both A and R are true, but R is not the correct explanation of A.(C) A is true, but R is false.(D) A is false, but R is true.
Answer:
(A) Both A and R are true, and R is the correct explanation of A.
Explanation:
Question 21:
"Excessive use of chemicals and pesticides in agriculture adversely affects the environment." Justify this statement.
Answer:
Soil Pollution: Excessive use of chemicals depletes soil fertility and disrupts natural microbial balance.
Water Contamination: Pesticides and fertilizers seep into groundwater and water bodies, harming aquatic life.
Bioaccumulation: Harmful chemicals enter the food chain, causing toxic accumulation in organisms.
Loss of Biodiversity: Chemicals kill beneficial insects, soil organisms, and can disrupt ecosystems.
Air Pollution: Volatilization of pesticides contributes to air pollution and harms human health.
Question 22 (a):
Consider the following circuits:
(i) A single 5 Ω resistor connected across 12 V battery.(ii) Two 5 Ω resistors in series across 12 V battery.(iii) Two 5 Ω resistors in parallel across 12 V battery.
In which circuit will the power dissipated be (i) minimum and (ii) maximum? Justify your answer.
Answer:
(i) P1=1225=28.8WP_1 = \frac{12^2}{5} = 28.8 WP1=5122=28.8W
(ii) P2=12210=14.4WP_2 = \frac{12^2}{10} = 14.4 WP2=10122=14.4W (Minimum Power Dissipation)
(iii) P3=1222.5=57.6WP_3 = \frac{12^2}{2.5} = 57.6 WP3=2.5122=57.6W (Maximum Power Dissipation)
Final Answer:
Question 22 (b) [Alternative Question]:
Two lamps (100 W, 220 V) and (60 W, 220 V) are connected in parallel to a 220 V electric main supply. Find the current drawn from the supply.
Answer:
Power formula: P=VIP = VIP=VI
Current calculation for each lamp:
Total current drawn: I=I1+I2=0.4545+0.2727=0.727AI = I_1 + I_2 = 0.4545 + 0.2727 = 0.727 AI=I1+I2=0.4545+0.2727=0.727A
Final Answer:
Total current drawn from the supply = 0.727 A.
Question 23:
An object is placed at a distance of 30 cm in front of a concave mirror of focal length 20 cm. Use mirror formula to determine the position of the image formed.
Answer:
1f=1v−1u\frac{1}{f} = \frac{1}{v} - \frac{1}{u}f1=v1−u1
Given:f=−20f = -20f=−20 cm (Concave mirror), u=−30u = -30u=−30 cm
1v=1−20+130\frac{1}{v} = \frac{1}{-20} + \frac{1}{30}v1=−201+3011v=−3+260=−160\frac{1}{v} = \frac{-3 + 2}{60} = \frac{-1}{60}v1=60−3+2=60−1v=−60 cmv = -60 \text{ cm}v=−60 cm
Final Answer:
The image is formed at 60 cm behind the mirror. It is real, inverted, and magnified.
Question 24 (a):
Besides minimizing the loss of blood, why is it essential to plug any leak in a blood vessel? Name the component of blood that helps in this process and state how they perform their function.
Answer:**
Importance of plugging a blood leak: Prevents excessive blood loss, maintains blood pressure, and prevents infection.
Component involved: Platelets
Function of platelets:
Platelets clump together at the site of injury.
They release clotting factors to form a fibrin network.
This stops bleeding and forms a clot.
Question 24 (b) [Alternative Question]:
(i) The transport system in plants is relatively slower than in animals. Give reasons.
Answer:
No pump: Plants lack a heart-like organ to speed up circulation.
Passive transport: Water and nutrients move via diffusion, osmosis, and transpiration pull, which is slow.
Less energy demand: Unlike animals, plants do not require rapid nutrient transport.
(ii) State the role of phloem in the transport of materials in plants.
Answer:
Phloem transports organic nutrients (sugars, amino acids) from leaves to other parts of the plant.
This movement is called translocation, and it occurs bidirectionally (both upward and downward).
It provides energy for growth, storage, and metabolism.
Question 25:
Draw labeled diagrams to show different stages of budding in Hydra.
Answer:
To show the different stages of budding in Hydra, draw and label the following:
Parent Hydra – Initial Hydra with no bud.
Bud Formation – A small outgrowth appears due to repeated mitotic divisions.
Bud Growth – The bud enlarges, developing tentacles and a mouth.
Mature Bud – The bud attains the shape of an adult Hydra.
Separation – The new Hydra detaches from the parent and lives independently.
Question 26:
What happens when:
(a) Lead nitrate is thermally decomposed? Write a balanced chemical equation.
Answer:
When lead nitrate is heated, it decomposes into lead oxide (PbO), nitrogen dioxide (NO₂), and oxygen (O₂).
2Pb(NO3)2→Heat2PbO+4NO2+O22Pb(NO_3)_2 \xrightarrow{\text{Heat}} 2PbO + 4NO_2 + O_22Pb(NO3)2Heat2PbO+4NO2+O2
(b) Natural gas burns in oxygen (or air)?
Answer:
Natural gas (mostly methane, CH₄) burns in oxygen to produce carbon dioxide and water.
CH4+2O2→CO2+2H2OCH_4 + 2O_2 \rightarrow CO_2 + 2H_2OCH4+2O2→CO2+2H2O
Question 27:
(a) Define one ampere.
Answer:
One ampere (A) is the amount of electric current when one coulomb (C) of charge flows through a conductor in one second (s).
1A=1C1s1 A = \frac{1 C}{1 s}1A=1s1C
(b) The resistance of a wire of 0.01 cm radius is 14 Ω. If the resistivity of the material of the wire is 44×10−844 \times 10^{-8}44×10−8 Ωm, find the length of the wire.
Answer:
Using the formula for resistance:
R=ρlAR = \rho \frac{l}{A}R=ρAl
R=14R = 14R=14 Ω
ρ=44×10−8\rho = 44 \times 10^{-8}ρ=44×10−8 Ωm
r=0.01r = 0.01r=0.01 cm = 1×10−41 \times 10^{-4}1×10−4 m
A=πr2=3.14×(10−4)2=3.14×10−8A = \pi r^2 = 3.14 \times (10^{-4})^2 = 3.14 \times 10^{-8}A=πr2=3.14×(10−4)2=3.14×10−8 m²
l=R×Aρl = \frac{R \times A}{\rho}l=ρR×Al=14×3.14×10−844×10−8l = \frac{14 \times 3.14 \times 10^{-8}}{44 \times 10^{-8}}l=44×10−814×3.14×10−8l=14×3.1444=1ml = \frac{14 \times 3.14}{44} = 1 ml=4414×3.14=1m
Final Answer: Length of the wire is 1 meter.
Question 28:
Consider the given electric circuit:
(a) Find the total resistance of the circuit.
Answer:
Parallel combination of 10 Ω and 15 Ω:
1R1=110+115=3+230=530\frac{1}{R_1} = \frac{1}{10} + \frac{1}{15} = \frac{3 + 2}{30} = \frac{5}{30}R11=101+151=303+2=305R1=6ΩR_1 = 6 ΩR1=6Ω
Series combination of 40 Ω and 60 Ω:
R2=40+60=100ΩR_2 = 40 + 60 = 100 ΩR2=40+60=100Ω
Rtotal=R1+R2=6+100=106ΩR_{\text{total}} = R_1 + R_2 = 6 + 100 = 106 ΩRtotal=R1+R2=6+100=106Ω
Final Answer: Total resistance = 106 Ω
(b) Find the total current drawn from the source.
V=IRV = IRV=IRI=VR=15106=0.14AI = \frac{V}{R} = \frac{15}{106} = 0.14 AI=RV=10615=0.14A
Final Answer: Total current drawn = 0.14 A
(c) Find the potential difference across the parallel combination of 10 Ω and 15 Ω resistors.
Voltage across parallel combination = Voltage drop across R1R_1R1:
V1=IR1=0.14×6=0.84VV_1 = I R_1 = 0.14 \times 6 = 0.84 VV1=IR1=0.14×6=0.84V
Final Answer: Potential difference = 0.84 V.
Q29. (a) Draw ray diagrams to show the nature, position, and relative size of the image formed by a convex mirror when the object is placed at (i) infinity and (ii) between infinity and the pole P of the mirror.
Answer:
(i) When the object is at infinity:
The image is formed at the focus (F) of the convex mirror.
The image is virtual, erect, and highly diminished (point-sized).
(ii) When the object is between infinity and the pole (P):
The image is formed between the focus (F) and the pole (P).
The image is virtual, erect, and diminished in size.
Ray Diagrams:(You can draw the standard ray diagrams for convex mirrors showing these conditions.)
Q30. (a) With the help of an activity, explain the conditions under which iron articles get rusted.
Answer:
(b) OR
(i) Name two metals which react violently with cold water. List any three observations which a student notes when these metals are dropped in a beaker containing water.
Answer:
(ii) Write a test to identify the gas evolved (if any) during the reaction of these metals with water.
Answer:
The gas evolved is hydrogen gas (H₂).
Test for Hydrogen: Bring a burning splinter near the gas; it will burn with a ‘pop’ sound, confirming the presence of hydrogen gas.
Q31. (a) "Displacement reactions also play a key role in extracting metals in the middle of the reactivity series." Justify this statement with two examples.
Answer:
Explanation:
Examples:
Zinc displaces Copper from Copper Sulphate solution: Zn+CuSO4→ZnSO4+CuZn + CuSO_4 → ZnSO_4 + CuZn+CuSO4→ZnSO4+Cu
Iron displaces Copper from Copper Sulphate solution: Fe+CuSO4→FeSO4+CuFe + CuSO_4 → FeSO_4 + CuFe+CuSO4→FeSO4+Cu
(b) Why can metals high up in the reactivity series not be obtained by reduction of their oxides by carbon?
Answer:
Highly reactive metals such as Sodium (Na), Potassium (K), Calcium (Ca), and Aluminium (Al) form very stable oxides.
Carbon is not reactive enough to reduce these metal oxides.
Instead, these metals are extracted using electrolysis, which provides the high energy needed to break the strong bonds between metal and oxygen.
Q32. In one of Mendelian experiments, when F₁ generation pea plants with round yellow seeds were self-pollinated, pea seeds with the following combinations were obtained in F₂ generation:
Analyse the result and describe the mechanism of inheritance of traits which explains the above results.
Answer:
Q33. (a) Name the glands that secrete:
(i) Adrenaline – Adrenal glands(ii) Thyroxine – Thyroid gland
(b) Explain with example how the timing and amount of hormone released are regulated in the human body.
Answer:
Hormones are regulated by feedback mechanisms.
Example:
Thyroxine regulation:
If thyroxine levels are low, the pituitary gland releases TSH (Thyroid Stimulating Hormone), signaling the thyroid gland to produce more thyroxine.
If thyroxine levels are high, the secretion of TSH is inhibited, stopping excess thyroxine production.
Insulin regulation:
When blood sugar levels are high, the pancreas releases insulin to lower it.
When blood sugar is low, insulin secretion stops, maintaining balance.
Question 29:
(a) A saturated organic compound ‘A’ with two carbon atoms belongs to the homologous series of alcohols. On oxidation, it forms an organic acid ‘B’ with a molecular mass of 60 u. On heating ‘A’ with excess concentrated sulphuric acid at 443 K, an unsaturated hydrocarbon ‘C’ is formed.
Answers:
(i) Name A, B, and C.
(ii) Calculate the molecular mass of C.
(iii) What happens when a pinch of sodium carbonate is added to compound B? Write the chemical equation for the reaction.
2CH3COOH+Na2CO3→2CH3COONa+H2O+CO22CH_3COOH + Na_2CO_3 \rightarrow 2CH_3COONa + H_2O + CO_22CH3COOH+Na2CO3→2CH3COONa+H2O+CO2
(iv) Draw the electron dot structure of compound B (Ethanoic acid, CH₃COOH).
Answer:The electron dot structure shows:
Structure:
OR (Option for Q29)
(b) What is a homologous series of carbon compounds? Write the name and formula of three successive members of the homologous series of compounds having the functional group –COOH.
Answer:
A homologous series is a group of organic compounds having the same functional group and similar chemical properties, with each successive member differing by –CH₂.
Three successive members of the –COOH series:
Methanoic acid (HCOOH)
Ethanoic acid (CH₃COOH)
Propanoic acid (C₂H₅COOH)
(ii) Write the name of two carbon compounds in which carbon atoms are arranged in the form of a ring. Draw the structure of any one.
Answer:
Examples of ring-shaped carbon compounds:
Cyclohexane (C₆H₁₂)
Benzene (C₆H₆)
(Benzene is represented as a hexagon with alternating double bonds.)
Question 30:
(a) Write the functions of the following parts of the human female reproductive system:
Answer:
(i) Ovary – Produces eggs (ova) and secretes female hormones (estrogen and progesterone).
(ii) Fallopian Tube – Transfers the ovum from the ovary to the uterus; also the site of fertilization.
(iii) Uterus – Nurtures the developing embryo/fetus during pregnancy.
(b) State briefly two contraceptive methods used by human males.
Answer:
Barrier method (Condoms) – Prevent sperm from reaching the egg.
Surgical method (Vasectomy) – The sperm ducts (vas deferens) are cut and sealed to prevent the release of sperm during ejaculation.
Q29. (a) Draw ray diagrams to show the nature, position, and relative size of the image formed by a convex mirror when the object is placed at (i) infinity and (ii) between infinity and the pole P of the mirror.
Answer:
(i) When the object is at infinity:
The image is formed at the focus (F) of the convex mirror.
The image is virtual, erect, and highly diminished (point-sized).
(ii) When the object is between infinity and the pole (P):
The image is formed between the focus (F) and the pole (P).
The image is virtual, erect, and diminished in size.
Ray Diagrams:(You can draw the standard ray diagrams for convex mirrors showing these conditions.)
Q30. (a) With the help of an activity, explain the conditions under which iron articles get rusted.
Answer:
(b) OR
(i) Name two metals which react violently with cold water. List any three observations which a student notes when these metals are dropped in a beaker containing water.
Answer:
(ii) Write a test to identify the gas evolved (if any) during the reaction of these metals with water.
Answer:
The gas evolved is hydrogen gas (H₂).
Test for Hydrogen: Bring a burning splinter near the gas; it will burn with a ‘pop’ sound, confirming the presence of hydrogen gas.
Q31. (a) "Displacement reactions also play a key role in extracting metals in the middle of the reactivity series." Justify this statement with two examples.
Answer:
Explanation:
Examples:
Zinc displaces Copper from Copper Sulphate solution: Zn+CuSO4→ZnSO4+CuZn + CuSO_4 → ZnSO_4 + CuZn+CuSO4→ZnSO4+Cu
Iron displaces Copper from Copper Sulphate solution: Fe+CuSO4→FeSO4+CuFe + CuSO_4 → FeSO_4 + CuFe+CuSO4→FeSO4+Cu
(b) Why can metals high up in the reactivity series not be obtained by reduction of their oxides by carbon?
Answer:
Highly reactive metals such as Sodium (Na), Potassium (K), Calcium (Ca), and Aluminium (Al) form very stable oxides.
Carbon is not reactive enough to reduce these metal oxides.
Instead, these metals are extracted using electrolysis, which provides the high energy needed to break the strong bonds between metal and oxygen.
Q32. In one of Mendelian experiments, when F₁ generation pea plants with round yellow seeds were self-pollinated, pea seeds with the following combinations were obtained in F₂ generation:
Analyse the result and describe the mechanism of inheritance of traits which explains the above results.
Answer:
Q33. (a) Name the glands that secrete:
(i) Adrenaline – Adrenal glands(ii) Thyroxine – Thyroid gland
(b) Explain with example how the timing and amount of hormone released are regulated in the human body.
Answer:
Hormones are regulated by feedback mechanisms.
Example:
Thyroxine regulation:
If thyroxine levels are low, the pituitary gland releases TSH (Thyroid Stimulating Hormone), signaling the thyroid gland to produce more thyroxine.
If thyroxine levels are high, the secretion of TSH is inhibited, stopping excess thyroxine production.
Insulin regulation:
When blood sugar levels are high, the pancreas releases insulin to lower it.
When blood sugar is low, insulin secretion stops, maintaining blance.
sQ37 (b) What should be the current rating of the electric circuit (220 V) so that an electric iron of 1 kW power rating can be operated?
Answer:
We use the formula:
P=V×IP = V \times IP=V×I
Where,
Solving for III:
I=PV=1000220≈4.55AI = \frac{P}{V} = \frac{1000}{220} \approx 4.55 AI=VP=2201000≈4.55A
So, the electric circuit should have a current rating of at least 4.55 A (rounded to 5 A for safety).
(c) (i) What is the function of the earth wire? State the advantage of the earth wire in domestic electric appliances such as an electric iron.
Answer:
Function of Earth Wire: The earth wire provides a safe path for excess electric current to flow into the ground in case of leakage or a short circuit.
Advantage: It prevents electric shocks and protects appliances from damage by directing excess current away from the device.
OR
(c) (ii) List two precautions to be taken to avoid electrical accidents. State how these precautions prevent possible damage to the circuit/appliance.
Answer:
Precautions to avoid electrical accidents:
Do not touch electrical appliances with wet hands – Water is a good conductor, and it can cause electric shocks.
Use proper earthing in electrical wiring – This prevents electric shocks and protects appliances from damage due to current leakage.
How these precautions help:
Q38. The maintenance functions of all living organisms must go on even when they are not doing anything particular. Even when we are just sitting in a class or even asleep, this maintenance job has to go on. These maintenance processes require energy to prevent damage and breakdown of cells and tissues, which is obtained by the individual organism from the food prepared by the autotrophs, called producers.
(a) Name and define the process by which green plants prepare food.
Answer:
Process Name: Photosynthesis.
Definition: It is the process by which green plants synthesize their food (glucose) using sunlight, carbon dioxide, and water, releasing oxygen as a by-product.
(b) Write the chemical equation involved in the above process.
Answer:
6CO2+6H2O→Sunlight, ChlorophyllC6H12O6+6O26CO_2 + 6H_2O \xrightarrow{\text{Sunlight, Chlorophyll}} C_6H_{12}O_6 + 6O_26CO2+6H2OSunlight, ChlorophyllC6H12O6+6O2
(Carbon dioxide + Water → Glucose + Oxygen)
(c) (i) State in proper sequence the events that occur in the synthesis of food by desert plants.
Answer:
Opening of stomata at night to take in carbon dioxide (CAM Photosynthesis).
Storage of CO₂ as an intermediate compound (like malic acid).
Closing of stomata during the daytime to prevent water loss.
Using stored CO₂ during daylight for photosynthesis, when sunlight is available.
OR
(c) (ii) Explain, giving reasons, what happens to the rate at which green plants will prepare food
(I) During cloudy weather
Answer:
Photosynthesis decreases because less sunlight is available.
Reason: Sunlight is essential for photosynthesis, and reduced light intensity slows down the process.
(II) When stomata get blocked due to dust
Answer:
Photosynthesis decreases significantly.
Reason: Stomata allow gas exchange (CO₂ intake and O₂ release). If blocked, CO₂ cannot enter, reducing the photosynthesis rate.
Q39. Seawater contains many salts dissolved in it. Common salt is separated from these salts. Deposits of solid salt are also found in several parts of the world. These large crystals are often brown due to impurities. This is called rock salt and is mined like coal. The common salt is an important raw material for chemicals of daily use.
(a) Write balanced chemical equations to show the products formed during electrolysis of brine.
Answer:
The electrolysis of brine (NaCl\text{NaCl}NaCl solution) produces sodium hydroxide (NaOH), chlorine gas (Cl₂), and hydrogen gas (H₂).
2NaCl+2H2O→Electrolysis2NaOH+H2+Cl22NaCl + 2H_2O \xrightarrow{\text{Electrolysis}} 2NaOH + H_2 + Cl_22NaCl+2H2OElectrolysis2NaOH+H2+Cl2
Chlorine gas (Cl₂) is liberated at the anode.
Hydrogen gas (H₂) is liberated at the cathode.
Sodium hydroxide (NaOH) remains in solution.
(b) List two uses of any one product obtained during electrolysis of brine.
Answer:
Used in water purification to kill bacteria.
Used in the production of PVC (polyvinyl chloride), disinfectants, and bleaching agents.
Uses of Sodium Hydroxide (NaOH):
Used in the manufacture of soap and detergents.
Used in the paper and textile industries.
(c) (i) A mild non-corrosive basic salt ‘A’, used for faster cooking, is strongly heated to produce a compound ‘B’, that is used for removing permanent hardness of water. Identify A and B and also write the equation for the reaction that occurs when A is heated.
Answer:
A = Baking soda (NaHCO3NaHCO_3NaHCO3, Sodium bicarbonate)
B = Washing soda (Na2CO3Na_2CO_3Na2CO3, Sodium carbonate)
2NaHCO3→HeatNa2CO3+CO2+H2O2NaHCO_3 \xrightarrow{\text{Heat}} Na_2CO_3 + CO_2 + H_2O2NaHCO3HeatNa2CO3+CO2+H2O
Baking soda (A) is used in cooking to make food fluffy.
When heated, it decomposes into washing soda (B), which is used to remove permanent hardness of water.
OR
(c) (ii) Define water of crystallisation. Give two examples of salts that have water of crystallisation.
Answer:
