▶️ Answer/Explanation
Green plants detect and respond to stimuli (e.g. phototropism), making statement 1 correct.
They respire aerobically, breaking down glucose to release energy for metabolism, making statement 2 correct.
They absorb \( \text{CO}_2 \), water, and mineral ions for photosynthesis, growth, and development, making statement 3 correct.
✅ Answer: (A)
They respire aerobically, breaking down glucose to release energy for metabolism, making statement 2 correct.
They absorb \( \text{CO}_2 \), water, and mineral ions for photosynthesis, growth, and development, making statement 3 correct.
✅ Answer: (A)
▶️ Answer/Explanation
A high temperature increases the kinetic energy of particles, speeding up their random movement and thus the rate of diffusion.
A large concentration gradient means there is a greater difference in particle concentration across the barrier, driving faster net movement of particles.
A short diffusion distance means particles have less far to travel, so they reach the other side more quickly.
✅ Answer: (A)
A large concentration gradient means there is a greater difference in particle concentration across the barrier, driving faster net movement of particles.
A short diffusion distance means particles have less far to travel, so they reach the other side more quickly.
✅ Answer: (A)
▶️ Answer/Explanation
Benedict’s solution is initially blue; when heated with a reducing sugar such as glucose, the \( \text{Cu}^{2+} \) ions are reduced to \( \text{Cu}^{+} \) ions, forming an insoluble brick-red precipitate of copper(I) oxide \( (\text{Cu}_2\text{O}) \).
A blue-black colour indicates starch (iodine test), a purple colour indicates protein (Biuret test), and a white emulsion indicates fat (emulsion test).
✅ Answer: (B)
A blue-black colour indicates starch (iodine test), a purple colour indicates protein (Biuret test), and a white emulsion indicates fat (emulsion test).
✅ Answer: (B)
▶️ Answer/Explanation
Denaturation refers to a permanent alteration of the enzyme’s tertiary structure, specifically the shape of its active site — it is the enzyme that is denatured, not the substrate.
Extremes of pH (above or below the optimum) break ionic bonds and hydrogen bonds within the enzyme, distorting the active site so the substrate can no longer bind.
Option A correctly identifies the enzyme’s active site and states a pH above the optimum as the cause — both elements are accurate.
✅ Answer: (A)
Extremes of pH (above or below the optimum) break ionic bonds and hydrogen bonds within the enzyme, distorting the active site so the substrate can no longer bind.
Option A correctly identifies the enzyme’s active site and states a pH above the optimum as the cause — both elements are accurate.
✅ Answer: (A)
▶️ Answer/Explanation
Since the leaf floats on water, the lower epidermis is in contact with water, so placing stomata there would block gas exchange; instead, stomata are located in the upper epidermis, which is exposed to the air.
This allows \( \text{CO}_2 \) to diffuse in for photosynthesis and \( \text{O}_2 \) to diffuse out — essential gas exchanges — directly with the atmosphere.
Guard cells can open and close (they are not always closed), and the cuticle on the upper surface prevents excess water loss rather than water entry.
✅ Answer: (C)
This allows \( \text{CO}_2 \) to diffuse in for photosynthesis and \( \text{O}_2 \) to diffuse out — essential gas exchanges — directly with the atmosphere.
Guard cells can open and close (they are not always closed), and the cuticle on the upper surface prevents excess water loss rather than water entry.
✅ Answer: (C)
▶️ Answer/Explanation
Incisors are the flat, chisel-shaped teeth at the very front of the mouth, specifically adapted for biting off pieces of food.
Canines are pointed and used for tearing food, while premolars and molars have broad, ridged surfaces used for grinding and chewing food.
✅ Answer: (B)
Canines are pointed and used for tearing food, while premolars and molars have broad, ridged surfaces used for grinding and chewing food.
✅ Answer: (B)
▶️ Answer/Explanation
Higher humidity means more water vapour is already present in the surrounding air, which reduces the water vapour concentration gradient between the leaf’s air spaces and the outside air.
A less steep diffusion gradient means water vapour diffuses out of the stomata more slowly, reducing the overall rate of transpiration.
As a result, the plant loses less water and the decrease in mass on the balance is smaller.
✅ Answer: (D)
A less steep diffusion gradient means water vapour diffuses out of the stomata more slowly, reducing the overall rate of transpiration.
As a result, the plant loses less water and the decrease in mass on the balance is smaller.
✅ Answer: (D)
▶️ Answer/Explanation0
During exercise, muscle cells respire faster, producing more \( \text{CO}_2 \) and consuming more \( \text{O}_2 \), which raises the \( \text{CO}_2 \) concentration in the blood.
This is detected by receptors, which send signals to increase both the rate (breaths per minute) and depth (volume per breath) of breathing to expel \( \text{CO}_2 \) and take in more \( \text{O}_2 \).
✅ Answer: (D)
This is detected by receptors, which send signals to increase both the rate (breaths per minute) and depth (volume per breath) of breathing to expel \( \text{CO}_2 \) and take in more \( \text{O}_2 \).
✅ Answer: (D)
▶️ Answer/Explanation
To focus on a near object like a book, the ciliary muscles contract, which reduces the tension on the suspensory ligaments, causing them to slacken.
With less tension, the elastic lens becomes fatter (more convex), increasing its refractive power so that light from the near object is focused correctly onto the retina.
✅ Answer: (A)
With less tension, the elastic lens becomes fatter (more convex), increasing its refractive power so that light from the near object is focused correctly onto the retina.
✅ Answer: (A)
▶️ Answer/Explanation
Asexual reproduction requires only one parent organism and does not involve the fusion of gamete nuclei, so all offspring are genetically identical to the parent (clones).
This means the correct row must show: only one parent — yes; fusion of nuclei — no; genetically identical offspring — yes, which corresponds to row C.
✅ Answer: (C)
This means the correct row must show: only one parent — yes; fusion of nuclei — no; genetically identical offspring — yes, which corresponds to row C.
✅ Answer: (C)
▶️ Answer/Explanation
The parents are both unaffected yet produce an affected child, which means the disease allele must be recessive — unaffected parents can carry a hidden recessive allele.
Since each parent is unaffected but passes on the disease allele, they must both be heterozygous carriers (one dominant allele, one recessive disease allele), written as \( Aa \times Aa \).
This cross gives a \( \frac{1}{4} \) probability of an affected child (\( aa \)), consistent with the pedigree shown.
✅ Answer: (B)
Since each parent is unaffected but passes on the disease allele, they must both be heterozygous carriers (one dominant allele, one recessive disease allele), written as \( Aa \times Aa \).
This cross gives a \( \frac{1}{4} \) probability of an affected child (\( aa \)), consistent with the pedigree shown.
✅ Answer: (B)
▶️ Answer/Explanation
The first trophic level consists of producers — green plants and algae — which capture light energy from the Sun and convert it into chemical energy stored in organic molecules through photosynthesis.
The equation for photosynthesis is: \( 6\text{CO}_2 + 6\text{H}_2\text{O} \xrightarrow{\text{light}} \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \), showing how solar energy is fixed into glucose.
Option B describes decomposers, C describes predators (higher trophic levels), and D describes herbivores (second trophic level).
✅ Answer: (A)
The equation for photosynthesis is: \( 6\text{CO}_2 + 6\text{H}_2\text{O} \xrightarrow{\text{light}} \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \), showing how solar energy is fixed into glucose.
Option B describes decomposers, C describes predators (higher trophic levels), and D describes herbivores (second trophic level).
✅ Answer: (A)
▶️ Answer/Explanation
Fertiliser washes into the lake, providing excess nitrates and phosphates, which causes a rapid increase in algal growth (algal bloom) — this is step 2.
The algae block sunlight from reaching aquatic plants, causing the aquatic plants to die (step 4); decomposer bacteria then break down the dead plant material (step 1), using up dissolved oxygen in the process.
This high demand by decomposers removes dissolved oxygen from the water, causing oxygen levels in the lake to fall (step 3), potentially killing other aquatic organisms.
✅ Answer: (A)
The algae block sunlight from reaching aquatic plants, causing the aquatic plants to die (step 4); decomposer bacteria then break down the dead plant material (step 1), using up dissolved oxygen in the process.
This high demand by decomposers removes dissolved oxygen from the water, causing oxygen levels in the lake to fall (step 3), potentially killing other aquatic organisms.
✅ Answer: (A)
▶️ Answer/Explanation
Reading the thermometer carefully: the initial temperature is \( 21.4 \, ^\circ\text{C} \) and the final temperature is \( 29.0 \, ^\circ\text{C} \) (the scale spans from 21 to 22 with subdivisions of 0.2 °C for initial, and 29 to 30 for final).
The temperature change is \( \Delta T = 29.0 – 21.4 = 7.6 \, ^\circ\text{C} \).
✅ Answer: (D)
The temperature change is \( \Delta T = 29.0 – 21.4 = 7.6 \, ^\circ\text{C} \).
✅ Answer: (D)
▶️ Answer/Explanation
The temperature crosses the boiling point of ammonia \( (-33 \, ^\circ\text{C}) \), so the liquid ammonia boils and becomes gaseous ammonia — this is a change of state (liquid → gas).
A change of state is a physical change because the chemical identity of ammonia \( (\text{NH}_3) \) is preserved; no new substances such as hydrogen or nitrogen are formed.
✅ Answer: (C)
A change of state is a physical change because the chemical identity of ammonia \( (\text{NH}_3) \) is preserved; no new substances such as hydrogen or nitrogen are formed.
✅ Answer: (C)
▶️ Answer/Explanation
Methanol has the formula \( \text{CH}_3\text{OH} \): carbon forms four covalent bonds (three C–H and one C–O), and oxygen forms two covalent bonds (one C–O and one O–H) while retaining two lone pairs.
The correct dot-and-cross diagram (A) shows all bonding pairs accurately and includes the two lone pairs on the oxygen atom, with electrons from different atoms distinguished by dots and crosses.
✅ Answer: (A)
The correct dot-and-cross diagram (A) shows all bonding pairs accurately and includes the two lone pairs on the oxygen atom, with electrons from different atoms distinguished by dots and crosses.
✅ Answer: (A)
▶️ Answer/Explanation
Silicon(IV) oxide \( (\text{SiO}_2) \) has a giant covalent (macromolecular) structure in which each silicon atom is covalently bonded to four oxygen atoms arranged tetrahedrally.
Each oxygen atom acts as a bridge between two silicon atoms, so every oxygen is bonded to exactly two silicon atoms — this accounts for the empirical formula \( \text{SiO}_2 \) (ratio 1 Si : 2 O, consistent with Si bonded to 4 O and each O shared between 2 Si).
✅ Answer: (B)
Each oxygen atom acts as a bridge between two silicon atoms, so every oxygen is bonded to exactly two silicon atoms — this accounts for the empirical formula \( \text{SiO}_2 \) (ratio 1 Si : 2 O, consistent with Si bonded to 4 O and each O shared between 2 Si).
✅ Answer: (B)
▶️ Answer/Explanation
The relative atomic masses are \( A_r(\text{Mg}) = 24 \) and \( A_r(\text{O}) = 16 \), giving \( M_r(\text{MgO}) = 40 \).
From the equation, \( 2 \) mol Mg (mass \( 2 \times 24 = 48 \, \text{g} \)) produces \( 2 \) mol MgO (mass \( 2 \times 40 = 80 \, \text{g} \)), so the ratio of Mg to MgO by mass is \( 48 : 80 \).
For \( 12 \, \text{g} \) of Mg: \( \text{mass of MgO} = \frac{12}{48} \times 80 = 20 \, \text{g} \).
✅ Answer: (A)
From the equation, \( 2 \) mol Mg (mass \( 2 \times 24 = 48 \, \text{g} \)) produces \( 2 \) mol MgO (mass \( 2 \times 40 = 80 \, \text{g} \)), so the ratio of Mg to MgO by mass is \( 48 : 80 \).
For \( 12 \, \text{g} \) of Mg: \( \text{mass of MgO} = \frac{12}{48} \times 80 = 20 \, \text{g} \).
✅ Answer: (A)
▶️ Answer/Explanation
In Experiment 1 (aqueous \( \text{CuSO}_4 \)): \( \text{Cu}^{2+} \) is preferentially discharged at the cathode (copper deposited), and \( \text{OH}^- \) from water is discharged at the anode, releasing oxygen gas.
In Experiment 2 (molten \( \text{LiBr} \)): only \( \text{Li}^+ \) and \( \text{Br}^- \) ions are present; lithium metal is deposited at the cathode, and bromine gas is produced at the anode.
✅ Answer: (C)
In Experiment 2 (molten \( \text{LiBr} \)): only \( \text{Li}^+ \) and \( \text{Br}^- \) ions are present; lithium metal is deposited at the cathode, and bromine gas is produced at the anode.
✅ Answer: (C)
▶️ Answer/Explanation
The reaction \( 2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2 \) is vigorous and releases heat (observed as fizzing and sometimes ignition of the hydrogen), confirming it is exothermic.
In an exothermic energy level diagram, the products (\( \text{NaOH} + \text{H}_2 \)) are drawn at a lower energy level than the reactants (\( \text{Na} + \text{H}_2\text{O} \)), with energy released shown as a downward drop — this matches diagram D.
✅ Answer: (D)
In an exothermic energy level diagram, the products (\( \text{NaOH} + \text{H}_2 \)) are drawn at a lower energy level than the reactants (\( \text{Na} + \text{H}_2\text{O} \)), with energy released shown as a downward drop — this matches diagram D.
✅ Answer: (D)
▶️ Answer/Explanation
Increasing the volume of the acid (while keeping its concentration constant) does not increase the frequency of collisions between acid particles and the magnesium surface — it simply provides more acid molecules, but they collide at the same rate per unit area.
In contrast, increasing concentration raises the number of acid particles per unit volume (more frequent collisions), increasing temperature gives particles more kinetic energy (more energetic and frequent successful collisions), and using powdered magnesium increases the surface area exposed to the acid.
✅ Answer: (C)
In contrast, increasing concentration raises the number of acid particles per unit volume (more frequent collisions), increasing temperature gives particles more kinetic energy (more energetic and frequent successful collisions), and using powdered magnesium increases the surface area exposed to the acid.
✅ Answer: (C)
▶️ Answer/Explanation
The test for nitrate ions \( (\text{NO}_3^-) \) uses aluminium powder and concentrated sodium hydroxide solution, which reduces the nitrate ions to ammonia gas \( (\text{NH}_3) \).
The ammonia produced turns moist red litmus paper blue, confirming the presence of nitrate ions; aluminium acts as the reducing agent in this specific test.
✅ Answer: (A)
The ammonia produced turns moist red litmus paper blue, confirming the presence of nitrate ions; aluminium acts as the reducing agent in this specific test.
✅ Answer: (A)
▶️ Answer/Explanation
Dense solids with high melting points are characteristic of transition metals (e.g. iron, copper, nickel), which occupy the central d-block of the Periodic Table between Groups II and III.
This region corresponds to position B in the diagram; non-metals (right side) and reactive metals in Groups I and II (left side) do not share these physical properties.
✅ Answer: (B)
This region corresponds to position B in the diagram; non-metals (right side) and reactive metals in Groups I and II (left side) do not share these physical properties.
✅ Answer: (B)
▶️ Answer/Explanation
Helium has only 2 electrons in its (full) outer shell, so option A is false — not all noble gases have 8 outer electrons.
Noble gases are monatomic (they exist as single atoms, not diatomic molecules), so option B is false; and it is helium (not argon) that fills weather balloons due to its very low density, so option C is false.
All noble gases (helium, neon, argon, krypton, xenon, radon) are naturally present as trace components in clean air, making option D correct.
✅ Answer: (D)
Noble gases are monatomic (they exist as single atoms, not diatomic molecules), so option B is false; and it is helium (not argon) that fills weather balloons due to its very low density, so option C is false.
All noble gases (helium, neon, argon, krypton, xenon, radon) are naturally present as trace components in clean air, making option D correct.
✅ Answer: (D)
▶️ Answer/Explanation
In the Haber process, nitrogen is obtained from the air, but hydrogen comes from natural gas (methane) — not from the air — so statement 1 is false.
The pressure used is approximately 200 atmospheres (not twice atmospheric pressure), so statement 2 is also false.
Statements 3 and 4 are both correct: an iron catalyst is used, and the operating temperature is around \( 450 \, ^\circ\text{C} \), which is a compromise between reaction rate and yield.
✅ Answer: (D)
The pressure used is approximately 200 atmospheres (not twice atmospheric pressure), so statement 2 is also false.
Statements 3 and 4 are both correct: an iron catalyst is used, and the operating temperature is around \( 450 \, ^\circ\text{C} \), which is a compromise between reaction rate and yield.
✅ Answer: (D)
▶️ Answer/Explanation
Compound X is ethene \( (\text{C}_2\text{H}_4) \), an alkene, which belongs to the homologous series with general formula \( \text{C}_n\text{H}_{2n} \).
Checking option C: \( \text{C}_3\text{H}_6 \) gives \( 3 \times 2 = 6 \) ✓ and \( \text{C}_5\text{H}_{10} \) gives \( 5 \times 2 = 10 \) ✓ — both fit the alkene general formula.
The other options include alkanes (\( \text{C}_n\text{H}_{2n+2} \)) or mixed series, which are different homologous families from ethene.
✅ Answer: (C)
Checking option C: \( \text{C}_3\text{H}_6 \) gives \( 3 \times 2 = 6 \) ✓ and \( \text{C}_5\text{H}_{10} \) gives \( 5 \times 2 = 10 \) ✓ — both fit the alkene general formula.
The other options include alkanes (\( \text{C}_n\text{H}_{2n+2} \)) or mixed series, which are different homologous families from ethene.
✅ Answer: (C)
▶️ Answer/Explanation
Molecule 1 contains a \( \text{C=C} \) double bond and can undergo addition polymerisation (like ethene forming polyethene).
Molecule 3 contains both an amine group \( (-\text{NH}_2) \) and a carboxylic acid group \( (-\text{COOH}) \), enabling it to undergo condensation polymerisation to form a polyamide (nylon-type).
Molecules 2 and 4 do not possess the required reactive groups for either type of polymerisation.
✅ Answer: (B)
Molecule 3 contains both an amine group \( (-\text{NH}_2) \) and a carboxylic acid group \( (-\text{COOH}) \), enabling it to undergo condensation polymerisation to form a polyamide (nylon-type).
Molecules 2 and 4 do not possess the required reactive groups for either type of polymerisation.
✅ Answer: (B)
▶️ Answer/Explanation
For a system to be in equilibrium, two conditions must both be satisfied simultaneously: the vector sum of all forces acting on the system must be zero (no resultant force), and the sum of all moments about any point must also be zero (no resultant turning effect).
If either condition is not met, the system will either accelerate linearly or rotate, meaning it is not in equilibrium.
✅ Answer: (D)
If either condition is not met, the system will either accelerate linearly or rotate, meaning it is not in equilibrium.
✅ Answer: (D)
▶️ Answer/Explanation
The volume of the aluminium equals the volume of water displaced: \( V = 51 – 43 = 8 \, \text{cm}^3 \).
Using the formula \( m = \rho \times V \): \( m = 2.7 \, \text{g/cm}^3 \times 8 \, \text{cm}^3 = 21.6 \, \text{g} \).
✅ Answer: (C)
Using the formula \( m = \rho \times V \): \( m = 2.7 \, \text{g/cm}^3 \times 8 \, \text{cm}^3 = 21.6 \, \text{g} \).
✅ Answer: (C)
▶️ Answer/Explanation
In a vacuum there is no air resistance, so all gravitational potential energy converts to kinetic energy: \( mgh = \frac{1}{2}mv^2 \), which simplifies to \( v = \sqrt{2gh} \).
Substituting: \( v = \sqrt{2 \times 10 \times 13} = \sqrt{260} \approx 16 \, \text{m/s} \).
✅ Answer: (D)
Substituting: \( v = \sqrt{2 \times 10 \times 13} = \sqrt{260} \approx 16 \, \text{m/s} \).
✅ Answer: (D)
▶️ Answer/Explanation
Natural gas, hydroelectric power (via the water cycle driven by solar evaporation), and wind are all ultimately powered by the Sun’s radiation.
Geothermal energy, however, comes from the heat generated by radioactive decay deep within the Earth’s mantle and core — a process entirely independent of the Sun.
✅ Answer: (A)
Geothermal energy, however, comes from the heat generated by radioactive decay deep within the Earth’s mantle and core — a process entirely independent of the Sun.
✅ Answer: (A)
▶️ Answer/Explanation
The process described — warm fluid rising and cool fluid falling in a cycle — is convection, which transfers thermal energy by the bulk movement of the fluid itself.
Hot air expands when heated, so its particles spread further apart, giving it a lower density than the surrounding cool air; this lower density causes the hot air to rise while denser cool air sinks to replace it.
✅ Answer: (D)
Hot air expands when heated, so its particles spread further apart, giving it a lower density than the surrounding cool air; this lower density causes the hot air to rise while denser cool air sinks to replace it.
✅ Answer: (D)
▶️ Answer/Explanation
Amplitude (definition 1) is the maximum displacement of a particle from its rest (equilibrium) position — it indicates the energy carried by the wave.
Frequency (definition 2) is the number of complete wave cycles (or wavefronts) that pass a fixed point each second, measured in hertz \( (\text{Hz}) \).
Wavelength (definition 3) is the distance between two consecutive points that are in phase, such as two adjacent crests or two adjacent troughs — this matches row A in the table.
✅ Answer: (A)
Frequency (definition 2) is the number of complete wave cycles (or wavefronts) that pass a fixed point each second, measured in hertz \( (\text{Hz}) \).
Wavelength (definition 3) is the distance between two consecutive points that are in phase, such as two adjacent crests or two adjacent troughs — this matches row A in the table.
✅ Answer: (A)
▶️ Answer/Explanation
A real image formed by a converging lens is always inverted (upside down) because the refracted rays cross over on the other side of the lens.
As the object moves further from the lens (object distance increases), the image distance decreases and approaches the focal length, so the image becomes smaller — but it remains inverted throughout.
✅ Answer: (C)
As the object moves further from the lens (object distance increases), the image distance decreases and approaches the focal length, so the image becomes smaller — but it remains inverted throughout.
✅ Answer: (C)
▶️ Answer/Explanation
Soft iron is a magnetically soft material — its magnetic domains align easily in an external field but return to random orientations once the field is removed, so it loses its magnetism quickly.
Steel is a magnetically hard material — once its domains are aligned by the external field, they remain in alignment even after the magnet is removed, making steel a permanent magnet.
✅ Answer: (C)
Steel is a magnetically hard material — once its domains are aligned by the external field, they remain in alignment even after the magnet is removed, making steel a permanent magnet.
✅ Answer: (C)
▶️ Answer/Explanation
Using \( Q = I \times t \): the time must be converted to seconds, so \( t = 1 \, \text{min} = 60 \, \text{s} \).
Substituting: \( Q = 1.5 \, \text{A} \times 60 \, \text{s} = 90 \, \text{C} \).
✅ Answer: (D)
Substituting: \( Q = 1.5 \, \text{A} \times 60 \, \text{s} = 90 \, \text{C} \).
✅ Answer: (D)
▶️ Answer/Explanation
In a series circuit, there is only one path for the current, so the same current flows through every component — statement C is correct.
In a parallel circuit, all branches are connected directly across the same supply, so they share an equal potential difference, but the currents through each branch differ if the resistances differ — statements A and B are both false.
In a series circuit, the p.d. is shared (split) across resistors in proportion to their resistance, so D is also false.
✅ Answer: (C)
In a parallel circuit, all branches are connected directly across the same supply, so they share an equal potential difference, but the currents through each branch differ if the resistances differ — statements A and B are both false.
In a series circuit, the p.d. is shared (split) across resistors in proportion to their resistance, so D is also false.
✅ Answer: (C)
▶️ Answer/Explanation
In a metal wire, the charge carriers are free electrons (negatively charged); in a.c., these electrons oscillate back and forth, periodically reversing direction rather than flowing steadily one way.
This is distinct from d.c. (option D, which flows steadily in one direction) and from the incorrect ideas of simultaneous bidirectional flow (C) or positively charged particles in a wire (A).
✅ Answer: (B)
This is distinct from d.c. (option D, which flows steadily in one direction) and from the incorrect ideas of simultaneous bidirectional flow (C) or positively charged particles in a wire (A).
✅ Answer: (B)
▶️ Answer/Explanation
First, find the secondary voltage using the turns ratio: \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \Rightarrow V_s = 240 \times \frac{10}{120} = 20 \, \text{V} \).
Since the transformer is 100% efficient, output power equals input power: \( P = V_s \times I_s \Rightarrow I_s = \frac{P}{V_s} = \frac{48}{20} = 2.4 \, \text{A} \).
✅ Answer: (C)
Since the transformer is 100% efficient, output power equals input power: \( P = V_s \times I_s \Rightarrow I_s = \frac{P}{V_s} = \frac{48}{20} = 2.4 \, \text{A} \).
✅ Answer: (C)
▶️ Answer/Explanation
An alpha particle is identical to a helium-4 nucleus \( \left( {}^4_2\text{He} \right) \), consisting of 2 protons and 2 neutrons.
Its total charge is \( 2 \times q = 2q \) (from the two protons) and its total mass is \( 2m + 2m = 4m \) (two protons each of mass \( m \) and two neutrons each of mass \( m \)).
✅ Answer: (B)
Its total charge is \( 2 \times q = 2q \) (from the two protons) and its total mass is \( 2m + 2m = 4m \) (two protons each of mass \( m \) and two neutrons each of mass \( m \)).
✅ Answer: (B)