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MathematicsClass 6CBSE

Are all whole numbers also natural numbers?

No, all whole numbers are not natural numbers. (0 is a whole number but not a natural number.)

MathematicsClass 6CBSE

How many whole numbers are there between 32 and 53?

There are 20 whole numbers between 32 and 53. These are 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51 and 52.

PhysicsClass 12CBSE

Bullet fires with velocity u; after 24 cm velocity = u/3. Then comes to rest at end of block. Find total length.

Step-1: Find retardation (v² = u² − 2as):(u/3)² = u² − 2a(0.24)u²/9 = u² − 0.48a0.48a = u²(1 − 1/9) = 8u²/9 ⇒ a = 50u²/27Step-2: Additional distance to stop from v = u/3:0 = (u/3)² − 2a·s₂s₂ = (u²/9) / (2 × 50u²/27) = (u²/9) × (27/100u²) = 0.03 m = 3 cmStep-3: Total length:L = 24 cm + 3 cm = 27 cmAnswer: Total length of the block = 27 cm

PhysicsClass 11CBSE

λ₁ = 350 nm, λ₂ = 450 nm; maximum velocities differ by factor 2 (v₁ = 2v₂). Find work function φ.

Einstein’s equations:hc/λ₁ = φ + ½mᵉv₁² ...(1)hc/λ₂ = φ + ½mᵉv₂² ...(2)Substitute v₁ = 2v₂ ⇒ v₁² = 4v₂² and subtract (2) from (1):hc(1/λ₁ − 1/λ₂) = (3/2)mᵉv₂²Calculate left side:1/350nm − 1/450nm = 100/(350×450) nm⁻¹ = 6.349×10⁵ m⁻¹hc × 6.349×10⁵ = 1.262×10⁻¹⁹ JSolve for v₂²:v₂² = (2/3) × 1.262×10⁻¹⁹ / 9.1×10⁻³¹ = 9.245×10¹⁰ m²/s²Substitute into equation (2):hc/λ₂ = (6.626×10⁻³⁴×3×10⁸) / (450×10⁻⁹) = 4.417×10⁻¹⁹ J½mᵉv₂² = 0.5 × 9.1×10⁻³¹ × 9.245×10¹⁰ = 4.206×10⁻²⁰ Jφ = 4.417×10⁻¹⁹ − 0.4206×10⁻¹⁹ ≈ 3.996×10⁻¹⁹ J ≈ 2.5 eVAnswer: Work function φ ≈ 2.5 eV

PhysicsClass 11CBSE

Charges −q, +q, −q at corners of equilateral triangle (side a). Find net force on +q at centroid O.

Distance from centroid to vertex: r = a/√3Force magnitude from each corner on +q at O:F = kq²/r² = kq²/(a²/3) = 3kq²/a²Directions (place +q corner at top, −q corners at bottom-left and bottom-right):From +q at top: repels downward (0, −F)From −q at bottom-L: attracts toward B-L (−F√3/2, −F/2)From −q at bottom-R: attracts toward B-R (+F√3/2, −F/2)Sum of components:Ex = 0 − F√3/2 + F√3/2 = 0Ey = −F − F/2 − F/2 = −2FF_net = 2F = 6kq²/a²Answer: F_net = 6kq²/a², directed away from the +q corner (toward midpoint of opposite side)Symmetry only guarantees x-cancellation here. The y-components all add never assume the force is zero without resolving components.

PhysicsClass 11CBSE

Two identical charges q₁ = q₂ = 5 μC. Find largest charge q to transfer so force decreases to 1/2.5 of original.

Step-1: After transferring charge q: new charges = (5−q) and (5+q) μCStep-2: New force = F₀ / 2.5:k(5−q)(5+q)/r² = k×25/(2.5×r²) = 10k/r²(5−q)(5+q) = 1025 − q² = 10 ⇒ q² = 15 ⇒ q = √15 μCAnswer: q = √15 μC ≈ 3.87 μCNote: (5−q)(5+q) = 25−q² is the difference-of-squares identity. Transferring charge always decreases the product q₁q₂ for charges of the same sign.

PhysicsClass 11CBSE

Charges +Q, −Q, +Q, −Q at corners of a square (side 5 cm). Find field at centre. (Q = 1 μC)

With alternating charges, the two +Q charges are at diagonally opposite corners, and the two −Q charges are at diagonally opposite corners.Field at centre from each pair:Both +Q fields: exactly opposite directions ⇒ cancelBoth −Q fields: exactly opposite directions ⇒ cancelEvery field vector has an equal and opposite partner by symmetry.Answer: Electric field at centre = 0 (zero)

GeneralClass 12CBSE

Electric Field at Centre of Ring with Non-Uniform Charge Distribution. Four quadrants carry linear charge densities: +2λ, −2λ, +λ, −λ. Find electric field at centre.

For a quarter-circle arc of linear charge density λ and radius R, the field at the centre has magnitude:E_quadrant = (√2 λ) / (4πε₀R)Direction: along the bisector of the quadrant, pointing away from arc (for +λ) or toward arc (for −λ).Resolving all four quadrant fields into x and y components:Ex = [−2 − 2 + 1 + 1] × (λK/√2) = −2λK/√2Ey = [−2 + 2 + 1 − 1] × (λK/√2) = 0Magnitude:|E| = 2λK/√2 = λ / (2πε₀R)Answer: E = λ / (2πε₀R), directed along the negative x-axis

PhysicsClass 12CBSE

Magnetic Moment of a Current-Carrying Wire Bent into a CircleWire length = 2 m, current = 3.14 A (≈ π A). Find magnetic moment M.

Step-1: Radius of circle:2πr = 2 ⇒ r = 1/π mStep-2: Area:A = πr² = π × (1/π)² = 1/π m²Step 3 — Magnetic moment:M = I × A = π × (1/π) = 1 A·m²Answer: M = 1 A·m²

GeneralClass 12CBSE

P°(hexane) = 408 Torr, P°(heptane) = 141 Torr. x(hexane) = 0.300. Find Y₆ and Y₇.

Step-1: Partial pressures:P(hexane) = 0.300 × 408 = 122.4 TorrP(heptane) = 0.700 × 141 = 98.7 TorrStep-2: Total pressure:P_total = 122.4 + 98.7 = 221.1 TorrStep-3: Vapour mole fractions:Y₆ (hexane) = 122.4 / 221.1 ≈ 0.554Y₇ (heptane) = 98.7 / 221.1 ≈ 0.446Answer: Y₆ ≈ 0.554, Y₇ ≈ 0.446

GeneralClass 12CBSE

2N₂O₅(g) → 4NO₂(g) + O₂(g). Initial P = 50 mmHg; P at 30 min = 87.5 mmHg. Find P at 60 min.

Step-1: Let N₂O₅ decrease by 2p. Total pressure = 50 + 3p:50 + 3p = 87.5 ⇒ p = 12.5 mmHgP(N₂O₅) at 30 min = 50 − 2(12.5) = 25 mmHgStep-2: Rate constant (first order):k = (1/30) × ln(50/25) = ln(2)/30 = 0.02310 min⁻¹Step-3: P(N₂O₅) at 60 min (= two half-lives = 25% of initial):P(N₂O₅) at 60 min = 50/4 = 12.5 mmHgStep-4: Total pressure at 60 min:Decrease = 50 − 12.5 = 37.5 = 2p ⇒ p = 18.75P_total = 50 + 3(18.75) = 106.25 mmHgAnswer: Total pressure after 60 minutes = 106.25 mm Hg

GeneralClass 11CBSE

P°A = 7×10³ Pa, P°B = 12×10³ Pa at 350 K. Liquid contains 40 mol% A. Find vapour composition.

Step-1: Partial pressures (Raoult’s Law):PA = 0.40 × 7000 = 2800 PaPB = 0.60 × 12000 = 7200 PaStep-2: Total pressure:P_total = 2800 + 7200 = 10000 PaStep-3: Vapour mole fractions (Dalton’s Law):yA = 2800 / 10000 = 0.28yB = 7200 / 10000 = 0.72Answer: Mole fraction of A in vapour = 0.28, B in vapour = 0.72The more volatile component (B, higher P°) is enriched in the vapour. This is the principle behind fractional distillation.

GeneralClass 11CBSE

O₂ at 293 K, partial pressure = 0.98 bar, KH = 34 kbar. Find solubility in g/L.

Step-1: Mole fraction:x(O₂) = 0.98 / 34000 = 2.882×10⁻⁵Step-2: Moles of O₂ in 1 L:n(O₂) ≈ 2.882×10⁻⁵ × 55.5 = 1.60×10⁻³ molStep-3: Mass (Mᵣ of O₂ = 32 g/mol):mass = 1.60×10⁻³ × 32 = 0.0512 gAnswer: Solubility of O₂ ≈ 0.0512 g/LO₂ has a smaller KH than N₂ (34 vs 76.48 kbar) meaning O₂ is more soluble, which is why aquatic life can survive using dissolved O₂.

GeneralClass 11CBSE

Find density (g/mL) of 3.60 M H₂SO₄ that is 29% by mass. (Mᵣ = 98 g/mol)

Formula linking molarity, density, and mass percent:M = (10 × d × w%) / Molar massSubstituting values:3.60 = (10 × d × 29) / 98d = (3.60 × 98) / (10 × 29) = 352.8 / 290 = 1.216 g/mLAnswer: Density = 1.216 g/mLNote: Memorise M = (10×d×w%)/Mᵣ. Always keep w% as a plain number (29, not 0.29) and density in g/mL.

GeneralClass 11CBSE

N₂ gas at 293 K, partial pressure = 0.987 bar, KH = 76.48 kbar. Find millimoles dissolved in 1 L water.

Step 1: Apply Henry’s Law (p = KH × x):x(N₂) = 0.987 / 76480 = 1.290×10⁻⁵Step 2: Moles of water in 1 L = 1000/18 = 55.5 molStep 3: Moles of N₂ (since x << 1):n(N₂) ≈ x × n(H₂O) = 1.290×10⁻⁵ × 55.5 = 7.16×10⁻⁴ mol = 0.716 mmolAnswer: ≈0.716 millimoles of N₂ dissolveKey Point: Henry’s Law: gas solubility ∝ partial pressure. The large KH for N₂ means it is only sparingly soluble a key concept in JEE solutions.

GeneralClass 11CBSE

50 g of ethylene glycol (CH₂OH)₂ dissolved in 200 g water; cooled to −9.3°C. Find mass of ice that separates. (Kf = 1.86 K·kg/mol)

Molar mass of ethylene glycol (C₂H₆O₂) = 62 g/molStep 1 — Moles of solute:n = 50 / 62 = 0.8065 molStep 2 — Freezing point depression formula: ΔTf = Kf × m9.3 = 1.86 × (0.8065 / W_solvent_kg)Step 3 — Mass of liquid water remaining:W_solvent = (1.86 × 0.8065) / 9.3 = 0.1613 kg = 161.3 gStep 4 — Ice separated = total water − liquid water:Ice = 200 − 161.3 = 38.7 gAnswer: ≈38.71 g of ice separates outWhen a solution cools, only pure water freezes. As ice forms, the remaining solution becomes more concentrated, lowering its freezing point further.

ChemistryClass 11CBSE

2.8×10⁻³ mol of CO₂ is left after removing 10²¹ molecules from its ‘x’ mg sample. Find x. (Nₐ = 6.02×10²³ mol⁻¹)

Molar mass of CO₂ = 44 g/molStep 1 — Convert removed molecules to moles:Moles removed = 10²¹ / (6.02×10²³) = 1.661×10⁻³ molStep 2 — Initial moles = remaining + removed:n_initial = 2.8×10⁻³ + 1.661×10⁻³ = 4.461×10⁻³ molStep 3 — Initial mass:mass = 4.461×10⁻³ × 44 = 0.19629 g = 196.3 mgAnswer: x ≈ 196.3 mgDividing molecule count by Nₐ gives moles; multiplying by molar mass gives mass the two-step core of all JEE mole-concept problems.

ChemistryClass 11CBSE

Total number of isomers, considering both structural and stereoisomers of cyclic ethers with the molecular formula C₄H₈O is

Degree of unsaturation = (2×4 + 2 − 8) / 2 = 1 → One ring, no double bonds.A cyclic ether contains one O atom inside the ring. Possible ring sizes: 3-membered (oxirane), 4-membered (oxetane), 5-membered (tetrahydrofuran).StructureIsomer Count2,2-dimethyloxirane1(R)-ethyloxirane1(S)-ethyloxirane1cis-2,3-dimethyloxirane (meso)1(R,R)-2,3-dimethyloxirane1(S,S)-2,3-dimethyloxirane13-methyloxetane1Tetrahydrofuran (THF)1Answer: 8 isomersAlways check for stereocentres in epoxide rings. The meso compound of 2, 3-dimethyloxirane is a frequent JEE trap.

ChemistryClass 10CBSE

Why is NaOH's molar mass exactly 40?

Atomic masses: Na=23, O=16, H=1 → 23+16+1 = 40 g/mol exactly. This means 1 mole of NaOH weighs 40 g, making it easy to prepare 1M solutions (40 g per litre).

ChemistryClass 10CBSE

Molecular Mass Calculations

NaOH molecular mass = 40 g/mol:Na(23) + O(16) + H(1) = 40 g/mol FeSO₄·7H₂O (Green Vitriol) = 278 g/mol:Fe(56) + S(32) + 4×O(16) + 7×H₂O(18) = 56 + 32 + 64 + 126 = 278 g/mol Na₂SO₄·10H₂O (Washing Soda / Glauber's Salt) = 322 g/mol:2×Na(23) + S(32) + 4×O(16) + 10×H₂O(18) = 46 + 32 + 64 + 180 = 322 g/mol

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