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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

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.

GeneralClass 10CBSE

Natural vs Synthetic&nbsp;

Natural = plant-based (litmus from lichens, turmeric, china rose); Synthetic = lab-made chemicals (phenolphthalein, methyl orange). Both detect acids/bases by colour change.Type 2 indicator = Reversible indicator returns to original colour when pH conditions change (e.g., phenolphthalein reverts to colourless when pH drops below 8.2).

GeneralClass 10CBSE

Favourable factors for cation formation

Low ionisation energy, large atomic radius, high electron shielding, stable noble-gas configuration after electron loss.

GeneralClass 10CBSE

Why is 7 lucky?

Cultural (7 days, 7 colours, 7 notes, 7 seas), religious (7 heavens, 7 sacred rivers), mathematical (prime number), and psychological (most chosen number between 1–10 in studies).

GeneralClass 10CBSE

Did Japan ever have wild tigers?

No tigers are not native to Japan. They exist on continental Asia (India, Russia, China, Southeast Asia).

GeneralClass 10CBSE

Which is the national animal of India?

National symbols:SymbolNameYearNational AnimalBengal Tiger1973National BirdIndian Peacock1963National FlowerLotus—National TreeBanyan—National RiverGanga2008National EmblemLion Capital of Ashoka1950National AnthemJana Gana Mana1950National SongVande Mataram1950

GeneralClass 10CBSE

POCSO Act&nbsp;Main&nbsp;Facts

Enacted 2012, amended 2019 (stricter penalties including death penalty)Protects all persons below 18 years of age (not 16 it's 18)Gender-neutral, covers all forms of sexual offencesMandates Special Courts and mandatory reportingFailure to report an offence is itself a crime under this Act

GeneralClass 10CBSE

POCSO vs POSCO

POCSO = Protection of Children from Sexual Offences Act (2012) India's child protection lawPOSCO = POSCO Steel Company (South Korean steel firm) completely unrelated

GeneralClass 10CBSE

What information does electron configuration provide?

Distribution of electrons in shells/subshellsNumber of valence electrons → chemical reactivityMagnetic properties (paired = diamagnetic, unpaired = paramagnetic)Element's block, period, and group in the periodic table

GeneralClass 10CBSE

Is Gorakhpur worth visiting?

Yes for religious, cultural, and heritage tourism. Also a gateway to Kushinagar (Buddhist circuit) and Nepal.

GeneralClass 10CBSE

Why is Gorakhpur famous?

Gorakhnath Temple major Hindu pilgrimage siteGita Press world's largest publisher of Hindu religious textsWorld's longest railway platform (1,366 m)Home constituency of CM Yogi AdityanathBRD Medical College serving eastern UP and Nepal

GeneralClass 10CBSE

Where is Gorakhpur?

Eastern Uttar Pradesh, near the Nepal border (~270 km from Lucknow).

GeneralClass 10CBSE

Is CaO or Ca(OH)₂ used for whitewashing?

Ca(OH)₂ (slaked lime) is used CaO is too reactive to apply directly. After application, Ca(OH)₂ reacts with atmospheric CO₂ to form hard, white CaCO₃:Ca(OH)₂ + CO₂ → CaCO₃ + H₂O

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