A body falls freely from rest. The ratio of distances covered in 1st, 2nd and 3rd seconds is
- A) 1 : 2 : 3
- B) 1 : 4 : 9
- C) 1 : 3 : 5 ✓
- D) 1 : 5 : 9
❌ Regular method
- s = ½ g t². So s(0→1) = 5 m, s(0→2) = 20 m, s(0→3) = 45 m.
- 1st sec = 5 m. 2nd sec = 20 − 5 = 15 m. 3rd sec = 45 − 20 = 25 m.
- Ratio 5 : 15 : 25 = 1 : 3 : 5.
✓ Shortcut method
Galileo's odd-number rule: in equal time intervals, distance ratio = 1 : 3 : 5 : 7…
Read off the answer in 1 second.
💡 Simple intuitionWhen a ball drops, each second it falls more than the second before — by an odd number more. First second 1 step, next 3 steps, then 5 steps, then 7 steps. Always odd jumps.
Two stones are projected with same speed: one at 15° and another at 75°. Their horizontal ranges are
- A) Equal ✓
- B) In ratio 1 : 5
- C) In ratio 1 : 3
- D) Different (cannot tell)
❌ Regular method
- R = u² sin(2θ)/g.
- At 15°: sin 30° = 0.5 → R₁ = u²(0.5)/g.
- At 75°: sin 150° = sin 30° = 0.5 → R₂ = u²(0.5)/g.
- R₁ = R₂. Equal.
✓ Shortcut method
15° + 75° = 90° → they are complementary.
Same range for any θ and (90°−θ). Answer in 2 seconds.
💡 Simple intuitionWhen you throw a ball, there is a pair of angles that always land at the same spot — they add to 90°. So a high arc throw and a flat throw can land together if the angles match this rule.
A car moving at 30 km/h stops in 8 m. The same car, with the same braking, moving at 90 km/h will stop in
- A) 24 m
- B) 48 m
- C) 72 m ✓
- D) 144 m
❌ Regular method
- v² = u² − 2as → for stopping, s = u²/(2a) (deceleration a same).
- s₂/s₁ = (u₂/u₁)² = (90/30)² = 9.
- s₂ = 9 × 8 = 72 m.
✓ Shortcut method
Stopping distance ∝ v². Speed ×3 → distance × 9.
9 × 8 = 72 m.
💡 Simple intuitionIf a bike is going twice as fast, it takes 4 times as much road to stop. Three times as fast → 9 times as much road. Going fast is dangerous because stopping space grows really quickly.
If the distance between two masses is increased by 5%, the gravitational force between them
- A) Increases by 10%
- B) Decreases by 5%
- C) Decreases by 10% ✓
- D) Decreases by 25%
❌ Regular method
- F = G m₁m₂ / r². F₁/F₂ = (r₂/r₁)² = (1.05)² = 1.1025.
- F₂/F₁ = 1/1.1025 = 0.907 → drop of 9.3%.
- Closest option: ~10% decrease.
✓ Shortcut method
For small Δr: ΔF% ≈ −2 × Δr%.
= −2 × 5% = −10%. Pick 10% decrease.
💡 Simple intuitionWhen two friends pull on a rubber band and step apart a tiny bit, the pull gets weaker. The 'inverse-square' rule means: pull weakens by about twice the % they moved apart.
In a Wheatstone bridge P = 100 Ω, Q = 10 Ω, R = 300 Ω, S = 30 Ω, galvanometer G = 50 Ω. Current through G is
- A) Maximum
- B) Zero ✓
- C) Same as battery
- D) Cannot be determined
❌ Regular method
- Solve Kirchhoff loops with G in middle… ugly mesh equations.
- After algebra: I_G works out to zero.
✓ Shortcut method
Check balance: P/Q = 100/10 = 10; R/S = 300/30 = 10. Balanced.
In a balanced bridge, no current flows through G regardless of G's value. Answer in 5 seconds.
💡 Simple intuitionImagine four water pipes forming a diamond shape. If both halves carry the same pressure, no water flows across the middle connecting pipe. Same with the bridge: if the resistor ratios match, the middle wire stays quiet.
A symmetric biconvex lens of focal length f is cut into two equal pieces along its principal axis. The focal length of each piece is
❌ Regular method
- Lens-maker: 1/f = (n−1)(1/R₁ − 1/R₂) = (n−1)(2/R) for biconvex with R₁ = R, R₂ = −R.
- After cutting along principal axis → one side becomes flat (R₂ = ∞), other R₁ = R.
- 1/f' = (n−1)(1/R) → f' = 2f.
✓ Shortcut method
Cut along principal axis → biconvex becomes two plano-convex → f doubles to 2f.
(Cut perpendicular → f stays same; intensity halves.)
💡 Simple intuitionA magnifying glass is round and curved on both sides. Slice it the long way and each thinner piece bends light only half as much, so the focus moves out twice as far.
A radioactive isotope has half-life 10 years. After 40 years, the fraction of original undecayed is
- A) 1/2
- B) 1/4
- C) 1/8
- D) 1/16 ✓
❌ Regular method
- N = N₀ · e^(−λt), λ = ln 2 / 10.
- N/N₀ = e^(−ln 2 × 40/10) = e^(−4 ln 2) = 2⁻⁴ = 1/16.
✓ Shortcut method
n = 40/10 = 4 half-lives. Fraction = (1/2)ⁿ = (1/2)⁴ = 1/16.
💡 Simple intuitionImagine a stack of 16 cookies. Each half-life, you eat half. After 4 rounds: 16 → 8 → 4 → 2 → 1. Only 1 cookie out of the original 16 is left.
The time period T of a pendulum depends on its length L and acceleration g. By dimensional analysis, T equals
- A) √(Lg)
- B) √(L/g) × 2π ✓
- C) L/g
- D) g/L
❌ Regular method
- Assume T = k · Lᵃ · gᵇ.
- Dimensions: [T] = [L]ᵃ [LT⁻²]ᵇ → 0 = a+b for L, 1 = −2b for T.
- b = −½, a = ½ → T ∝ √(L/g). With k = 2π → T = 2π√(L/g).
✓ Shortcut method
Time has dimension [T]. Test each option:
- (a) √(Lg) → √(L · LT⁻²) = LT⁻¹ — wrong (velocity!).
- (b) √(L/g) → √(L / LT⁻²) = √(T²) = T ✓
- (c) L/g → L / LT⁻² = T² — wrong.
- (d) g/L → LT⁻² / L = T⁻² — wrong.
Only (b) has units of seconds.
💡 Simple intuitionEvery answer has 'units' baked in (like meters, seconds, kg). If the answer should be in seconds, only one option's units come out to seconds. The others are immediately wrong — no calculation needed.
A spring with constant k = 100 N/m has 1 kg mass attached. The angular frequency ω of oscillation is
- A) 1 rad/s
- B) 10 rad/s ✓
- C) 100 rad/s
- D) 0.1 rad/s
❌ Regular method
- F = −kx = ma. → m·d²x/dt² = −kx.
- Standard SHM solution → ω² = k/m.
- ω = √(k/m) = √(100/1) = 10 rad/s.
✓ Shortcut method
Memorised: ω = √(k/m). = √100 = 10 rad/s.
💡 Simple intuitionStiffer spring = faster wiggle. Heavier mass = slower wiggle. The square root of (stiffness ÷ heaviness) tells you how fast it wiggles per second.
The energy of a photon of wavelength 620 nm in eV is approximately
- A) 1 eV
- B) 2 eV ✓
- C) 3 eV
- D) 10 eV
❌ Regular method
- E = hc/λ.
- = (6.626×10⁻³⁴ × 3×10⁸) / (620×10⁻⁹) J = 3.2×10⁻¹⁹ J.
- Convert to eV: 3.2×10⁻¹⁹ / 1.6×10⁻¹⁹ = 2 eV.
✓ Shortcut method
Memorise: E (eV) = 1240 / λ (nm).
= 1240 / 620 = 2 eV. Done.
💡 Simple intuitionLight has tiny energy packets. The shorter the colour's 'wave', the more energy each packet has. The magic number 1240 divided by the wavelength in nanometres gives the answer straight away.
A car takes a turn of radius 50 m at 36 km/h. The banking angle for no friction is (g = 10)
- A) tan⁻¹(0.1)
- B) tan⁻¹(0.2) ✓
- C) tan⁻¹(0.5)
- D) tan⁻¹(1.0)
❌ Regular method
- For frictionless banking, N sin θ = mv²/r, N cos θ = mg → tan θ = v²/rg.
- v = 36 km/h = 10 m/s.
- tan θ = 100/(50 × 10) = 0.2 → θ = tan⁻¹ 0.2.
✓ Shortcut method
Direct formula: tan θ = v²/(r g).
= 100/500 = 0.2.
💡 Simple intuitionOn a curve, the road tilts inward so the car doesn't slip. Faster cars or tighter curves need more tilt. The formula just compares 'how fast' with 'how tight'.
pH of 0.01 M acetic acid (Ka = 1.8 × 10⁻⁵) is approximately
- A) 2.4
- B) 3.4 ✓
- C) 4.7
- D) 5.9
❌ Regular method
- Set up ICE table: HA ⇌ H⁺ + A⁻. Initial 0.01, change −x, +x, +x.
- Ka = x²/(0.01−x) ≈ x²/0.01 = 1.8×10⁻⁵.
- x² = 1.8×10⁻⁷ → x = 4.24×10⁻⁴.
- pH = −log(4.24×10⁻⁴) = 3.37.
✓ Shortcut method
pH = ½ [pKa − log C].
pKa = −log(1.8×10⁻⁵) = 4.74. log C = log(0.01) = −2.
pH = ½ (4.74 − (−2)) = ½ × 6.74 = 3.37.
💡 Simple intuitionA weak acid only sort-of falls apart in water. Instead of doing a long table, you can take the acid's 'strength number' (pKa) plus the 'how-much number', cut in half, and you have the pH.
A buffer contains 0.1 M CH₃COOH and 0.1 M CH₃COONa. pKa(HA) = 4.76. pH of the buffer is
- A) 3.76
- B) 4.76 ✓
- C) 5.76
- D) 7.00
❌ Regular method
- Set up Ka equilibrium with salt providing common ion (acetate).
- Ka = [H⁺][A⁻]/[HA]. With [A⁻]=[HA]=0.1, [H⁺] = Ka.
- pH = pKa = 4.76.
✓ Shortcut method
Henderson: pH = pKa + log([salt]/[acid]).
= 4.76 + log(0.1/0.1) = 4.76 + 0 = 4.76.
💡 Simple intuitionA buffer is a 'shock absorber' for acids/bases. When acid and salt are equal, the pH is just the acid's strength number (pKa). When they're not equal, you add a small log correction.
A reaction's rate at 27 °C is r. The approximate rate at 67 °C (temperature coefficient = 2) is
❌ Regular method
- Use ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂). Without Ea, use temp-coefficient rule.
- Each 10°C rise → rate × 2. ΔT = 40°C → 4 doublings.
- Factor = 2⁴ = 16.
✓ Shortcut method
(67 − 27)/10 = 4 doublings → 2⁴ = 16.
💡 Simple intuitionReactions go faster when it's hotter. A rule of thumb: every 10 degrees hotter, the speed roughly doubles. 40 degrees hotter = double, double, double, double = 16 times.
A first-order reaction has half-life 30 minutes. How long until 75% of the reactant has decomposed?
- A) 30 min
- B) 45 min
- C) 60 min ✓
- D) 90 min
❌ Regular method
- For first-order: t = (2.303/k) log(N₀/N). k = 0.693/30 = 0.0231 min⁻¹.
- N/N₀ = 0.25 → t = (2.303/0.0231) × log 4 = 99.7 × 0.602 = 60 min.
✓ Shortcut method
75% gone → 25% left → 1/4 = (1/2)² → 2 half-lives.
2 × 30 = 60 min.
💡 Simple intuitionIf half disappears in 30 minutes, then half of what's left disappears in the next 30 minutes. After two rounds, only a quarter remains — that means three-quarters are gone.
❌ Regular method
- O₂⁺ has 15 electrons.
- Fill MO diagram: σ1s², σ*1s², σ2s², σ*2s², σ2p_z², π2p² (×2), π*2p¹ (one electron, removed from O₂).
- Bonding = 10, antibonding = 5. BO = (10−5)/2 = 2.5.
✓ Shortcut method
Anchor: 14 e⁻ = BO 3.0. O₂⁺ has 15 e⁻ → drop by 0.5 → 2.5.
💡 Simple intuitionImagine a perfect tower of 14 building blocks giving bond strength 3. Add or remove any block and the strength changes by half a step. O₂⁺ has 15, so 3 − 0.5 = 2.5.
Which of the following is paramagnetic?
- A) N₂ (14 e⁻)
- B) F₂ (18 e⁻)
- C) O₂ (16 e⁻) ✓
- D) CO (14 e⁻)
❌ Regular method
- Draw MO diagrams for each.
- O₂ MO config has 2 unpaired electrons in π*2p (Hund's rule) → paramagnetic.
- N₂, F₂, CO all paired → diamagnetic.
✓ Shortcut method
O₂ (16 e⁻) is an anomaly: even electrons but still paramagnetic. The two famous anomalies are B₂ (10 e⁻) and O₂ (16 e⁻).
💡 Simple intuitionUsually, even-numbered teams pair up (boring). Odd-numbered teams have a lonely one (= magnetic). But O₂ and B₂ break the rule and act like odd teams even though they have even numbers. Memorise these two oddballs.
Hybridisation of central atom in SF₆ is
- A) sp³
- B) sp³d
- C) sp³d² ✓
- D) sp³d³
❌ Regular method
- S has 6 valence electrons.
- 6 F atoms each share 1 electron → 6 bond pairs around S.
- Total = 6 pairs → sp³d² → octahedral.
✓ Shortcut method
VSEP = ½ (V + M − C + A) = ½ (6 + 6) = 6 → sp³d².
💡 Simple intuitionCount the central atom's outer electrons plus the number of stuck-on simple atoms. Cut in half. That number tells you the shape: 4 → tetrahedral, 5 → trigonal bipyramid, 6 → octahedral.
Spin-only magnetic moment (BM) of [Fe(H₂O)₆]³⁺ is
- A) 1.73
- B) 2.83
- C) 3.87
- D) 5.92 ✓
❌ Regular method
- Fe³⁺ has [Ar] 3d⁵ configuration.
- H₂O is weak-field ligand → high-spin → all 5 electrons unpaired.
- μ = √(n(n+2)) = √35 = 5.916 BM.
✓ Shortcut method
Fe³⁺ = d⁵, H₂O weak field → n = 5 → 5.92 BM.
First digit of μ = n. Pick the answer starting with 5.
💡 Simple intuitionLonely electrons make a complex 'magnetic'. Count the lonely ones — that single digit IS the start of the answer. 5 lonely → 5.92. 3 lonely → 3.87. No need to calculate the square root.
Solubility of Ag₂CrO₄ (Ksp = 1.1 × 10⁻¹²) in water is approximately
- A) 6.5 × 10⁻⁵ ✓
- B) 1.0 × 10⁻⁴
- C) 1.0 × 10⁻⁶
- D) 1.1 × 10⁻¹²
❌ Regular method
- Ag₂CrO₄ ⇌ 2 Ag⁺ + CrO₄²⁻. Let s = solubility.
- Ksp = [Ag⁺]²[CrO₄²⁻] = (2s)²(s) = 4s³.
- 4s³ = 1.1×10⁻¹² → s³ = 2.75×10⁻¹³ → s = (2.75)^⅓ × 10⁻¹³/³ ≈ 6.5×10⁻⁵.
✓ Shortcut method
A₂B type: Ksp = 4s³ → s = (Ksp/4)^⅓.
= (2.75×10⁻¹³)^⅓ ≈ 6.5 × 10⁻⁵ M.
💡 Simple intuitionWhen salt dissolves, the recipe for Ksp depends on the formula: AB type → s = √Ksp; A₂B → cube-root with a 4. Memorise the recipe; skip deriving it.
For Zn|Zn²⁺(0.01 M)||Cu²⁺(1 M)|Cu, E° = 1.10 V. The EMF at 25 °C is
- A) 1.10 V
- B) 1.16 V ✓
- C) 1.04 V
- D) 0.91 V
❌ Regular method
- E = E° − (RT/nF) ln Q.
- Q = [Zn²⁺]/[Cu²⁺] = 0.01/1 = 0.01.
- (RT/nF) at 298 K → (0.0591/2) log 0.01 = 0.02955 × (−2) = −0.0591.
- E = 1.10 − (−0.0591) = 1.1591 V ≈ 1.16 V.
✓ Shortcut method
E = E° − (0.0591/n) log Q. n = 2, Q = 0.01.
= 1.10 − (0.0591/2)(−2) = 1.10 + 0.0591 = 1.16 V.
💡 Simple intuitionAt room temperature, every change of 1 in the 'log Q' number shifts voltage by about 0.06 ÷ n volts. Skip the gas constant and Faraday — just use 0.0591/n directly.
A reaction has K = 10⁵ at 298 K. ΔG° (kJ/mol) is approximately
- A) −5.7
- B) −11.4
- C) −28.5 ✓
- D) +28.5
❌ Regular method
- ΔG° = −RT ln K = −2.303 RT log K.
- 2.303 RT at 298 K = 2.303 × 8.314 × 298 = 5705 J/mol ≈ 5.7 kJ/mol.
- = −5.7 × 5 = −28.5 kJ/mol.
✓ Shortcut method
At 298 K: ΔG° (kJ/mol) ≈ −5.7 × log K.
= −5.7 × 5 = −28.5.
💡 Simple intuitionBigger equilibrium constant K means a more 'wanted' reaction — bigger negative free-energy. The rule of thumb: at room temperature, multiply −5.7 by log K to get kilojoules.
For ΔH = −50 kJ and ΔS = −100 J/K, reaction is spontaneous at
- A) All temperatures
- B) Low T only ✓
- C) High T only
- D) Never
❌ Regular method
- ΔG = ΔH − TΔS.
- ΔS = −0.1 kJ/K. ΔG = −50 − T(−0.1) = −50 + 0.1T.
- ΔG < 0 when T < 500 K → spontaneous at low temperatures.
✓ Shortcut method
Sign table: ΔH(−), ΔS(−) → spontaneous only at low T.
(High T makes TΔS overpower the negative ΔH.)
💡 Simple intuitionWhen something gives off heat (−ΔH) but becomes more tidy (−ΔS), it happens by itself only when it's cool. Heat it up too much and the 'tidiness' loss overrides the heat release.
Osmotic pressure of 0.1 M NaCl (i ≈ 2) at 27 °C is (R = 0.0821 L·atm/mol·K)
- A) 2.46 atm
- B) 4.92 atm ✓
- C) 9.84 atm
- D) 0.246 atm
❌ Regular method
- π = i × C × R × T.
- = 2 × 0.1 × 0.0821 × 300.
- = 4.926 atm ≈ 4.92 atm.
✓ Shortcut method
Direct: π = i·CRT. NaCl → i = 2.
= 2 × 0.1 × 0.0821 × 300 = 4.92 atm.
💡 Simple intuitionSalt water 'pulls' on pure water through a thin wall — that pull is osmotic pressure. The pull is bigger when concentration is bigger, the temperature is hotter, and (for salt) it splits into 2 pieces (so multiply by 2).
A current of 9.65 A passes through molten AgNO₃ for 1000 s. Mass of Ag deposited (M = 108) is
- A) 10.8 g ✓
- B) 21.6 g
- C) 5.4 g
- D) 108 g
❌ Regular method
- Q = I × t = 9.65 × 1000 = 9650 C.
- Moles of e⁻ = 9650/96500 = 0.1.
- Ag⁺ + e⁻ → Ag, so n = 1 → moles Ag = 0.1.
- Mass = 0.1 × 108 = 10.8 g.
✓ Shortcut method
m = (M I t)/(n F) = (108 × 9.65 × 1000)/(1 × 96500) = 10.8 g.
💡 Simple intuitionElectricity coats metal onto a wire. Every 96 500 units of charge deposits one 'pack' of metal. Bigger current or longer time = more coating, in direct proportion.
Ratio of rates of diffusion of H₂ and O₂ is
- A) 1 : 16
- B) 16 : 1
- C) 1 : 4
- D) 4 : 1 ✓
❌ Regular method
- Graham's law: r ∝ 1/√M.
- r(H₂)/r(O₂) = √(M(O₂)/M(H₂)) = √(32/2) = √16 = 4.
- 4 : 1.
✓ Shortcut method
r₁/r₂ = √(M₂/M₁) = √(32/2) = 4. Pick 4 : 1.
💡 Simple intuitionA light gas runs around much faster than a heavy gas. The exact speedup is the square root of how much heavier the slow one is. Oxygen is 16× heavier than hydrogen, so hydrogen is √16 = 4× faster.
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