Problem Types & Solved Examples
Six categories of problems. Every type that has ever appeared in CBSE, NEET, and JEE — solved with examiner mindset.
Strategy for Direct Formula Problems
Identify given quantities → Write formula → Substitute. These are guaranteed marks. Never lose them. Typical in CBSE 2M, NEET 1M. Mistake: wrong substitution of r (surface to surface vs centre to centre).
📋 Given
m₁ = 100 kg, m₂ = 200 kg, r = 2 m, G = 6.67 × 10⁻¹¹
🎓 What Examiner Tests: Direct recall of Newton's law. Watch: r = 2m (not squared error).
Apply Newton's Law of Gravitation
Substitute values
Calculate numerator and denominator
Final answer
⚡ Shortcut Insight
Numerically: G × m₁ × m₂ first, then divide by r². Don't rearrange formula — substitution errors are most common here.
📋 Given
g_h = g/4, R = 6400 km = 6.4 × 10⁶ m
🎓 Examiner tests: g_h formula, algebraic manipulation. Trap: Don't use approximate formula here — h is not << R.
Use exact formula: g_h = g × R²/(R+h)²
Given g_h = g/4: g/4 = g × R²/(R+h)²
Taking square root: R+h = 2R
⚡ Shortcut Insight
g_h = g/n² → h = (n−1)R. Here n=2 → h = R. Memorize this pattern for NEET speed.
Strategy for Conceptual Problems
These test whether you UNDERSTAND, not just know formulas. Ask: what changes, what stays constant? These are the highest-yield questions in NEET (1 correct MCQ = 4 marks). Think before solving.
🎓 This tests understanding of satellite energy relationships — not formula substitution. "JEE twists this by asking which quantity decreases when orbit energy increases."
v₀ = √(GM/r) → as r increases, v₀ decreases. ❌ (a) wrong
T = 2π√(r³/GM) → as r increases, T increases. ✓ (b) correct
KE = GMm/2r → as r increases, KE decreases. ❌ (c) wrong
E_total = −GMm/2r → as r increases, E becomes less negative (increases). ✓ (d) correct
Answer: (b) Time period and (d) Total mechanical energy increase.
⚡ Shortcut Insight
Moving to higher orbit: speed ↓, T ↑, KE ↓, PE ↑, Total E ↑ (less negative). This seems paradoxical — you add energy but speed decreases. That's the trap.
🎓 CBSE 3M, NEET concept MCQ, JEE paragraph. Tests definition of weightlessness vs absence of gravity.
Gravity still acts. g ≈ 9.8 m/s² for the person (and the lift).
Both the person and lift fall with the same acceleration g (free fall).
Normal reaction N: In lift at rest → N = mg. In free fall → N = m(g−g) = 0.
Since normal force = 0, person feels weightless. Apparent weight = 0, actual weight (mg) still exists.
⚡ Shortcut Insight
Weightlessness = apparent weight = 0. Apparent weight = m(g − a). When a = g (free fall), apparent weight = 0. Gravity hasn't disappeared.
Strategy for Multi-Step Problems
Break into sub-problems. Identify what you need → what you're given → bridge the gap step by step. "If this step is wrong, the entire solution fails." Always write intermediate results clearly.
📋 Given
Mass of satellite = m, Initial orbit = r, Final orbit = 2r, Earth mass = M
🎓 3-part problem: energy, energy difference, speed ratio. Very common in JEE Main Paper 2 and NEET long-answer type.
Part (a): Total Energy
KE = ½mv² = ½m(GM/r) = GMm/2r
PE = −GMm/r
Total E = KE + PE = GMm/2r − GMm/r = −GMm/2r
Part (b): Energy for Transfer
E₁ = −GMm/2r, E₂ = −GMm/(2×2r) = −GMm/4r
ΔE = E₂ − E₁ = −GMm/4r − (−GMm/2r) = GMm/4r
Part (c): Speed Ratio
v₁ = √(GM/r), v₂ = √(GM/2r)
⚡ Shortcut Insight
E ∝ 1/r (but negative). Speed ratio v₁/v₂ = √(r₂/r₁). Energy needed = E₂ − E₁ (always positive for outward transfer).
Strategy for Graph-Based Problems
First identify what the axes represent. Then use the shape of the graph (linear, inverse square, parabolic) to extract the physics. Slope and area under graph have physical meaning.
🎓 Kepler's 3rd law graph — T² vs r³ is a straight line through origin. Slope = 4π²/GM. Examiner tests: what does slope represent physically?
Kepler's 3rd law: T² = (4π²/GM) × r³
Comparing with y = mx: slope of T² vs r³ graph = 4π²/GM
Slope gives us the mass of the central body (Sun): M = 4π²/(G × slope)
⚡ Shortcut Insight
Slope of T² vs r³ = 4π²/GM. Graph is same for all planets around same star. Different stars → different slopes. If graph of log T vs log r: slope = 3/2 (Kepler's 3rd law exponent).
🎓 Classic graph question. NEET asks this almost every 2-3 years. Know both inside and outside behavior.
Inside (r < R): g = GMr/R³ → linearly increases from 0 to g₀ at r=R
At surface (r = R): g = GM/R² = g₀ (maximum)
Outside (r > R): g = GM/r² → decreases as inverse square
⚡ Shortcut Insight
Maximum g is at the surface. Inside: linear (slope = GM/R³). Outside: inverse square (area ∝ 1/r²). The graph is NOT smooth at r=R — slope changes abruptly (kink).
Strategy for Assertion-Reason
Step 1: Is Assertion (A) true? Step 2: Is Reason (R) true? Step 3: Does R correctly explain A? Don't assume R is wrong just because it sounds unusual. Most marks are lost by skipping step 3.
Standard Options (memorize these)
Reason (R): There is no gravitational force on the spacecraft when it is in orbit.
Assertion (A): TRUE. Astronauts DO feel weightless in orbit (apparent weight = 0).
Reason (R): FALSE. Gravity very much acts on the spacecraft (~8.7 m/s² at ISS altitude). That's what keeps it in orbit!
Weightlessness is due to free fall, not absence of gravity.
Answer: (C) — A is true, R is false.
⚡ Shortcut Insight
Whenever a reason says "no gravity in space/orbit" → it's always FALSE. Gravity never stops. Only apparent weight can become zero.
Reason (R): The Moon has smaller mass and radius compared to the Earth.
A: TRUE — Vₑ(Moon) = 2.4 km/s < 11.2 km/s (Earth).
R: TRUE — Moon has smaller mass AND smaller radius.
Vₑ = √(2GM/R). Both smaller M (reduces Vₑ) and smaller R (increases Vₑ partially). Net effect: Moon's Vₑ is less because M effect dominates. R is the reason? Partially yes — both mass and radius matter.
Answer: (A) — Both true, R correctly explains A.
⚡ Shortcut Insight
For Moon: even though smaller R would increase Vₑ, the much smaller M dominates and net Vₑ is smaller. Both mass and radius matter in Vₑ — don't consider only one.
Strategy for Case-Based Problems
Read the passage TWICE before solving. All answers are in the passage + concept combo. These appear in CBSE Class 11 final exams (4-5 marks for 4 sub-questions). Don't rush.
📖 Passage
Satellites are objects that orbit around a planet. A satellite can be natural (like the Moon) or artificial (like communication satellites). The motion of a satellite is governed by Newton's law of gravitation. Geostationary satellites orbit at a specific height where their time period equals Earth's rotation period (24 hours). They appear stationary from Earth and are used for communication and weather forecasting. The orbital speed decreases as the orbital radius increases, while the time period increases.
Q(i): (B) 24 hours — directly stated in passage.
Q(ii): (C) Decreases — v₀ = √(GM/r), as r increases, v₀ decreases. Also stated in passage.
⚡ Shortcut Insight
In case-based, 80% of answers are directly in the passage. Only 20% need formula application. Always read passage carefully — examiners explicitly include the answer in the text.