Class 11 Physics · Chapter 3
Short answer:
Class 11 Physics Chapter 3 "Motion in a Plane" NCERT Solutions — is chapter mein vectors (scalar vs vector, resolution, addition-subtraction), motion in a plane with constant acceleration, relative velocity in two dimensions, projectile motion aur uniform circular motion cover hota hai. Neeche saare in-chapter solved examples, exercise questions (step-by-step CBSE marking-scheme style working ke saath), formula table, common mistakes aur exam-important FAQs mile hain — same format jo humne Class 11 Physics Chapter 2 Motion in a Straight Line NCERT solutions mein use kiya tha.
Chapter 2 "Motion in a Straight Line" mein humne motion sirf ek dimension (seedhi line) mein padha tha — position, velocity, acceleration sab ek axis pe. Real duniya mein motion aise limited nahi hota: ek cricket ball ka projectile path, gaadi ka turn lena, ya Earth ka Sun ke around circular orbit — ye sab do dimensions (ek plane) mein hota hai. Chapter 3 "Motion in a Plane" us jump ko cover karta hai — 1D se 2D motion tak, aur iske liye pehle vectors ka poora mathematical toolkit seekhna padta hai.
Ye chapter rationalised syllabus (2026-27 session) ke under Class 11 Physics ka 3rd chapter hai — purane "Physical World" chapter ko standalone examinable chapter se hata diya gaya hai, isliye ab numbering "Units and Measurements" (Ch.1) → "Motion in a Straight Line" (Ch.2) → "Motion in a Plane" (Ch.3) se shuru hoti hai. Chapter ke andar bhi kuch trimming hui hai: relative velocity (two dimensions) ka derivation portion pehle se halka kar diya gaya hai — concept aur core formula syllabus mein hain, lekin lambi multi-case derivations nikal di gayi hain. (Note: ye confirmation registry-grade third-party sources se hai, official ncert.nic.in PDF se direct verify nahi ho paya is run mein — kisi bhi numerically precise claim se pehle final cross-check zaroor karo.) Agar tumhe poora class 11 physics all chapters pdf 2026-27 list chahiye ya class 11 physics deleted syllabus 2026-27 ka poora breakdown, wo alag resource me milega.
Iske foundation pe hi aage ke chapters khade hain — laws of motion class 11 ncert exemplar level ke force-vector problems, gravitation class 11 numericals ke orbital-velocity questions, aur system of particles and rotational motion ke torque-vector cross-product questions — sab yahi ke vector-resolution skills use karte hain. Isliye is chapter ko sirf "ratt" mat karo, vectors ka geometric intuition build karo.
Chapter 3 Summary — 5 Minute Revision
Chapter 3 "Motion in a Plane" ka core summary: scalar quantities (mass, speed, temperature) sirf magnitude se define hoti hain, vector quantities (displacement, velocity, force) ko magnitude + direction dono chahiye aur ye triangle/parallelogram law se add hote hain, ordinary algebra se nahi. Ek vector ko do perpendicular components (resolution) mein todna — Ax = A cosθ, Ay = A sinθ — 2D motion ke har problem ki key technique hai, kyunki x aur y direction independently solve ho sakti hain.
Position vector r = x î + y ĵ se displacement, velocity (v = dr/dt) aur acceleration (a = dv/dt) define hote hain — agar acceleration constant hai to seedhi-line ke wahi kinematic equations (v = v₀ + at, r = r₀ + v₀t + ½at²) x aur y components pe alag-alag apply hoti hain. Relative velocity (vAB = vA − vB) do dimensions mein rain-umbrella aur river-boat type problems solve karti hai. Projectile motion (u sinθ, g downward) se time of flight, max height, range nikalte hain, aur uniform circular motion mein speed constant rehte hue bhi velocity direction badalne ki wajah se centripetal acceleration (a꜀ = v²/r) centre ki taraf act karti hai.
In-Text Questions — Solutions
Example 3.1 — Ek scooter uttar (north) ki taraf 10 km jaata hai, phir turn karke poorab (east) ki taraf 5 km jaata hai. Resultant displacement ka magnitude aur direction nikalo.
North displacement A = 10 km (ĵ direction), East displacement B = 5 km (î direction). In dono ki koi common direction nahi, isliye Pythagoras se resultant nikalega kyunki angle 90° hai.
R = √(A² + B²) = √(10² + 5²) = √(100+25) = √125 = 11.18 km
Direction (east se North ki taraf angle θ):
tanθ = A/B = 10/5 = 2 ⟹ θ = tan⁻¹(2) ≈ 63.4° north of east
Resultant displacement = 11.18 km, 63.4° north of east direction se.
Example 3.2 — Ek force vector F ka magnitude 50 N hai aur x-axis se 37° ka angle banata hai. Iske x aur y components nikalo. (sin37° ≈ 0.6, cos37° ≈ 0.8)
Resolution formula use karte hain:
F_x = F cosθ = 50 × 0.8 = 40 N
F_y = F sinθ = 50 × 0.6 = 30 N
Check: F = √(F_x² + F_y²) = √(1600+900) = √2500 = 50 N ✓ (matches original magnitude)
Components: F_x = 40 N, F_y = 30 N.
Example 3.3 — Ek particle x-y plane mein a = (2î + 3ĵ) m/s² constant acceleration se move karta hai, initial velocity v₀ = (5î) m/s aur t=0 pe origin se shuru hota hai. t = 2 s pe position vector nikalo.
Constant acceleration ke liye har component pe alag se kinematic equation lagti hai: r = r₀ + v₀t + ½at²
x-component: x = 0 + 5(2) + ½(2)(2)² = 10 + 4 = 14 m
y-component: y = 0 + 0(2) + ½(3)(2)² = 0 + 6 = 6 m
r(2s) = (14î + 6ĵ) m
Position vector t=2s pe = (14î + 6ĵ) m, yaani (14 m, 6 m) point.
Example 3.4 — Baarish 30 m/s ki speed se vertically neeche gir rahi hai. Ek cyclist 10 m/s speed se seedhi line mein move karta hai. Cyclist ko umbrella vertical se kitne angle pe tilt karna chahiye taaki bheega na?
Rain ki velocity ground frame mein v_rain = 30 m/s (vertically down). Cyclist ki velocity v_cycle = 10 m/s (horizontal). Rain ki velocity relative to cyclist:
v_rain,cycle = v_rain − v_cycle
Yani horizontal component = −10 m/s (rain cyclist ki taraf backward relative move karti dikhti hai) aur vertical component = 30 m/s (down).
tanθ = horizontal/vertical = 10/30 = 1/3
θ = tan⁻¹(1/3) ≈ 18.4°
Cyclist ko umbrella vertical se ≈18.4°, apni motion ki direction ki taraf tilt karna chahiye.
Example 3.5 — Ek ball 20 m/s speed se ground se 30° angle pe project ki jaati hai (g = 10 m/s²). Time of flight, maximum height aur horizontal range nikalo.
u = 20 m/s, θ = 30°, sin30° = 0.5, cos30° = 0.866, sin60° = 0.866
Time of flight:
T = 2u sinθ / g = (2×20×0.5)/10 = 20/10 = 2 s
Maximum height:
H = u²sin²θ / 2g = (400×0.25)/20 = 100/20 = 5 m
Horizontal range:
R = u²sin2θ / g = (400×0.866)/10 = 346.4/10 = 34.64 m
Answers: T = 2 s, H = 5 m, R = 34.64 m
Example 3.6 — Ek cyclist 20 m radius wale circular track pe 5 m/s ki constant speed se move karta hai. Centripetal acceleration aur ek full round pura karne ka time (period) nikalo.
r = 20 m, v = 5 m/s
Centripetal acceleration:
a_c = v²/r = 25/20 = 1.25 m/s²
Direction: hamesha centre ki taraf.
Period (ek round ka time) — pehle angular velocity nikalo:
ω = v/r = 5/20 = 0.25 rad/s
T = 2π/ω = 2×3.14/0.25 = 25.12 s
Answers: a_c = 1.25 m/s² (centre ki taraf), T ≈ 25.1 s
Poore Class 11 Physics ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q14)
3.1 Neeche di gayi physical quantities ke liye batao ki wo scalar hai ya vector: volume, mass, speed, acceleration, density, number of moles, velocity, angular frequency, displacement, angular velocity.
Definition yaad rakho: scalar = sirf magnitude, vector = magnitude + direction, aur addition triangle/parallelogram law follow kare.
| Quantity | Type |
|---|---|
| Volume | Scalar |
| Mass | Scalar |
| Speed | Scalar |
| Acceleration | Vector |
| Density | Scalar |
| Number of moles | Scalar |
| Velocity | Vector |
| Angular frequency | Scalar |
| Displacement | Vector |
| Angular velocity | Vector |
↔ Table ko side me swipe karein
Note: angular frequency (ω, sirf rate) scalar hai jabki angular velocity (rotation axis ki direction ke saath) vector treat hoti hai.
3.2 Neeche di gayi list mein do scalar quantities pehchano: force, angle, work, current, average velocity, magnetic moment, relative velocity.
In saaton mein se sirf work aur current scalar hain. Work sirf magnitude se define hota hai (energy transfer), isliye clearly scalar hai. Current yahan trick hai — current ki ek 'direction' (flow ki taraf) hoti hai, phir bhi ise vector nahi maana jaata, kyunki current triangle/parallelogram law of vector addition follow nahi karta (ek junction pe milne wale currents simple algebra se add hote hain, vector addition se nahi). Baaki (force, angle, average velocity, magnetic moment, relative velocity) is exercise ke maksad se vector treat kiye jaate hain.
3.3 Ek particle P displacement 3 m poorab (east) taraf, phir 4 m north taraf move karta hai. Resultant displacement ka magnitude aur direction nikalo.
Do perpendicular displacements: A = 3 m (east), B = 4 m (north). Ye ek right-angled triangle banate hain.
R = √(3² + 4²) = √(9+16) = √25 = 5 m
tanθ = 4/3 ⟹ θ = tan⁻¹(4/3) ≈ 53.1°
Resultant displacement = 5 m, 53.1° north of east.
3.4 Do vectors A = 3î + 4ĵ aur B = 2î − ĵ diye gaye hain. (a) A+B aur (b) A−B nikalo, aur unka magnitude bhi batao.
(a) A + B:
A+B = (3+2)î + (4−1)ĵ = 5î + 3ĵ
|A+B| = √(25+9) = √34 ≈ 5.83
(b) A − B:
A−B = (3−2)î + (4−(−1))ĵ = 1î + 5ĵ
|A−B| = √(1+25) = √26 ≈ 5.10
Answers: A+B = 5î + 3ĵ (magnitude ≈ 5.83), A−B = î + 5ĵ (magnitude ≈ 5.10)
3.5 Ek vector A ka magnitude 10 units hai aur x-axis se 60° angle banata hai. Iske rectangular components (A_x, A_y) nikalo.
A_x = A cosθ = 10 × cos60° = 10 × 0.5 = 5
A_y = A sinθ = 10 × sin60° = 10 × 0.866 = 8.66
Components: A_x = 5, A_y = 8.66. Check: √(5²+8.66²) = √(25+75) = √100 = 10 ✓
3.6 Ek particle ki position r = (2t²)î + (3t)ĵ metre se di gayi hai (t seconds mein). t=2s pe velocity aur acceleration vectors nikalo.
Velocity = dr/dt:
v = (4t)î + 3ĵ
t=2s pe: v = (4×2)î + 3ĵ = (8î + 3ĵ) m/s, |v| = √(64+9) = √73 ≈ 8.54 m/s
Acceleration = dv/dt:
a = 4î + 0ĵ = 4î m/s²
Ye acceleration constant hai (t pe depend nahi karta), sirf x-direction mein.
3.7 Ek nadi 3 km/h ki speed se flow karti hai. Ek tairak (swimmer) still water mein 4 km/h ki speed se seedha (perpendicular to bank) tairta hai. Resultant velocity ka magnitude aur direction (bank se angle) nikalo.
Swimmer ki velocity relative to water v_s = 4 km/h (perpendicular to bank), river ki velocity v_r = 3 km/h (along bank). Resultant velocity ground frame mein dono ka vector sum hai — perpendicular hone se Pythagoras lagta hai:
v_resultant = √(4² + 3²) = √(16+9) = √25 = 5 km/h
tanθ = 3/4 ⟹ θ = tan⁻¹(0.75) ≈ 36.9° (bank ki normal se, downstream ki taraf)
Resultant velocity = 5 km/h, bank se 90°−36.9° ≈ 53.1° ka angle banate hue (ya perpendicular se 36.9° downstream shift).
3.8 Do vectors A aur B, jinka magnitude equal hai, ek doosre se 120° ka angle banate hain. Agar resultant A aur B dono ke saath equal angle banaye, to prove karo ki resultant ka magnitude bhi A ke barabar hoga.
Jab dono vectors ka magnitude equal ho (A = B = 'a') aur unke beech angle θ ho, resultant ka magnitude:
R = √(a² + a² + 2a·a·cosθ) = a√(2 + 2cosθ)
θ = 120° ke liye (cos120° = −0.5):
R = a√(2 + 2×(−0.5)) = a√(2−1) = a√1 = a
Isliye R = a = A, yaani resultant ka magnitude A (aur B) ke barabar hai — jo prove karna tha.
Equal-angle part: kyunki |A| = |B|, parallelogram ek rhombus banata hai, aur rhombus ka diagonal (resultant) angle ko symmetrically bisect karta hai. Isliye resultant, A aur B dono ke saath θ/2 = 60° ka equal angle banata hai.
3.9 Ek footballer ball ko 25 m/s speed se, horizontal se 37° ke angle pe kick karta hai (g=10 m/s², sin37°≈0.6, cos37°≈0.8). Time of flight, maximum height aur range nikalo.
u = 25 m/s, θ = 37°
Time of flight:
T = 2u sinθ/g = (2×25×0.6)/10 = 30/10 = 3 s
Maximum height:
H = u²sin²θ/2g = (625×0.36)/20 = 225/20 = 11.25 m
Range (sin2θ = 2 sinθ cosθ = 2×0.6×0.8 = 0.96):
R = u²sin2θ/g = (625×0.96)/10 = 600/10 = 60 m
Answers: T=3s, H=11.25 m, R=60 m
3.10 Prove karo ki projectile motion mein trajectory (path) ek parabola hoti hai.
Horizontal motion: x = (u cosθ)t ⟹ t = x/(u cosθ)
Vertical motion: y = (u sinθ)t − ½gt²
t ki value substitute karo:
y = (u sinθ)·x/(u cosθ) − ½g·x²/(u cosθ)²
y = x tanθ − (g x²)/(2u²cos²θ)
Ye equation y = Ax − Bx² form mein hai (A, B constants), jo ek parabola ka standard equation hai. Isliye projectile ka path parabolic hota hai.
3.11 Ek stone ko 15 m/s speed se horizontally throw kiya jaata hai ek 20 m unchi building se (g=10 m/s²). Stone ground pe kitni door (horizontal distance) girega, aur kitne time mein?
Ye horizontal projectile hai — initial vertical velocity = 0.
Vertical motion se time nikalo: h = ½gt²
20 = ½×10×t² ⟹ t² = 4 ⟹ t = 2 s
Horizontal distance:
x = u×t = 15×2 = 30 m
Answers: t = 2 s, horizontal distance = 30 m
3.12 Ek satellite Earth ke around 7 km/s ki constant speed se, 7000 km radius ke circular orbit mein ghoomta hai. Iski centripetal acceleration nikalo, aur batao ye direction mein kya hai.
v = 7 km/s = 7000 m/s, r = 7000 km = 7×10⁶ m
a_c = v²/r = (7000)²/(7×10⁶) = 4.9×10⁷/7×10⁶ = 7 m/s²
Ye acceleration hamesha satellite se Earth ke centre ki taraf directed hoti hai — isi wajah se satellite curved path pe rehta hai, seedhi line mein nahi udta.
3.13 Ek cyclist 10 m/s speed se ek horizontal circular track (radius 40 m) pe move kar raha hai. Uska angular speed aur ek chakkar (revolution) ka time period nikalo.
v = 10 m/s, r = 40 m
ω = v/r = 10/40 = 0.25 rad/s
T = 2π/ω = 2×3.14/0.25 = 25.12 s
Answers: ω = 0.25 rad/s, T ≈ 25.1 s
3.14 Do buses A aur B seedhi road pe same direction mein move kar rahe hain — A ki speed 60 km/h aur B ki speed 45 km/h hai. A ki velocity, B ke relative kya hai? Agar A, B se 500 m aage hai, to kya B kabhi A ko pakad payega — reasoning ke saath batao.
Same direction mein move kar rahe honge to relative velocity magnitude subtraction se aata hai:
v_AB = v_A − v_B = 60 − 45 = 15 km/h
Yani A, B se 15 km/h ki relative speed se aur aage badhta jaayega. A pehle se hi 500 m aage hai aur teji se move kar raha hai, isliye gap kam nahi hoga, badhta jaayega.
v_BA = v_B − v_A = 45 − 60 = −15 km/h
Answers: v_AB = 15 km/h (A ki taraf badhta relative), B kabhi A ko pakad nahi payega jab tak koi speed na badle — kyunki B, A se 15 km/h ki rate se peeche chhootta jaata hai.
Important Equations — Ek Nazar Me
| Concept | Formula |
|---|---|
| Resultant of two vectors (triangle/parallelogram law) | R = √(A² + B² + 2AB cosθ) |
| Direction of resultant | tanβ = B sinθ / (A + B cosθ) |
| Resolution of a vector into components | A_x = A cosθ, A_y = A sinθ |
| Unit vector | Â = A / |A| |
| Position vector | r = x î + y ĵ |
| Average velocity | v_avg = Δr / Δt |
| Instantaneous velocity | v = dr/dt |
| Average / instantaneous acceleration | a_avg = Δv/Δt, a = dv/dt |
| Motion in a plane, constant acceleration | v = v₀ + at ; r = r₀ + v₀t + ½at² |
| Relative velocity (2D) | v_AB = v_A − v_B |
| Projectile — time of flight | T = 2u sinθ / g |
| Projectile — maximum height | H = u² sin²θ / 2g |
| Projectile — horizontal range | R = u² sin2θ / g |
| Projectile — trajectory equation | y = x tanθ − g x² / (2u² cos²θ) |
| Uniform circular motion — speed-angular velocity relation | v = ωr |
| Centripetal acceleration | a_c = v²/r = ω²r |
| Angular velocity — period/frequency | ω = 2π/T = 2πν |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Vectors ko ordinary numbers ki tarah add karna — 3 units east + 4 units north kabhi 7 nahi hoga, triangle/parallelogram law use karo (yahan resultant 5 hoga, Pythagoras se), sirf same-direction vectors hi algebraically add hote hain.
- Resolution mein galat trig function lagana — angle kis reference se measure ho raha hai (x-axis se ya y-axis se) confirm kiye bina cosθ/sinθ swap kar dena, jisse A_x aur A_y ulte aa jaate hain.
- Relative velocity mein sign/direction bhool jaana — v_AB = v_A − v_B mein vector subtraction hai, sirf magnitude subtract karke direction ignore kar dena galat answer deta hai (khaaskar rain-umbrella ya boat-river type problems mein).
- Projectile motion mein g ka sign galat lagana — upward ko positive maankar bhi acceleration ko +g likh dena; g hamesha downward (motion ki upward direction ke opposite) hota hai, isliye equations mein −g use karo jab tak axis convention clearly define na ho.
- Maximum height aur range ke formulas confuse karna — H mein sin²θ hota hai jabki R mein sin2θ (= 2sinθcosθ), dono ko mix kar dena ek common calculation mistake hai jo 45° ke alawa har angle pe bada farak dalta hai.
- Uniform circular motion mein 'uniform' ka matlab galat samajhna — speed constant hone ka matlab acceleration zero nahi hota; velocity ki direction continuously badalti hai isliye centripetal acceleration (v²/r) hamesha centre ki taraf non-zero rehti hai.
Board-Style Important Questions
- 1 mark: Neeche di gayi list mein se sirf ek vector quantity chuno: temperature, pressure, impulse, time, energy.
- 2 marks: Do vectors A aur B ka magnitude equal hai. Prove karo ki (A+B) vector, (A−B) vector ke perpendicular hota hai.
- 3 marks: Ek particle ki position vector r = (3t² î + 2t ĵ) m se di gayi hai. t = 3s pe iski velocity aur acceleration nikalo.
- 3 marks: Ek ball 30 m/s speed se 45° ke angle pe project ki jaati hai. Time of flight, maximum height aur range nikalo (g = 10 m/s²).
- 5 marks: Projectile motion ki trajectory equation derive karo aur prove karo ki path parabolic hota hai. Isse maximum height aur range ke formulas bhi derive karo.
- 5 marks: Uniform circular motion mein centripetal acceleration ka expression a_c = v²/r derive karo, aur explain karo ki ye acceleration kyun hamesha velocity ke perpendicular hoti hai.
Aksar Poochhe Jaane Wale Sawaal
Class 11 Physics Chapter 3 'Motion in a Plane' mein kaunse topics aate hain?
Scalars vs vectors, vector addition/subtraction (graphical + analytical), resolution of vectors, motion in a plane with constant acceleration, relative velocity in two dimensions, projectile motion, aur uniform circular motion — ye poore class 11 physics ncert solutions series ka teesra chapter hai (Ch.1 Units and Measurements, Ch.2 Motion in a Straight Line ke baad).
Kya Motion in a Plane chapter se 2026-27 rationalised syllabus mein koi topic hataya gaya hai?
Poora topic nahi hataya gaya, lekin relative velocity in two dimensions ka derivation portion trim (halka) kar diya gaya hai — core concept aur formula (v_AB = v_A − v_B) syllabus mein hai. Poore chapter-wise breakdown ke liye class 11 physics deleted syllabus 2026-27 resource dekho, aur official ncert.nic.in PDF se cross-verify zaroor karo.
Class 11 Physics Chapter 2 (Motion in a Straight Line) ke NCERT solutions kahan milenge?
Uska poora solved set (position-time graphs, uniformly accelerated motion, relative velocity 1D) separate se available hai — search karo class 11 physics chapter 2 motion in a straight line ncert solutions, jo is Chapter 3 ke liye foundation ka kaam karta hai.
Motion in a Plane ke baad kaunsa chapter aata hai aur uski taiyari kaise karein?
Agla chapter Laws of Motion hai, jahan yahi vector-resolution skills force aur Newton's laws pe apply hoti hain. Uske baad Work-Energy-Power, phir System of Particles and Rotational Motion — jiske important formulas torque/angular-momentum cross-products pe based hain, aur ye sab yahi chapter ke vector basics maangte hain.
Motion in a Plane chapter CBSE board exam ke liye kitna important hai?
Ye Mechanics unit ka core part hai, aur projectile motion + uniform circular motion jaise numericals is chapter ki typical problem-solving style hain. Exact class 11 physics chapter wise weightage (kis chapter se kitne marks aate hain) CBSE ke official marking scheme/sample paper se hi confirm karo — humne yahan koi specific marks-weightage number claim nahi kiya hai kyunki ye session-wise vary karta hai aur is run mein third-party se verify nahi ho paya.
Motion in a Plane padhne ke baad kaunse chapters vector-skills use karte hain?
Bahut saare — gravitation class 11 numericals mein orbital velocity/escape velocity vectors, system of particles and rotational motion mein torque, oscillations aur waves class 11 physics doppler effect formula mein bhi direction-based reasoning, aur mechanical properties of fluids class 11 numericals mein force-vector balance — sab is chapter ke resolution/relative-velocity concepts pe base karte hain.
Class 11 Physics — Saare Chapters
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