18 public questions tagged with this topic.
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Ref: महाराष्ट्र राज्य मंडळ मराठी म्हणी व वाक्प्रचार पुस्तक म्हणींचा अर्थ उपयोग उदाहरणांसह स्पष्ट करते CUET तयारीसाठी उपयुक्त आहे.
गती ही योग्य म्हण आहे, कारण ती दिलेल्या अर्थाशी जुळते आणि वापरात आहे.
Ref: महाराष्ट्र राज्य मंडळ मराठी म्हणी व वाक्प्रचार पुस्तक म्हणींचा अर्थ उपयोग उदाहरणांसह स्पष्ट करते CUET तयारीसाठी उपयुक्त आहे.
**Matter waves** wave-like properties of moving particles, macroscopic objects not exhibit measurable wave-like because λ = h/(m v) extremely small for large m, e.g., 1 kg at 1 m/s λ=6.6×10⁻³⁴ m undetectable, for electron 9.11×10⁻³¹ kg at 2×10⁶ m/s λ=0.36 nm measurable via diffraction, de Broglie hypothesis λ = h/p proposed by de Broglie 1924, Davisson-Germer experiment confirmed. Momentum p = m v = 9.11 × 10⁻³¹ × 3.0 × 10⁶ = 2.733 × 10⁻²⁴ kg m/s . λ = (h/p) = (6.63 × 10⁻³⁴/2.733 × 10⁻²⁴) ≈ 2.425 × 10⁻¹⁰ m = 0.2425 nm . Applying E = h f = h c/λ, p
Ref: NCERT > Physics Book > Dual Nature of Matter > De Broglie Wavelength and Matter Waves
**Poynting vector** S = E×B/μ₀ gives energy flow (W/m²), magnitude S = E B/μ₀, average = E₀ B₀/(2μ₀) = intensity, direction of propagation, showing energy transport perpendicular to E and B. The speed in a medium is reduced due to the medium’s permittivity ( ε ) and permeability ( μ ), giving v = (1/√(μ ε)) , which is less than c because ε > ε₀ and μ > μ₀ . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields Medium’s electric properties, illustrating EM wave transverse nature and Maxwell's displacement current concept.
Ref: NCERT > Physics Book > Electromagnetic Waves > Energy, Intensity and Momentum of EM Waves
**Periodic motion** repeats after fixed period T, x(t+T)=x(t), while oscillatory motion involves to-and-fro about equilibrium. SHM is special periodic motion where restoring force proportional to displacement, F = -k x, acceleration a = -ω² x, leading to sinusoidal displacement x = A cos(ωt + φ). Velocity: v = ± ω √(A² - x²) . A = 3 m, ω = 2 s⁻¹, x = 1.5 m . v = 2 √(3² - 1.5²) = 2 √(9 - 2.25) = 2 √(6.75) ≈ 5.2 m/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA²
Ref: NCERT > Physics Book > Oscillations > Periodic Motion and SHM Basic Concepts
**Distinction between periodic and oscillatory** clarifies all SHM is periodic but not all periodic is SHM. SHM requires linear restoring force and inertia, a ∝ -x, with ω = √(k/m). Functions like sin²ωt have period π/ω but lack a = -ω² x, thus periodic not SHM, while uniform circular motion is periodic without linear oscillation. Velocity: v = ± ω √(A² - x²) . A = 4 m, ω = 3 s⁻¹, x = 2 m . v = 3 √(4² - 2²) = 3 √(16 - 4) = 3 √(12) = 6√(3) ≈ 10.39 m/s . Applying x = A cos(ωt + φ), v
Ref: NCERT > Physics Book > Oscillations > Periodic Motion and SHM Basic Concepts
**SHM kinematics** given by x = A cos(ωt + φ), velocity v = dx/dt = -ω A sin(ωt + φ), acceleration a = dv/dt = -ω² A cos(ωt + φ) = -ω² x, maxima v_max = ωA at mean position x=0, a_max = ω²A at extremes x=±A. Phase φ determines initial position, ω = 2π/T = √(k/m). Velocity: v = ± ω √(A² - x²) . A = 5 m, ω = 2 s⁻¹, x = 2.5 m . v = 2 √(5² - 2.5²) = 2 √(25 - 6.25) = 2 √(18.75) ≈ 8.66 m/s . Applying x = A cos(ωt + φ), v
Ref: NCERT > Physics Book > Oscillations > Displacement, Velocity and Acceleration in SHM
Given: A cyclist travels at a constant speed of 18 km/h for 10 minutes . What is the distance covered? These values define the system as per NCERT data. Formula: Convert speed: 18 km/h = 18 · 1000/3600 = 5 m/s. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Time = 10 min = 10 · 60 = 600 s . Distance x = v t = 5 · 600 = 3000 m = 3 km . The distance covered is 3 km . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kgâ»Â¹ Kâ»Â¹, m/s², 10â»âµ are properly used as per NCERT.
Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.
Centripetal acceleration a_c=v²/r=4π²r/T²=4π²rf² per NCERT circular motion. Substituting r and frequency/period gives a_c value matching 27 m/s. Other options use incorrect r or f, so C is correct.
Ref: NCERT Class 11 Physics > Chapter 4: Motion in a Plane > Topic: Motion in a Plane and Projectile Motion
Vertical velocity: vy=2gh=2×9.8×44.1=863.28≈29.38m/s. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 25 m/s as the result, so option A is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Acceleration and Velocity Analysis - Part 8
For free fall, s=½gt² and v²=2gh per NCERT. With g=10 m/s², given height or velocity yields time or distance via these equations. Substituting values gives result matching 25 m/s. This confirms option A as correct by uniformly accelerated motion equations.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Acceleration and Velocity Analysis - Part 8
Convert speed: 108km/h=108⋅10003600=30m/s. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 30 m/s as the result, so option A is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Distance, Displacement and Velocity-Time Relations
