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#distance

30 public questions tagged with this topic.

Diamagnetism, Paramagnetism and Ferromagnetism

A bar magnet’s field strength decreases with distance because:

**Ferromagnetism** shows large positive χ ≈ 10³ to 10⁵, strong attraction, domain structure with spontaneous magnetization, hysteresis, retentivity. Distinction based on sign and magnitude of χ and behaviour in non-uniform field, explaining attraction versus repulsion. The magnetic field strength of a bar magnet decreases with distance as the field lines spread out, reducing their density and thus the field intensity, following an inverse cube relationship ( B ∝ 1/r³ ) for a dipole at large distances. Substituting values gives The field lines spread out, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on m/r³ and torque relation τ = m B sinθ.

Ref: NCERT > Physics Book > Magnetism and Matter > Diamagnetism, Paramagnetism and Ferromagnetism

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PHYSICS

A cyclist travels at a constant speed of 18 km/h for 10 minutes . What is the distance covered?

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 kg⁻¹ K⁻¹, m/s², 10⁻⁵ 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.

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PHYSICS

A stone is dropped from rest. What is the distance covered in the 5th second of its fall? (Take g = 10 m/s² )

Given: A stone is dropped from rest. What is the distance covered in the 5th second of its fall? (Take g = 10 m/s² ) These values define the system as per NCERT data. Formula: Distance in n -th second = g · 2n - 1/2. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: For 5th second, n = 5 . Substitute: d = 10 · 2 · 5 - 1/2 = 10 · 9/2 = 45 m . The distance covered is 45 m . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ 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.

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Vector Product and Rotational Kinematics

Two particles of masses 3kg and 7kg are at (1,4) and (9,2) respectively. What is the distance of their center of mass fr

X = (3×1)+(7×9)3+7 = 3+6310 = 6.6. Y = (3×4)+(7×2)3+7 = 12+1410 = 2.6. Distance = (6.6)2+(2.6)2 = 43.56+6.76 = 50.32≈7.09m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 7.09 m. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Rotational Motion Applications

Two particles of masses 8kg and 2kg are at (0,5) and (10,0) respectively. What is the distance of their center of mass f

X = (8×0)+(2×10)8+2 = 2010 = 2. Y = (8×5)+(2×0)8+2 = 4010 = 4. Distance = (2)2+(4)2 = 4+16 = 20≈4.47m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 4.47 m. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Motion in a Straight Line - Advanced Problems

A body accelerates uniformly from rest to a velocity of 20m/s over a distance of 50m. What is the acceleration?

Use v2=v02+2ax. Here, v0=0, v=20m/s, x=50m. 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 4 m/s² as the result, so option C is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Motion in a Straight Line - Advanced Problems

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Distance, Displacement and Velocity-Time Relations

A particle moves at 18m/s and decelerates at 4m/s2 for 2s, then accelerates at 3m/s2 until its speed is 18m/s. What is t

Phase 1: v=18−4⋅2=10m/s, x1=18⋅2−12⋅4⋅(2)2=36−8=28m. 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 65 m 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

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