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In the figure the variation of potential energy of a particle of mass m = 2kg is represented w.r.t. its x-coordinate. The particle moves under the effect of this conservative force along the x-axis.Q.If the particle is released at x = 2 + Δ where Δ → 0 (it is positive) then its maximum speed in subsequent motion will be :a)√10m/sb)5 m/sc)5√2 m/sd)7.5 m/sCorrect answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about In the figure the variation of potential energy of a particle of mass m = 2kg is represented w.r.t. its x-coordinate. The particle moves under the effect of this conservative force along the x-axis.Q.If the particle is released at x = 2 + Δ where Δ → 0 (it is positive) then its maximum speed in subsequent motion will be :a)√10m/sb)5 m/sc)5√2 m/sd)7.5 m/sCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In the figure the variation of potential energy of a particle of mass m = 2kg is represented w.r.t. its x-coordinate. The particle moves under the effect of this conservative force along the x-axis.Q.If the particle is released at x = 2 + Δ where Δ → 0 (it is positive) then its maximum speed in subsequent motion will be :a)√10m/sb)5 m/sc)5√2 m/sd)7.5 m/sCorrect answer is option 'B'. Can you explain this answer?.

Solutions for In the figure the variation of potential energy of a particle of mass m = 2kg is represented w.r.t. its x-coordinate. The particle moves under the effect of this conservative force along the x-axis.Q.If the particle is released at x = 2 + Δ where Δ → 0 (it is positive) then its maximum speed in subsequent motion will be :a)√10m/sb)5 m/sc)5√2 m/sd)7.5 m/sCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.

Here you can find the meaning of In the figure the variation of potential energy of a particle of mass m = 2kg is represented w.r.t. its x-coordinate. The particle moves under the effect of this conservative force along the x-axis.Q.If the particle is released at x = 2 + Δ where Δ → 0 (it is positive) then its maximum speed in subsequent motion will be :a)√10m/sb)5 m/sc)5√2 m/sd)7.5 m/sCorrect answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of In the figure the variation of potential energy of a particle of mass m = 2kg is represented w.r.t. its x-coordinate. The particle moves under the effect of this conservative force along the x-axis.Q.If the particle is released at x = 2 + Δ where Δ → 0 (it is positive) then its maximum speed in subsequent motion will be :a)√10m/sb)5 m/sc)5√2 m/sd)7.5 m/sCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for In the figure the variation of potential energy of a particle of mass m = 2kg is represented w.r.t. its x-coordinate. The particle moves under the effect of this conservative force along the x-axis.Q.If the particle is released at x = 2 + Δ where Δ → 0 (it is positive) then its maximum speed in subsequent motion will be :a)√10m/sb)5 m/sc)5√2 m/sd)7.5 m/sCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of In the figure the variation of potential energy of a particle of mass m = 2kg is represented w.r.t. its x-coordinate. The particle moves under the effect of this conservative force along the x-axis.Q.If the particle is released at x = 2 + Δ where Δ → 0 (it is positive) then its maximum speed in subsequent motion will be :a)√10m/sb)5 m/sc)5√2 m/sd)7.5 m/sCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice In the figure the variation of potential energy of a particle of mass m = 2kg is represented w.r.t. its x-coordinate. The particle moves under the effect of this conservative force along the x-axis.Q.If the particle is released at x = 2 + Δ where Δ → 0 (it is positive) then its maximum speed in subsequent motion will be :a)√10m/sb)5 m/sc)5√2 m/sd)7.5 m/sCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice JEE tests.