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    a particle of mass 10g moves along a circle of radius 6.4 cm with a constant tangential acceleration

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    A particle of mass 10g moves along a circle pf radius 6.4 cm with a constant tangential accelaration. What is the magnitude of this accelaration if the kinetic energy of the particle becomes equal to 8×10-4 by tye end of the second revolution after the beginning of the motion?

    Open in App Solution Given

    mass of particle m = 0.01 kg

    radius of circle along which particle is moving , r= 6.4 cm

    kinetic energy of the particle K. E. = 8×10-4 J

    1/2 mv^2 = 8×10-4 J

    v^2= (16×10^-4 ) /0.01

    v^2= 16*10^-2 -------- (i)

    Given that K. E. Of particle is equal to 8×10-4

    by tye end of the second revolution after the beginning of the motion of particle.

    it means initial velocity (u ) 0 m/s at this moment.

    now using newton's 3rd equation of motion ,

    v^2 = u^2 + 2as

    v^2= 2as = 2 a (4 pi r)

    a= v^2/8 pi r

    a = (16*10^-2) / (8*3.14*6.4*10^-2)

    a = 0.1 m/s^r ans= 0.1 m/s^2 Intuition for COM

    Standard XII Physics

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    स्रोत : byjus.com

    A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to 8 × 10^

    Click here👆to get an answer to your question ✍️ A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to 8 × 10^-4 J boy the end of the second revolution after the beginning of the motion?

    A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to 8×10

    Question −4

    J boy the end of the second revolution after the beginning of the motion?

    A0.2m/s

    2

    B0.1m/s

    2

    C0.15m/s

    2

    D0.18m/s

    2 Hard Open in App

    Updated on : 2022-09-05

    Solution Verified by Toppr

    Correct option is B)

    kinetic energy =8×10

    −4 J or, 2 1 ​ mv²=8×10 −4 or, 2 1 ​ ×10×10 −3 v²=8×10 −4 or, v²=16×10 −2 =>v=0.4m/s

    initial velocity of particle,   u=0m/s

    we have to find Tangential acceleration at the end of 2nd revolution.

    total distance covered, s=2(2πr)=4πr

    so, v 2 =2as a= 2s v 2 ​ = 2(4πr) (0.4) 2 ​ = (8×3.14×6.4×10 − 2) 16×10 −2 ​ =0.0995m/s²≈0.1m/s²

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    स्रोत : www.toppr.com

    A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration. What is the magnitude of this acceleration, if the kinetic energy of the particle becomes equal to 8xx10^(

    Given, mass of particle, m = 0.01 kg. Radius of circle along which particle is moving, r=6.4 cm. because Kinetic energy of particle, KE=8xx10^(-4)J rArr" "(1)/(2)mv^(2)=8xx10^(-4)J rArr" "v^(2)=(16xx10^(-4))/(0.01)=16xx10^(-2)...(i) As it is given that KE of particle is equal to 8xx10^(-4) J by the end of second revolution after the beginning of motion of particle. It means, its initial velocity (u) is 0ms^(-1) at this moment. because By Newton's third equation of motion, rArr" "v^(2)=2a(t)s" or "v^(2)=2a(t)(4pir) (because "particle covers 2 revolutions") rArr" "a(t)=(v^(2))/(8pir)=(16xx10^(-2))/(8xx3.14xx6.4xx10^(-2)) [because"from Eq. (i), v"^(2)=16xx10^(-2)] therefore" "a(t)=0.1ms^(-2)

    Home > English > Class 11 > Physics > Chapter > Circular Motion >

    A particle of mass 10 g moves ...

    A particle of mass 10 g moves along a circle of radius

    6.4 6.4

    cm with a constant tangential acceleration. What is the magnitude of this acceleration. What is the magnitude of this acceleration, if the kinetic energy of the particle becomes equal to

    8× 10 −4 8×10-4

    J by the end of the second revolution after the beginning of the motion?

    Updated On: 27-06-2022

    00 : 30

    UPLOAD PHOTO AND GET THE ANSWER NOW!

    Text Solution Open Answer in App A 0.15m s −2 0.15ms-2 B 0.18m s −2 0.18ms-2 C 0.2m s −2 0.2ms-2 D 0.1m s −2 0.1ms-2 Answer

    The correct Answer is D

    Solution

    Given, mass of particle, m =

    0.01 0.01 kg.

    Radius of circle along which particle is moving,

    r=6.4 r=6.4 cm. ∵ ∵

    Kinetic energy of particle, KE=

    8× 10 −4 J 8×10-4J ⇒ 1 2 m v 2 =8× 10 −4 J ⇒ 12mv2=8×10-4J ⇒ v 2 = 16× 10 −4 0.01 =16× 10 −2

    ⇒ v2=16×10-40.01=16×10-2

    ...(i)

    As it is given that KE of particle is equal to

    8× 10 −4 8×10-4

    J by the end of second revolution after the beginning of motion of particle. It means, its initial velocity (u) is

    0m s −1 0ms-1 at this moment. ∵ ∵

    By Newton's third equation of motion,

    ⇒ v 2 =2 a t s or v 2 =2 a t (4πr)

    ⇒ v2=2ats or v2=2at(4πr)

    (∵particle covers 2 revolutions)

    (∵particle covers 2 revolutions)

    ⇒ a t = v 2 8πr = 16× 10 −2 8×3.14×6.4× 10 −2

    ⇒ at=v28πr=16×10-28×3.14×6.4×10-2

    [∵ from Eq. (i), v 2 =16× 10 −2 ]

    [∵from Eq. (i), v2=16×10-2]

    ∴ a t =0.1m s −2 ∴ at=0.1ms-2

    Answer

    Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.

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    10 ग्राम द्रव्यमान का एक कण एक नियत स्पर्शीय त्वरण के साथ 6.4 सेमी वृत्त की त्रिज्या के अनुदिश गति करता है | इस त्वरण का परिमाण क्या है ? गति शुरू करने के पश्चात द्वितीय चक्कर लगाने के बाद गतिज ऊर्जा

    8× 10 −4 8×10-4 जूल है | 645690662 100 5.6 K 4:00

    10g द्रव्यमान का कोई कण 6.4 सेमी लम्बी त्रिज्या के वृत्त के अनुदिश किसी नियत स्पर्श-रेखीय त्वरण से गति करता है, यदि गति आरम्भ करने के पश्चात दो परिक्रमाएँ पूरी करने पर कण की गतिज ऊर्जा

    8× 10 −4 8×10-4

    जूल हो जाती है, तो इस त्वरण का परिमाण क्या है ?।

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