# 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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## 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 8×10Question −4

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

**A**0.2m/s

2

**B**0.1m/s

2

**C**0.15m/s

2

**D**0.18m/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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## 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)

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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

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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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