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# a mass m moves in a circle on a smooth horizontal plane with velocity v0

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## A mass, m , moves in a circle on a smooth horizontal plane with velocity, V0 , at a radius, R0 . The mass is attached to a string which passes through a smooth hole in the plane as shown. The tension in the string is increased gradually and finally m moves in a circle of radius R02 . The final value of the kinetic energy is

Click here👆to get an answer to your question ✍️ A mass, m , moves in a circle on a smooth horizontal plane with velocity, V0 , at a radius, R0 . The mass is attached to a string which passes through a smooth hole in the plane as shown. The tension in the string is increased gradually and finally m moves in a circle of radius R02 . The final value of the kinetic energy is

A mass, m, moves in a circle on a smooth horizontal plane with velocity, V

Question 0 ​ , at a radius, R 0 ​

. The mass is attached to a string which passes through a smooth hole in the plane as shown. The tension in the string is increased gradually and finally m moves in a circle of radius

2 R 0 ​ ​

. The final value of the kinetic energy is

A2mv

0 2 ​

B

2 1 ​ mv 0 2 ​

Cmv

0 2 ​

D

4 1 ​ mv 0 2 ​ Medium NEET Open in App Solution Verified by Toppr

Correct option is A)

The angular momentum of mass about center will remain constant as the torque of tension is zero.

mv ∘ ​ r ∘ ​ =mv ′ 2 r ∘ ​ ​ v ′ =2v ∘ ​

Final value of K.E.=

2 1 ​ m(2v ∘ ​ ) 2 =2mv ∘ ​ 2 Video Explanation

144 10

स्रोत : www.toppr.com

Question A mass m

moves in a circle on a smooth horizontal plane with velocity

v 0 at a radius R 0

. The mass is attached to a string which passes through a smooth hole in the plane as shown.

The tension in the string is increased gradually and finally

m moves in a circle of radius R 0 / 2

. Then, the final value of the kinetic energy of particle is

Open in App Solution

The correct option is D

2 m v 0 2

Since no external torque is acting on the system, thus according to law of conservation of angular momentum,

Initial angular momentum = Final angular momentum

m v 0 R 0 = m v r ⇒ m v 0 R 0 = m v R 0 2 ( ∵ New radius , r = R 0 2 ) ⇒ v = 2 v 0 New kinetic energy = 1 2 m v 2 = 2 m v 2 0

COM in Collision and Explosion

Standard XII Physics

Suggest Corrections 10

स्रोत : byjus.com

## A mass m moves in a circles on a smooth horizontal plane with velocity v(0) at a radius R(0). The mass is atteched to string which passes through a smooth hole in the plane as shown. The tension in string is increased gradually and finally m moves in a cricle of radius (R(0))/(2). the final value of the kinetic energy is

( c) Initial angular momentum L(initial) = mv0 R Initial angular momentum L(initial) = mv ( R)/(2) Conservation of angular momentum m v0 R0 =m mv (R0)/(2) rArr v = 2 v0 KE = (1)/(2) mv^2 = 2 mv0^2.

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A mass m moves in a circles on...

A mass m m

moves in a circles on a smooth horizontal plane with velocity

v 0 v0 at a radius R 0 R0

. The mass is atteched to string which passes through a smooth hole in the plane as shown.

The tension in string is increased gradually and finally

m m

moves in a cricle of radius

R 0 2 R02

. the final value of the kinetic energy is

Updated On: 27-06-2022

00 : 30

Text Solution Open Answer in App A m v 2 0 mv02 B 2m v 2 0 2mv02 C 1 4 m v 2 0 14mv02 D 1 2 m v 2 0 12mv02 Answer

Solution

( c) Initial angular momentum

L ∈itial =m v 0 R L∈itial=mv0R

Initial angular momentum

L ∈itial =mv R 2 L∈itial=mvR2

Conservation of angular momentum

m v 0 R 0 =mmv R 0 2 ⇒v=2 v 0 mv0R0=mmvR02⇒v=2v0 KE= 1 2 m v 2 =2m v 2 0 KE=12mv2=2mv02 .

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

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Mohammed 6 day ago

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