HSC Science (Electronics) 12th Board ExamMaharashtra State Board
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# Solution for A particle rotates in U.C.M. with tangential velocity V along a horizontal circle of diameter ‘D' . Total angular displacement of the particle in time 't' is - HSC Science (Electronics) 12th Board Exam - Physics

#### Question

A particle rotates in U.C.M. with tangential velocity V along a horizontal circle of diameter ‘D' . Total angular displacement of the particle in time 't' is..........

(a) vt

(b) (v/D)-t

(c) vt/2D

(d) 2vt/D

#### Solution

2(vt)/D

Velocity= v ; Diameter= D →Radius= D/2

omega=theta/t andv=r*omega

therefore v=D/2*theta/t

therefore theta=(2vt)/D

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Solution for question: A particle rotates in U.C.M. with tangential velocity V along a horizontal circle of diameter ‘D' . Total angular displacement of the particle in time 't' is concept: Uniform Circular Motion. For the courses HSC Science (Electronics), HSC Science (General) , HSC Science (Computer Science)
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