Q126 / 160 · TS EAMCET 2019 (03 May Shift 2) · Physics

A particle of mass m is moving along a circle of radius R such that its tangential acceleration at varies with distance covered x as at=αx2, where α is a constant. The kinetic energy K of the particle varies with the distance as K=βxc, where β and c are constants. The values of β and c are

  • Correct answer
    A
    β=mα3;c=3
  • B
    β=mα4;c=4
  • C
    β=mα2;c=4
  • D
    β=mα2;c=3

Solution

Given,

Tangential acceleration, at=αx2 and

Kinetic energy, K=βxc.

If F is the tangential force acting on the particle thenF=mαx2

Now applying work energy theorem,

W=∆KE

⇒  ∫F dx=K

⇒  ∫mαx2dx=K

⇒  K=mαx33=βxc

∴  β=mα3; c=3.

Hence, the correct answer is option: 'A'.

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