A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a plane. It follows

its velocity is constant

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its K.E. is constant

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its acceleration is constant

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it moves in a straight line

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Solution

The correct option is B its K.E. is constant
The force vector is always perpendicular to the displacement vector (or velocity vector) i.e $\to F ⊥ \to S$
$∴$ Work done by the force $W = \to F . \to S = 0$
From work-energy theorem, $W = Δ K . E$
$⟹$ $Δ K . E = 0$ i.e Kinetic energy of the particle is constant.

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A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a plane. It follows

The correct option is B its K.E. is constantThe force vector is always perpendicular to the displacement vector (or velocity vector) i.e →F⊥→S∴ Work done by the force W=→F.→S=0From work-energy theorem, W=ΔK.E⟹ ΔK.E=0 i.e Kinetic energy of the particle is constant.