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JEE PHYSICS MCQ: WORK-ENERGY-POWER

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Application +4 | -1
A force of constant magnitude F=10 NF=10 \mathrm{~N} acts on a particle moving in a plane such that it is perpendicular to the velocity v⃗(∣v⃗∣=v=5 m/s)\vec{v}(|\vec{v}|=v=5 \mathrm{~m} / \mathrm{s}) of the body, and the force is always directed towards a fixed point. Then the angle (in radian) turned by the velocity vector of the particle as it covers a distance S=10 mS=10 \mathrm{~m} is (take: mass of the particle as m=2 kg)m=2 \mathrm{~kg})
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Correct option: A
2
Option A (2) is correct because the angle turned by the velocity vector is calculated using the formula derived from circular motion and the given conditions.

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