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MHT CET 2026 (11 Apr, Shift 1) · Angular Variables

Angular Variables question from MHT CET 2026 (11 Apr, Shift 1)

Q12MHT CET 2026 · 11 APR · SHIFT 1Medium

Angular Variables

The position vector of a particle is r⃗=acos⁡ωt i^+asin⁡ωt j^\vec{r} = a\cos\omega t\,\hat{i} + a\sin\omega t\,\hat{j}. Calculate the angle between its position vector and the velocity vector.

(cos⁡0=1,cos⁡90=0,cos⁡60=0.5,sin⁡30=0.5,sin⁡0=0\cos 0 = 1, \cos 90 = 0, \cos 60 = 0.5, \sin 30 = 0.5, \sin 0 = 0)

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Answer: (c)

  1. v⃗=dr⃗dt=−aωsin⁡ωt i^+aωcos⁡ωt j^\vec v = \frac{d\vec r}{dt} = -a\omega\sin\omega t\,\hat i + a\omega\cos\omega t\,\hat j.
  2. r⃗⋅v⃗=−a2ωsin⁡ωtcos⁡ωt+a2ωsin⁡ωtcos⁡ωt=0\vec r \cdot \vec v = -a^2\omega\sin\omega t\cos\omega t + a^2\omega\sin\omega t\cos\omega t = 0.
  3. Zero dot product means r⃗⊥v⃗\vec r \perp \vec v.
  4. So the answer is (c) 90∘90^\circ.
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