JEE Main 2022 · 28 Jun · Shift 1 · Physics Q33

← JEE Main 2022 · 28 Jun · Shift 1

PhysicsSingle correct+4 / -1Question 33 of 90

Motion of a particle in $x$ - $y$ plane is described by a set of following equations $x=4\sin\left(\dfrac\pi2-\omega t\right)$ m and $y=4\sin(\omega t)$ m. The path of the particle will be :

circular

helical

parabolic

elliptical

Answer and worked solution

Official answer: A

Rewrite. $x=4\cos\omega t$ , $y=4\sin\omega t$ , so $x^2+y^2=16$ : a circle.

(A)

Topics:Motion in a plane

← Previous question Next question →

JEE Main 2022 · 28 Jun · Shift 1, Physics question 33