The displacement of a particle moving on a circular path when it makes 60° at the centre is

Question

The displacement of a particle moving on a circular path when it makes 60° at the centre is (A) 2 r (B) r (C) √2 r (D) None of these

Tardigrade Admin · Accepted Answer

In the figure, A B is the required displacement of the particle. In triangle O A B, O A=O B and angle A O B=60° <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/e72ae4ee4a55d3cb150e0c33a4bdff34-.png" /> Therefore, triangle A O B is an equilateral triangle, so O A=O B=r=A B .

Q. The displacement of a particle moving on a circular path when it makes $6 0^{\circ}$ at the centre is

$2 r$

$r$

$2 r$

None of these

In the figure, $A B$ is the required displacement of the particle.
In triangle $O A B, O A = OB$ and $∠ A OB = 6 0^{\circ}$

Therefore, $△ A OB$ is an equilateral triangle, so
$O A = OB = r = A B$ .

Q. The displacement of a particle moving on a circular path when it makes 60∘ at the centre is

2r

r

2​r

None of these

In the figure, AB is the required displacement of the particle. In triangle OAB,OA=OB and ∠AOB=60∘ Therefore, △AOB is an equilateral triangle, so OA=OB=r=AB.

Q. The displacement of a particle moving on a circular path when it makes $6 0^{\circ}$ at the centre is

$2 r$

$r$

$2 r$

In the figure, $A B$ is the required displacement of the particle.
In triangle $O A B, O A = OB$ and $∠ A OB = 6 0^{\circ}$

Therefore, $△ A OB$ is an equilateral triangle, so
$O A = OB = r = A B$ .