AIEEE 2010 | Circular Motion Question 57 | Physics

AIEEE 2010

MCQ (Single Correct Answer)

-1

For a particle in uniform circular motion the acceleration $$\overrightarrow a $$ at a point P(R, θ) on the circle of radius R is (here θ is measured from the x–axis)

$$ - {{{v^2}} \over R}\cos \theta \widehat i + {{{v^2}} \over R}\sin \theta \widehat j$$

$$ - {{{v^2}} \over R}\sin \theta \widehat i + {{{v^2}} \over R}\cos \theta \widehat j$$

$$ - {{{v^2}} \over R}\cos \theta \widehat i - {{{v^2}} \over R}\sin \theta \widehat j$$

$${{{v^2}} \over R}\widehat i + {{{v^2}} \over R}\widehat j$$

AIEEE 2010

MCQ (Single Correct Answer)

-1

A point $$P$$ moves in counter-clockwise direction on a circular path as shown in the figure. The movement of $$P$$ is such that it sweeps out a length $$s = {t^3} + 5,$$ where $$s$$ is in metres and $$t$$ is in seconds. The radius of the path is $$20$$ $$m.$$ The acceleration of $$'P'$$ when $$t=2$$ $$s$$ is nearly.

AIEEE 2010 Physics - Circular Motion Question 56 English

AIEEE 2010 Physics - Circular Motion Question 56 English

$$13m/{s_2}$$

$$12m/{s^2}$$

$$7.2m{s^2}$$

$$14m/{s^2}$$

AIEEE 2004

MCQ (Single Correct Answer)

-1

Which of the following statements is FALSE for a particle moving in a circle with a constant angular speed?

The velocity vector is tangent to the circle.

The acceleration vector is tangent to the circle.

The acceleration vector points to the centre of the circle.

The velocity and acceleration vectors are perpendicular to each other.

AIEEE 2002

MCQ (Single Correct Answer)

-1

The minimum velocity (in $$m{s^{ - 1}}$$) with which a car driver must traverse a flat curve of radius 150 m and coefficient of friction $$0.6$$ to avoid skidding is

$$60$$

$$30$$

$$15$$

$$25$$

JEE Main Subjects