Practice free Motion in a Lane (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
If the x-component ($A_x$) and y-component ($A_y$) of a vector $\vec{A}$ are known, which expression correctly gives the angle $\theta$ that $\vec{A}$ makes with the x-axis?
From the summary section, point 10, it is given that '$\tan \theta = \frac{A_y}{A_x}$'.
Consider a vector $\vec{A}$. If $\vec{A_1}$ and $\vec{A_2}$ are its component vectors along the x and y axes respectively, then the component vectors are given by:
The NCERT text states: 'Since $\vec{A_1}$ is parallel to $\hat{i}$ and $\vec{A_2}$ is parallel to $\hat{j}$, we have : $\vec{A_1} = A_x \hat{i}$, $\vec{A_2} = A_y \hat{j}$ where $A_x$ and $A_y$ are real numbers.'
An object undergoes uniform circular motion. Which of the following statements about its acceleration is true?
According to the NCERT text, 'Thus, we find that the acceleration of an object in uniform circular motion is always directed towards the centre of the circle.' Also, it states that 'The direction of velocity v is along the tangent to the circle at every point. The acceleration is directed towards the centre of the circle. Since this direction changes continuously, acceleration here is not a constant vector.' Therefore, the acceleration is always directed towards the center.
For an object in uniform circular motion, if v is its linear speed and R is the radius of the circular path, what is the magnitude of its centripetal acceleration?
The NCERT text states, 'The magnitude of a is... given by $a_c = v^2/R$'. This formula directly relates the centripetal acceleration to the linear speed and radius of the circular path.
A car is taking a circular turn on a horizontal road. What force provides the necessary centripetal force for this motion?
The NCERT text explicitly states, 'For a car taking a circular turn on a horizontal road, the centripetal force is the force of friction.' It further clarifies that 'The centripetal force required for circular motion is along the surface of the road, and is provided by the component of the contact force between road and the car tyres along the surface. This by definition is the frictional force.'
A cyclist completes 7 revolutions in 100 s in a circular groove of radius 12 cm. What is the angular speed of the cyclist?
Given R = 12 cm. The number of revolutions is 7 in 100 s. So, frequency $\nu = 7/100$ Hz. The angular speed $\omega$ is given by $\omega = 2\pi\nu$. Therefore, $\omega = 2\pi \times (7/100) = 14\pi/100 \approx 0.44$ rad/s.
What is the relationship between the linear speed (v) and angular speed ($\omega$) of a particle in circular motion with radius (r)?
The NCERT text states, 'We know from our study of circular motion that the magnitude of linear velocity v of a particle moving in a circle is related to the angular velocity of the particle $\omega$ by the simple relation $v = \omega r$, where r is the radius of the circle.'
For a car on a level road, what is the maximum speed ($v_{max}$) possible for circular motion without skidding, given the coefficient of static friction ($\mu_s$) and radius of the turn (R)?
The NCERT text provides the formula for the maximum speed on a level road as '$v_{max} = \sqrt{\mu_s R g}$'.
Which of the following forces provides the centripetal force for a planet orbiting the Sun?
As per the NCERT text, 'The centripetal force for motion of a planet around the sun is the gravitational force on the planet due to the sun.'
The kinematic equations for uniform acceleration do not apply to uniform circular motion because:
The 'POINTS TO PONDER' section in NCERT clearly states, 'The kinematic equations for uniform acceleration do not apply to the case of uniform circular motion since in this case the magnitude of acceleration is constant but its direction is changing.'
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