Physics MCQs for NEET — Practice Questions with Answers

The introductory paragraph explains elasticity: 'The property of a body, by virtue of which it tends to regain its original size and shape when the applied force is removed, is known as elasticity.' Materials like steel are excellent for springs because they possess high elasticity and exhibit a well-defined elastic region, ensuring they return to their original form.

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.

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.

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.'

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.

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.'

The NCERT text provides the formula for the maximum speed on a level road as '$v_{max} = \sqrt{\mu_s R g}$'.

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 '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.'

From the NCERT text, we have $v = 2\pi R \nu$. Substituting this into $a_c = v^2/R$, we get $a_c = (2\pi R \nu)^2 / R = 4\pi^2 R^2 \nu^2 / R = 4\pi^2 \nu^2 R$.