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NEET Practice Questions (MCQs) with Answers & Solutions

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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?

A. $a_c = vR$ B. $a_c = v^2R$ C. $a_c = v^2/R$ D. $a_c = v/R$

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Explanation

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.

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A car is taking a circular turn on a horizontal road. What force provides the necessary centripetal force for this motion?

A. Gravitational force B. Normal reaction force C. Frictional force D. Tension in the string

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Explanation

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.'

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A cyclist completes 7 revolutions in 100 s in a circular groove of radius 12 cm. What is the angular speed of the cyclist?

A. 07 rad/s B. 14 rad/s C. 44 rad/s D. 88 rad/s

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Explanation

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.

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What is the relationship between the linear speed (v) and angular speed ($\omega$) of a particle in circular motion with radius (r)?

A. $v = \omega^2 r$ B. $v = r/\omega$ C. $v = \omega r$ D. $\omega = vr$

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Explanation

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.'

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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)?

A. $v_{max} = \sqrt{\mu_s g / R}$ B. $v_{max} = \mu_s g R$ C. $v_{max} = \sqrt{\mu_s R g}$ D. $v_{max} = \sqrt{R g / \mu_s}$

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Explanation

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

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Which of the following forces provides the centripetal force for a planet orbiting the Sun?

A. Frictional force B. Normal force C. Gravitational force D. Tension

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Explanation

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.'

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The kinematic equations for uniform acceleration do not apply to uniform circular motion because:

A. The speed is constant in uniform circular motion. B. The path is circular, not straight. C. The magnitude of acceleration is constant but its direction is changing. D. There is no net force acting on the object.

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Explanation

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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In uniform circular motion, if 'T' is the time period, 'R' is the radius, and '$\nu$' is the frequency, which of the following expressions for centripetal acceleration ($a_c$) is correct?

A. $a_c = 2\pi R / T$ B. $a_c = vT / R$ C. $a_c = 4\pi^2 \nu^2 R$ D. $a_c = v^2 T / (2\pi R)$

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Explanation

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

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What is the optimum speed ($v_o$) for a car on a banked road of radius R and banking angle $\theta$ to avoid wear and tear on its tires?

A. $v_o = \sqrt{Rg / \tan\theta}$ B. $v_o = R g \tan\theta$ C. $v_o = \sqrt{R g \tan\theta}$ D. $v_o = \sqrt{R g \mu_s \tan\theta}$

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Explanation

The NCERT text states that at optimum speed, frictional force is not needed at all. The formula given is '$v_o = (R g \tan\theta)^{1/2}$'.

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A car is moving on a banked road. For the optimum speed, what is the role of frictional force?

A. Frictional force provides the entire centripetal force. B. Frictional force acts up the slope. C. Frictional force acts down the slope. D. Frictional force is not needed at all.

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Explanation

According to the NCERT text, 'At this speed [optimum speed], frictional force is not needed at all to provide the necessary centripetal force. Driving at this speed on a banked road will cause little wear and tear of the tyres.'

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