NEET Circular Motion MCQs & PYQs

Q: When milk is churned, cream gets separated due to: (1999)

Concept: Understanding apparent forces in a non-inertial rotating frame of reference. Solution: When milk is churned, the denser skim milk experiences a larger centripetal force and moves towards the outer edge, while the less dense cream moves towards the center of rotation due to centrifugal force. Thus, they separate.

Practice NEET Circular Motion Questions

Question 1:

easy

A car of mass $1000\text{ kg}$ negotiates a banked curve of radius $90\text{ m}$ on a frictionless road. If the banking angle is $45^\circ$, the speed of the car is:

(2012 Pre)

For an ideal frictionless banked curve, the optimum speed is given by $v = \sqrt{gRtan\theta}$. Given R = 90 m, $\theta = 45^\circ$, and assuming $g = 10\text{ m/s}^2$, $v = \sqrt{10 \times 90 \times tan 45^\circ} = \sqrt{900 \times 1} = 30\text{ m/s}$.

💬 Discuss this Question

Question 2:

easy

A roller coaster is designed such that riders experience “weightlessness” as they go round the top of a hill whose radius of curvature is $20\text{ m}$. The speed of the car at the top of the hill is between:

(2008)

At the top of a hill, "weightlessness" implies the normal force is zero. The centripetal force is provided solely by gravity: (mg = \frac{mv^2}{R}). Thus, (v = \sqrt{gR}). Using (g = 9.8\text{ m/s}^2) and (R = 20\text{ m}), (v = \sqrt{9.8 \times 20} = \sqrt{196} = 14\text{ m/s}). This speed lies between $14\text{ m/s}$ and $15\text{ m/s}$.

💬 Discuss this Question

Question 3:

moderate

A tube of length $L$ is filled completely with an incompressible liquid of mass $M$ and closed at both ends. The tube is then rotated in a horizontal plane about one of its ends with a uniform angular velocity $omega$. The force exerted by the liquid at the other end is:

(2006)

Concept: Centrifugal force in a rotating system.
Formula: The force $F$ is the integral of centrifugal force elements $dF = dm \cdot r \cdot \omega^2$ from $0$ to $L$. Here, $dm = (M/L)dr$.
Solution: $F = \int_0^L \frac{M}{L} \omega^2 r dr = \frac{M\omega^2}{L} \left[\frac{r^2}{2}\right]_0^L = \frac{ML\omega^2}{2}$.

💬 Discuss this Question

Question 4:

moderate

A stone tied to the end of a string of $1 \text{ m}$ long is whirled in a horizontal circle with a constant speed. If the stone makes 22 revolutions in 44 seconds, what is the magnitude and direction of acceleration of the stone?

(2005)

Concept: Centripetal acceleration in uniform circular motion.
Formula: $a = \omega^2 r$, where $\omega = 2\pi f$ and $f$ is frequency.
Solution: Frequency $f = 22 \text{ rev}/44 \text{ s} = 0.5 \text{ Hz}$. Angular velocity $omega = 2\pi(0.5) = \pi \text{ rad/s}$. Radius $r = 1 \text{ m}$. So, $a = (\pi)^2 (1) = \pi^2 \text{ m/s}^2$. Direction is always towards the centre.

💬 Discuss this Question

Question 5:

easy

A particle of mass $m$ is tied to a string of length $l$ and whirled into a horizontal plane. If tension in the string is $T$ then the speed of the particle will be:

(1999)

Concept: Centripetal force is provided by the tension in the string.
Formula: Centripetal force $F_c = \frac{mv^2}{l}$. Here, $F_c = T$.
Solution: $T = \frac{mv^2}{l}$. Rearranging for $v$, we get $v^2 = \frac{Tl}{m}$, so $v = \sqrt{\frac{Tl}{m}}$.

💬 Discuss this Question

Question 6:

easy

When milk is churned, cream gets separated due to:

(1999)

Concept: Understanding apparent forces in a non-inertial rotating frame of reference.
Solution: When milk is churned, the denser skim milk experiences a larger centripetal force and moves towards the outer edge, while the less dense cream moves towards the center of rotation due to centrifugal force. Thus, they separate.

💬 Discuss this Question

Question 7:

easy

When a body moves with a constant speed along a circle:

(1994)

In uniform circular motion, the centripetal force is always perpendicular to the instantaneous displacement. Work $W = F.d cos\theta$. Since $\theta = 90^{\circ}$, $cos\theta = 0$. Therefore, $W = 0$.

💬 Discuss this Question

Question 8:

moderate

A point mass ‘$m$’ is moved in a vertical circle of radius ‘$r$’ with the help of a string. The velocity of the mass is $\sqrt{7gr}$ at the lowest point. The tension in the string at the lowest point is

(2020-Covid)

Tension at the lowest point is given by $T = mg + \frac{mv^2}{r} = mg + \frac{m(\sqrt{7gr})^2}{r} = mg + 7mg = 8mg$.

💬 Discuss this Question

Question 9:

moderate

A mass $m$ is attached to a thin wire and whirled in a vertical circle. The wire is most likely to break when:

(2019)

Tension in the string/wire is maximum at the lowest point of the vertical circle ($T = mg + \frac{mv^2}{r}$), making it most likely to break there.

💬 Discuss this Question

Question 10:

moderate

What is the minimum velocity with which a body of mass $m$ must enter a vertical loop of radius $R$ so that it can complete the loop?

(2016-I)

To complete a vertical loop, the minimum velocity required at the lowest point is $\sqrt{5gR}$ to ensure zero tension at the highest point.

💬 Discuss this Question