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JEE • PHYSICS • SYSTEM OF PARTICLES

System of particles practice · Hard

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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1 Hard

1 of 6

Application +4 | -1

A solid cylinder of mass 4 kg and radius 8 cm is rotating about its axis at the rate of 6 rpm. The torque required to stop after

4 \pi

revolutions is:

View answer and explanation

Correct option: D

3.2 \times 10^{-5} \mathrm{~N} \mathrm{~m}

Option D (

3.2 \times 10^{-5} \mathrm{~N} \mathrm{~m}

) is correct because it is calculated by finding the angular deceleration and then using

\tau = I\alpha

Solution approach

Correct Answer + Core Reason: $3.2 \times 10^{-5} \mathrm{~N} \mathrm{~m}$ . The torque is calculated by finding the angular deceleration and then using $\tau = I\alpha$ .

Step-by-Step Solution:

Step 1: Calculate the moment of inertia: $I = \frac{1}{2}MR^2 = \frac{1}{2} (4 \mathrm{~kg}) (0.08 \mathrm{~m})^2 = 0.0128 \mathrm{~kg} \mathrm{~m}^2$ .

Step 2: Convert the initial angular velocity from rpm to rad/s: $\omega_i = 6 \frac{\mathrm{rev}}{\mathrm{min}} \times \frac{2\pi \mathrm{rad}}{1 \mathrm{rev}} \times \frac{1 \mathrm{min}}{60 \mathrm{s}} = \frac{\pi}{5} \mathrm{rad/s}$ .

Step 3: Convert the number of revolutions to radians: $\theta = 4\pi \mathrm{rev} \times \frac{2\pi \mathrm{rad}}{1 \mathrm{rev}} = 8\pi^2 \mathrm{rad}$ .

Step 4: Use the kinematic equation to find the angular acceleration: $\omega_f^2 = \omega_i^2 + 2\alpha\theta \Rightarrow 0 = (\frac{\pi}{5})^2 + 2\alpha(8\pi^2) \Rightarrow \alpha = -\frac{1}{400} \mathrm{rad/s^2}$ .

Step 5: Calculate the torque: $\tau = I\alpha = (0.0128 \mathrm{~kg} \mathrm{~m}^2) (-\frac{1}{400} \mathrm{rad/s^2}) = -3.2 \times 10^{-5} \mathrm{~N} \mathrm{~m}$ . The magnitude of the torque is $3.2 \times 10^{-5} \mathrm{~N} \mathrm{~m}$ .

Why other options are wrong:
- **Option A ( $3.2 \times 10^{-4} \mathrm{~N} \mathrm{~m}$ ): This is incorrect because it uses an incorrect value for the angular acceleration or an incorrect formula for the torque.
- Option B ( $2 \times 10^{-6} \mathrm{~N} \mathrm{~m}$ ): This is incorrect because it uses an incorrect value for the angular acceleration or an incorrect formula for the torque.
- Option C ( $4 \times 10^{-6} \mathrm{~N} \mathrm{~m}$ ): This is incorrect because it uses an incorrect value for the angular acceleration or an incorrect formula for the torque.

Key Concepts:
- Rotational motion
- Torque
- Moment of inertia
- Angular acceleration

Memory Aids:**
- Remember the formula for torque: $\tau = I\alpha$ .

Quick Tips:
- Always convert units to SI units before substituting into the formulas.

2 Hard

2 of 6

Understanding +4 | -1

Moment of inertia of a uniform annular disc of internal radius

r

and external radis

R

and mass

M

about an axis through its centre and perpendicular to its plane is

View answer and explanation

Correct option: B

\frac{1}{2} M\left(R^2+r^2\right)

Option B (

\frac{1}{2} M(R^2+r^2)

) is correct because it represents the moment of inertia of an annular disc about an axis through its center and perpendicular to its plane.

Solution approach

Correct Answer + Core Reason: $\frac{1}{2} M(R^2+r^2)$ . This is the standard formula for the moment of inertia of an annular disc.

Step-by-Step Solution:

Step 1: Define the problem: Find the moment of inertia of a uniform annular disc with internal radius $r$ , external radius $R$ , and mass $M$ about an axis through its center and perpendicular to its plane.

Step 2: Consider an elementary ring: Consider an elementary ring of radius $x$ and thickness $dx$ .

Step 3: Calculate the surface mass density: $\sigma = \frac{M}{{\pi (R^2 - r^2)}}$

Step 4: Calculate the mass of the elementary ring: $dm = \sigma \times 2 \pi x dx = \frac{{2Mxdx}}{{(R^2 - r^2)}}$

Step 5: Calculate the moment of inertia of the elementary ring: $dI = (dm)x^2 = \frac{{2Mx^3dx}}{{(R^2 - r^2)}}$

Step 6: Integrate to find the total moment of inertia: $I = \int_r^R dI = \int_r^R {\frac{{2Mx^3dx}}{{(R^2 - r^2)}}} = \frac{{2M}}{{(R^2 - r^2)}}\left[ {\frac{{x^4}}{4}} \right]_r^R = \frac{{2M}}{{(R^2 - r^2)}}\left[ {\frac{{R^4 - r^4}}{4}} \right] = \frac{1}{2}M(R^2 + r^2)$

Why other options are wrong:
- **Option A ( $\frac{1}{2} M(R^2-r^2)$ ): This is incorrect because it involves subtraction instead of addition of the squares of the radii.
- Option C ( $\frac{M(R^4+r^4)}{2(R^2+r^2)}$ ):** This is incorrect as it has $R^4 + r^4$ in the numerator instead of $R^2 + r^2$ .
- **Option D ( $\frac{1}{2} \frac{M(R^4+r^4)}{(R^2-r^2)}$ ):** This is incorrect as it has $R^4 + r^4$ in the numerator and $R^2 - r^2$ in the denominator.

Key Concepts:
- Moment of inertia
- Annular disc
- Integration

Memory Aids:
- Remember the formula for the moment of inertia of a disc: $\frac{1}{2}MR^2$ .

Quick Tips:
- When dealing with continuous mass distributions, use integration to find the moment of inertia.

3 Hard

3 of 6

Application +4 | -1

The rotors velocity of a helicopter engine changes from

300 \,\mathrm{rev} / \mathrm{min}

110 \,\mathrm{rev} / \mathrm{min}

in 2 minutes. How long does the rotor blades take to stop

View answer and explanation

Correct option: A

3 min

Option A (3 min) is correct because it is calculated by first finding the angular acceleration and then the time it takes for the rotor to stop.

Solution approach

Correct Answer + Core Reason: 3 min. It is calculated by first finding the angular acceleration and then the time it takes for the rotor to stop.

Step-by-Step Solution:

Step 1: Understand the problem. The rotor's velocity changes from 330 rev/min to 110 rev/min in 2 minutes. We need to find the time it takes for the rotor to stop.

Step 2: Convert the given values to appropriate units. Initial angular velocity, $\omega_i = 330 \text{{ rev/min}} = \frac{{2 \pi \times 330}}{{60}} \text{{ rad/s}}$ . Final angular velocity, $\omega_f = 110 \text{{ rev/min}} = \frac{{2 \pi \times 110}}{{60}} \text{{ rad/s}}$ . Time, $t = 2 \text{{ min}} = 120 \text{{ s}}$ .

Step 3: Calculate the angular acceleration. $\alpha = \frac{{\omega_f - \omega_i}}{t} = \frac{{\frac{{2 \pi \times 110}}{{60}} - \frac{{2 \pi \times 330}}{{60}}}}{{120}} = \frac{{2 \pi (110 - 330)}}{{60 \times 120}} = \frac{{2 \pi (-220)}}{{7200}} = -\frac{{11\pi}}{{180}} \text{{ rad/s}}^2$ .

Step 4: Calculate the time it takes for the rotor to stop. Use the equation $\omega_f = \omega_i + \alpha t$ , where $\omega_f = 0$ . So, $0 = \frac{{2 \pi \times 330}}{{60}} + \left( -\frac{{11\pi}}{{180}} \right)t$ .
$t = \frac{{\frac{{2 \pi \times 330}}{{60}}}}{{\frac{{11\pi}}{{180}}}} = \frac{{2 \pi \times 330 \times 180}}{{60 \times 11\pi}} = \frac{{330 \times 3}}{{11}} = 30 \times 3 = 90 \text{{ s}} = 3 \text{{ min}}$ .

Why other options are wrong:
- Option B (4 min): This is incorrect. It might be a result of incorrect calculation of angular acceleration or using the wrong formula.
- Option C (5 min): This is incorrect. It might be a result of incorrect calculation of angular acceleration or using the wrong formula.
- Option D (6 min): This is incorrect. It might be a result of incorrect calculation of angular acceleration or using the wrong formula.

Key Concepts:
- Angular velocity
- Angular acceleration
- Rotational kinematics

Memory Aids:
- Remember the rotational kinematic equations.

Quick Tips:
- Always convert the units to be consistent before substituting into the formula.
- Pay attention to the signs of the quantities (angular velocity and angular acceleration).

4 Hard

4 of 6

Analyzing +4 | -1

Assertion (A): A disc is rolling on a rough horizontal surface without slipping. The velocity of centre of mass is

v .

Then all the other points on the disc lying on a circular arc with point of contact as the center and this arc passing through center of mass of disc will have the same speed

v.

Reason (R): When a disc is rotating without sliding on a rough horizontal surface the magnitude of velocities of all the points at a distance

r

from point of contact is same.

View answer and explanation

Correct option: A

Both (A) & (R) are true and the (R) is the correct explanation of the (A)

Option A is correct because both the assertion and reason are true, and the reason correctly explains why points on a circular arc have the same speed.

Solution approach

Correct Answer + Core Reason: Both (A) & (R) are true and the (R) is the correct explanation of the (A). The assertion is correct because all points on the arc have the same distance from the point of contact, and the reason correctly explains why this leads to the same speed.

Step-by-Step Solution:

Step 1: Understand the rolling without slipping condition.

Step 2: Recall that in rolling without slipping, the velocity of any point on the disc is the vector sum of the translational velocity of the center of mass ( $v$ ) and the rotational velocity ( $r\omega$ ), where $r$ is the distance from the center of the disc.

Step 3: Consider a circular arc centered at the point of contact and passing through the center of mass. All points on this arc are at the same distance from the point of contact.

Step 4: The magnitude of the velocity of any point at a distance $r$ from the point of contact is $v = r\omega$ . Since all points on the arc are at the same distance from the point of contact, their speeds are the same.

Step 5: The velocity of the center of mass is $v$ . Therefore, all points on the arc passing through the center of mass have the same speed $v$ .

Why other options are wrong:
- Option B: The reason is the correct explanation of the assertion.
- Option C: The reason is true, so the assertion cannot be true while the reason is false.
- Option D: Both the assertion and reason are true.

Key Concepts:
- Rolling without slipping
- Velocity of a point on a rolling disc
- Vector addition of velocities

Memory Aids:
- Rolling without slipping: $v = r\omega$

Quick Tips:
- Always consider the point of contact as the instantaneous center of rotation in rolling without slipping.

5 Hard

5 of 6

Application +4 | -1

A body rolls down an inclined plane without slipping. The kinetic energy of rotation is 50% of its translational kinetic energy. The body is

View answer and explanation

Correct option: B

Solid cylinder

Option B (Solid cylinder) is correct because the moment of inertia derived from the kinetic energy relationship matches that of a solid cylinder.

Solution approach

Correct Answer + Core Reason: Solid cylinder. The moment of inertia derived from the kinetic energy relationship matches that of a solid cylinder.

Step-by-Step Solution:

Step 1: State the given information: ${K_{rotational}} = 50\% \,\,\mathrm{of}\,\,{K_{translational}}$

Step 2: Write the expressions for rotational and translational kinetic energies:
${K_{rotational}} = \frac{1}{2}I{\omega ^2}$ and ${K_{translational}} = \frac{1}{2}m{v^2}$

Step 3: Use the given relationship: $\frac{1}{2}I{\omega ^2} = \frac{1}{2}\left( {\frac{1}{2}m{v^2}} \right)$

Step 4: Use the pure rolling condition: $v = \omega R$

Step 5: Substitute $v = \omega R$ into the kinetic energy equation:
$\frac{1}{2}I{\omega ^2} = \frac{1}{4}m{R^2}{\omega ^2}$

Step 6: Solve for the moment of inertia $I$ :
$I = \frac{1}{2}m{R^2}$

Step 7: Compare the result with known moments of inertia for different shapes. The moment of inertia $I = \frac{1}{2}m{R^2}$ corresponds to a solid cylinder.

Why other options are wrong:
- Option A: Solid sphere - The moment of inertia for a solid sphere is $\frac{2}{5}mR^2$ .
- Option C: Hollow cylinder - The moment of inertia for a hollow cylinder is $mR^2$ .
- Option D: Ring - The moment of inertia for a ring is $mR^2$ .

Key Concepts:
- Rotational Kinetic Energy
- Translational Kinetic Energy
- Moment of Inertia
- Pure Rolling Condition

Memory Aids:
- Remember the moments of inertia for common shapes (sphere, cylinder, ring).

Quick Tips:
- Relate rotational and translational motion using $v = \omega R$ for pure rolling.

6 Hard

6 of 6

Understanding +4 | -1

Which of the following statements are correct:

(a) Centre of mass of a body always coincides with the centre of gravity of the body.

(b) A couple on a body produce pure rotational motion in a body.

(c) A couple on a body produce both translational and rotational motion in a body.

(d) Mechanical advantage greater than one means that small effort can be used to lift a large load.

View answer and explanation

Correct option: B

b and d

Option B (b and d) is correct because a couple produces pure rotational motion, and a mechanical advantage greater than one allows a small effort to lift a large load.

Solution approach

Correct Answer + Core Reason: b and d. A couple produces pure rotational motion, and a mechanical advantage greater than one means a small effort can lift a large load.

Step-by-Step Solution:

Step 1: Analyze statement (a): Centre of mass of a body always coincides with the centre of gravity of the body. This is not always true. It is true when the gravitational field is uniform.

Step 2: Analyze statement (b): A couple on a body produces pure rotational motion in a body. This is true because a couple is defined as two equal and opposite forces acting on a body, creating a torque without a net force.

Step 3: Analyze statement (c): A couple on a body produces both translational and rotational motion in a body. This is false because a couple produces only rotational motion.

Step 4: Analyze statement (d): Mechanical advantage greater than one means that small effort can be used to lift a large load. This is true by the definition of mechanical advantage.

Step 5: Identify the correct statements: (b) and (d).

Why other options are wrong:
- Option A: c and d. Statement c is incorrect.
- Option C: a and b. Statement a is incorrect.
- Option D: b and c. Statement c is incorrect.

Key Concepts:
- Centre of mass and centre of gravity
- Couple and torque
- Mechanical advantage

Memory Aids:
- Remember that a couple produces pure rotation.

Quick Tips:
- Understand the definitions of centre of mass, centre of gravity, couple, and mechanical advantage.

Revision notes and key takeaways

Review SYSTEM OF PARTICLES in your textbook or class notes, then use this set to check your understanding.

Key Takeaways

✓ This set contains 6 single-correct-answer MCQs on SYSTEM OF PARTICLES.
✓ Try each question before opening its answer. Read the explanation where one is available.
✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.

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