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NEET • PHYSICS • SYSTEM OF PARTICLES

System of particles practice · Mixed difficulty

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Showing 61-90 of 239 questions

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61 Medium
61 of 239
Analyzing +4 | -1
Two solid cylinders P and Q of same mass and same radius start rolling down a fixed inclined plane from the same height at the same time. Cylinder P has most of its mass concentrated near its surface, while Q has most of its mass concentrated near the axis. Which statement (s) is (are) correct
View answer and explanation
Correct option: D
Cylinder Q reaches the ground with larger angular speed
Option D (Cylinder Q reaches the ground with larger angular speed) is correct because cylinder Q has a smaller moment of inertia, leading to a larger angular acceleration and thus a larger angular speed at the bottom.
62 Easy
62 of 239
Application +4 | -1
A fly wheel of M.I 6×10−2 kg m26 \times {10^{ - 2}}\,kg\,{m^2} is rotating with an angular velocity of  20 rad s−1\,20\,rad\,{s^{ - 1}}. The torque required to bring it to rest in 4 s is
View answer and explanation
Correct option: D
0.3 Nm
Option D (0.3 Nm) is correct because it is the torque required to produce the necessary angular deceleration to bring the flywheel to rest in 4 seconds.
63 Medium
63 of 239
Analyzing +4 | -1
A horizontal platform is rotating with uniform angular velocity around the vertical axis passing through its centre. At some instant of time a viscous fluid of mass ‘m’ is dropped at the centre and is allowed to spread out and finally fall. The angular velocity during this period
View answer and explanation
Correct option: B
Decreases initially and increases again
Option B (Decreases initially and increases again) is correct because the moment of inertia increases initially as the fluid spreads, then decreases as the fluid falls off, affecting the angular velocity.
64 Medium
64 of 239
Application +4 | -1
Two circular rings (of same material) are of same thickness. The diameter of first ring is twice that of second. The moment of inertia of first ring as compared to that of second is
View answer and explanation
Correct option: B
8 times
Option B (8 times) is correct because the moment of inertia is proportional to the mass and the square of the radius, and the mass is proportional to the area.
65 Medium
65 of 239
Analyzing +4 | -1
Assertion A Moment of inertia of a circular disc of mass MM and radius RR about X, Y−X,\,Y - axes (passing
through its plane) and Z−Z - axis which is perpendicular to its plane were found to be lx  ,   ly{l_x}\,\,,\,\,\,{l_y} and lz{l_z}
respectively. The respective radii of gyration about all the three axes will be the same.
Reason R A rigid body making rotational motion has fixed mass and shape.
In the light of the above statements, choose the most appropriate answer from the options given below.
View answer and explanation
Correct option: D
A is not correct but R is correct.
Option D (A is not correct but R is correct) is correct because the radii of gyration about different axes of a circular disc are not the same, but the statement about a rigid body having fixed mass and shape is correct.
66 Medium
66 of 239
Application +4 | -1
The distance b/w the carbon atom and the oxygen atom in a carbon monoxide molecule is 1.1A01.1{A^0}. Given mass of carbon atom is 12.a.m.u., and mass of oxygen atom is 16.a.m.u., calculate the position of the centre of mass of the carbon monoxide molecule
View answer and explanation
Correct option: C
0.63A00.63{A^0} from the carbon atom
Option C (0.63A00.63{A^0} from the carbon atom) is correct because it is calculated using the center of mass formula, considering the masses and distance between the carbon and oxygen atoms.
67 Medium
67 of 239
Remembering +4 | -1
Angular momentum is
View answer and explanation
Correct option: A
Moment of momentum
Option A (Moment of momentum) is correct because angular momentum is defined as the moment of momentum.
68 Medium
68 of 239
Understanding +4 | -1
A stone of mass m tied to string of length ll is rotating along a circular path with constant speed v. The torque on the stone is
View answer and explanation
Correct option: D
Zero
Option D (Zero) is correct because the force is always radial, so the torque (r⃗×F⃗\vec{r} \times \vec{F}) is zero.
69 Medium
69 of 239
Analyzing +4 | -1
A thin horizontal circular disc is rotating about a vertical axis passing through its centre. An insect is at rest at a point near the rim of the disc. The insect now moves along a diameter of the disc to reach its other end. During the journey of the insect, the angular speed of the disc
View answer and explanation
Correct option: C
first increases and then decreases
Option C (first increases and then decreases) is correct because as the insect moves towards the center, the moment of inertia of the system decreases, causing the angular speed to increase to conserve angular momentum. As the insect moves away from the center, the moment of inertia increases, causing the angular speed to decrease.
70 Medium
70 of 239
Application +4 | -1
A particle moves with constant speed v along a circular path of radius r and completes the circle in time T. The acceleration of the particle is
View answer and explanation
Correct option: A
2πv/T2\pi v/T
Option A (2πv/T) is correct because the acceleration of a particle in circular motion is given by v²/r, and v = 2πr/T.
71 Medium
71 of 239
Analyzing +4 | -1
A satellite of mass MM is in a circular orbit of radius RR about the centre of the earth. A meteorite of the same mass falling towards the earth collides with the satellite completely inelastically. The speeds of the satellite and the meteorite are the same just before the collision. The subsequent motion of the combined body will be
View answer and explanation
Correct option: B
In an elliptical orbit
Option B (In an elliptical orbit) is correct because the collision changes the velocity vector, and since the collision is inelastic, energy is lost, resulting in an elliptical orbit.
72 Hard
72 of 239
Application +4 | -1
A sphere of mass M{M} and radius R{R} is released from the top of an inclined plane of inclination θ\theta. The minimum coefficient of friction between the plane and the sphere so that it rolls down the plane without sliding is given by
View answer and explanation
Correct option: D
μ=27tan⁡θ\mu=\frac{2}{7} \tan \theta
Option D (μ=27tan⁡θ\mu=\frac{2}{7} \tan \theta) is correct because it represents the minimum coefficient of friction required for a sphere to roll down an inclined plane without sliding.
73 Easy
73 of 239
Application +4 | -1
If RR is the radius of gyration of a body of mass MM and radius rr, then the ratio of its rotational to translational kinetic energy in the rolling condition is
View answer and explanation
Correct option: B
R2r2\frac{R^2}{r^2}
Option B (R2r2\frac{R^2}{r^2}) is correct. The ratio of rotational to translational kinetic energy is derived from the definition of radius of gyration.
74 Medium
74 of 239
Remembering +4 | -1
Angular momentum is
View answer and explanation
Correct option: A
Axial vector
Option A (Axial vector) is correct because angular momentum is an axial vector, meaning it's direction is along the axis of rotation.
75 Easy
75 of 239
Analyzing +4 | -1
A man standing on a platform holds weights in his outstretched arms. The system is rotated about a central vertical axis. If the man now pulls the weights inwards close to his body, then
View answer and explanation
Correct option: D
All of the above
Option D (All of the above) is correct because decreasing the radius increases angular velocity and kinetic energy while conserving angular momentum.
76 Easy
76 of 239
Application +4 | -1
A Mass of mass 40 kg in standing on a uniform planck of mass 60 kg lying on horizontal frictionless ice. The man walks from one end to the other end of the plank. The distance walked by the man relative to ice (given length of plank = 5 m)
View answer and explanation
Correct option: B
3 m
Option B is correct. Using the concept that the center of mass of the man-plank system remains constant, the distance walked by the man relative to the ice is calculated to be 3 m.
77 Medium
77 of 239
Understanding +4 | -1
The motion of the centre of mass is the result of
View answer and explanation
Correct option: B
External forces
Option B (External forces) is correct because the motion of the center of mass is solely determined by the net external force acting on the system.
78 Easy
78 of 239
Application +4 | -1
A ring and a disc of same mass roll without slipping along a horizontal surface with same velocity. If the K.E. of ring is 4J, then that of disc is
View answer and explanation
Correct option: B
3 J
Option B (3 J) is correct. The kinetic energy of a rolling object is the sum of translational and rotational kinetic energies. For a ring, KE=mv2KE = mv^2, and for a disc, KE=(3/4)mv2KE = (3/4)mv^2.
79 Medium
79 of 239
Analyzing +4 | -1
A solid homogeneous sphere is moving on a rough horizontal surface partly rolling and partly sliding. During this kind of motion of the sphere
View answer and explanation
Correct option: B
The angular momentum of the sphere about the point of contact with the plane is conserved
Option B (The angular momentum of the sphere about the point of contact with the plane is conserved) is correct because there is no net torque about the point of contact.
80 Medium
80 of 239
Application +4 | -1
A body is whirled in a horizontal circle of radius 20 cm. It has angular velocity of 10 rad/s. What is its linear velocity at any point on circular path
View answer and explanation
Correct option: B
2 m/s
Option B (2 m/s) is correct because linear velocity is the product of the radius and the angular velocity.
81 Medium
81 of 239
Remembering +4 | -1
The direction of the angular velocity vector is along
View answer and explanation
Correct option: D
The axis of rotation
Option D (The axis of rotation) is correct because the angular velocity vector points along the axis of rotation, according to the right-hand rule.
82 Medium
82 of 239
Application +4 | -1
The second’s hand of a watch has length 6 cm. Speed of end point and magnitude of difference of velocities at two perpendicular positions will be
View answer and explanation
Correct option: D
6.28 and 8.88 mm/s
Option D (6.28 and 8.88 mm/s) is correct because it correctly calculates the speed of the endpoint and the magnitude of the difference in velocities at two perpendicular positions.
83 Easy
83 of 239
Application +4 | -1
A stationary wheel starts rotating about its own axis at uniform angular acceleration 16 rad/s216\,rad/{s^2}. The time taken by it to complete 77 rotations is
View answer and explanation
Correct option: A
5.5 sec
Option A (5.5 sec) is correct because it is calculated using the rotational kinematic equation θ=ω0t+12αt2\theta = \omega_0 t + \frac{1}{2} \alpha t^2, where ω0=0\omega_0 = 0 and θ=77×2π\theta = 77 \times 2\pi radians.
84 Medium
84 of 239
Application +4 | -1
The angular velocity of a wheel is 70 rad/sec. If the radius of the wheel is 0.5 m, then linear velocity of the wheel is
View answer and explanation
Correct option: B
35 m/s
Option B (35 m/s) is correct because the linear velocity is calculated by multiplying the angular velocity by the radius of the wheel.
85 Easy
85 of 239
Application +4 | -1
Two particles whose masses are 10 kg10 \mathrm{~kg} and 30 kg30 \mathrm{~kg} and their position vectors are i^+j^+k^\hat{i}+\hat{j}+\hat{k} and −i^−j^−k^-\hat{i}-\hat{j}-\hat{k} respectively would have the centre of mass at
View answer and explanation
Correct option: A
−(i^+j^+k^)2-\frac{(\hat{i}+\hat{j}+\hat{k})}{2}
Option A (−(i^+j^+k^)2-\frac{(\hat{i}+\hat{j}+\hat{k})}{2}) is correct because it is the result of applying the center of mass formula with the given masses and position vectors.
86 Easy
86 of 239
Analyzing +4 | -1
When a body is projected at an angle with the horizontal in the uniform gravitational field of the earth, the angular momentum of the body about the point of projection, as it proceeds along the path
View answer and explanation
Correct option: B
Increases continuously
Option B (Increases continuously) is correct because the torque due to gravity increases as the projectile moves along its path, leading to an increase in angular momentum.
87 Medium
87 of 239
Analyzing +4 | -1
A solid sphere is rotating in free space. If the radius of sphere is increased keeping mass same which one of the following will not be affected?
View answer and explanation
Correct option: B
Angular momentum
Option B (Angular momentum) is correct because in free space, no external torque acts on the sphere, so its angular momentum remains constant.
88 Easy
88 of 239
Application +4 | -1
Two particles each of mass 1 kg1 {~kg} are located at the point having co-ordinates (1,0,1)(1,0,1) and (1,1,0)(1,1,0). The co-ordinates of the centre of mass will be
View answer and explanation
Correct option: B
(1,12,12)\left(1, \frac{1}{2}, \frac{1}{2}\right)
Option B ((1, 1/2, 1/2)) is correct because the center of mass is calculated by averaging the coordinates of the particles, weighted by their masses.
89 Hard
89 of 239
Analyzing +4 | -1
A uniform cubical box of side aa is placed on a rough floor and it is to be moved by applying minimum possible force FF at a point which is at a distance bb above its centre of mass. If the coeflicient of friction is μ=0.4\mu=0.4, then what is the maximum possible value of 100×ba100 \times \frac{b}{a} for which the box will not topple before moving?
View answer and explanation
Correct option: A
50
Option A is correct. The box will not topple if the torque due to the applied force is less than or equal to the torque due to the weight of the box. This condition leads to the maximum possible value of b/a being 0.5, hence 100 * (b/a) = 50.
90 Medium
90 of 239
Analyzing +4 | -1
Statement-I: A coin is placed on phonogram turn table. The motor is started, coin moves along the moving table.
Statement-II: The rotating table is providing necessary centripetal force to the coin.
View answer and explanation
Correct option: D
Both statements are false
Option D (Both statements are false) is correct because the coin moves due to friction (not centripetal force), and the table doesn't necessarily provide the required centripetal force.
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