Question 1

A satellite moving through a region of free space collects interplanetary dust that is initially at rest. If the dust is gathered at the rate dM/dt = αv, then the satellite’s acceleration is

Accepted Answer

-αv²/M. The correct option is right because the satellite's acceleration is derived from the change in momentum due to the mass being collected at a rate proportional to its velocity. As the satellite gathers mass, the force exerted by the collected dust results in a negative acceleration, which is accurately represented by the equation -αv²/M, where M is the mass of the satellite.

Question 2

A block of mass m lies at rest on a frictionless horizontal floor. A uniform rope of mass m/3 has one end attached to the block, and the free end is pulled horizontally by a force F. What is the tension at the midpoint of the rope?

Accepted Answer

7F/8. Total mass = m + m/3 = 4m/3, so a = F/(4m/3) = 3F/(4m). Mass behind the rope's midpoint = block + half the rope = m + m/6 = 7m/6. Tension there = (7m/6)*a = (7m/6)(3F/4m) = 7F/8.

Question 3

A block of mass m is attached to a second block of mass M by a light spring of spring constant k. Both blocks lie on a frictionless horizontal surface. At the beginning, the system is at rest and the spring is at its natural length. A constant force F is then applied to the block of mass M

Accepted Answer

mF/(m + M). The whole system accelerates at a = F/(m+M). The only horizontal force on block m is the spring force, which must equal m*a = mF/(m+M).

Question 4

A thin ring of mass M and radius R is spinning about its symmetry axis with angular speed ω. Two identical small masses, each of mass m, are then lightly fixed at the two opposite ends of a diameter of the ring. The decrease in kinetic energy is

Accepted Answer

(b) Mm/(M+2m) × ω² R². The correct option reflects the change in the moment of inertia of the system when the two small masses are added to the ring, which affects the total kinetic energy. The decrease in kinetic energy is calculated by considering the initial and final moments of inertia and their corresponding angular velocities, leading to the expression given in option (b).

Question 5

A sphere rolls on a surface without slipping. Its radius of gyration about an axis through its centre of mass is K, and its radius is R. What fraction of the total kinetic energy is due to rotation?

Accepted Answer

K²/(K² + R²). With I = mK^2 and v = wR, KE_rot/KE_total = (1/2 I w^2)/((1/2 I w^2)+(1/2 m v^2)) = K^2/(K^2 + R^2).

Question 6

A uniform circular ring has mass M and radius R. If a 90° arc of the ring is cut away, the moment of inertia of the remaining portion about an axis through the centre of the ring and perpendicular to its plane becomes kMR². What is the value of k?

Accepted Answer

3/4. When a 90° arc is removed from the uniform circular ring, the remaining portion consists of three-quarters of the original mass and retains the same radius. The moment of inertia of the full ring is MR², and removing the arc reduces the moment of inertia proportionally, resulting in a new moment of inertia of (3/4)MR².

Question 7

A uniform rod of length l can rotate freely in a vertical plane about a fixed horizontal axis passing through O. Starting from rest in its unstable equilibrium position, the rod turns through an angle θ. Its angular speed ω is expressed by

Accepted Answer

√(6g/l) sin(θ/2). With I = ml^2/3 about the end and the centre of mass falling (l/2)(1-cos th) from the inverted position: (1/2)(ml^2/3)w^2 = mg(l/2)(1-cos th). Thus w^2 = (3g/l)(1-cos th) = (6g/l)sin^2(th/2), so w = sqrt(6g/l) sin(th/2).

Question 8

A uniform rod PQ of length l is rotating in a horizontal plane about the axis YY'. If the rod has cross-sectional area A, density ρ, and angular speed ω, then its rotational kinetic energy is

Accepted Answer

(1/24) A l³ ρ ω². For rotation about the perpendicular axis YY' through the rod's centre, I = (1/12) M l^2 with M = A l rho. KE = (1/2) I w^2 = (1/2)(1/12)(A l rho) l^2 w^2 = (1/24) A l^3 rho w^2.

Question 9

A solid sphere of mass 2 kg is rolling on a frictionless horizontal floor with a speed of 10 m/s. It then moves up a frictionless inclined plane making an angle of 30° with the horizontal. The maximum vertical height reached by the sphere before coming to rest is

Accepted Answer

7.1 m. A solid sphere rolling without slipping has total KE = (7/10) m v^2 (translational + rotational). Setting (7/10) m v^2 = m g h gives h = 7 v^2/(10 g) = 7*100/98 ~ 7.14 m, i.e. about 7.1 m.

Question 10

What type of motion is produced by a couple?

Accepted Answer

Only rotational motion. A couple consists of two equal and opposite forces, so the net force is zero (no translation) but the net torque is nonzero, producing only rotational motion.

Question 11

Four point masses of 1 kg, 2 kg, 3 kg and 4 kg are placed at the coordinates (0, 0), (2, 0), (0, 3) and (−2, −2), respectively. What is the moment of inertia of this arrangement about the x-axis?

Accepted Answer

43 kgm². The moment of inertia about the x-axis is calculated by summing the products of each mass and the square of its perpendicular distance from the x-axis. For the given masses and their coordinates, the distances are 0, 3, 2, and 2, leading to a total moment of inertia of 43 kgm².

Question 12

A disc of radius R moves by rolling without slipping on a level surface with speed v0. Let A be a point on the rim of the disc. The acceleration of point A is

Accepted Answer

constant in magnitude. For rolling without slipping with the centre moving at constant v0, any rim point has only centripetal acceleration toward the centre of magnitude omega^2 R = v0^2/R. This magnitude is constant, but the direction continuously points from the rim point to the centre and so rotates. Thus the acceleration is constant in magnitude only.

Question 13

A rigid body has a moment of inertia of 1.2 kg m² about a specified axis and is initially stationary. If it is to acquire a rotational kinetic energy of 1500 J by applying a constant angular acceleration of 25 rad s⁻² about that axis, for how long should the acceleration act?

Accepted Answer

2 s. KE = (1/2)I w^2 = 1500 gives w^2 = 2500, w = 50 rad/s. With alpha = 25, t = w/alpha = 50/25 = 2 s.

Question 14

A pulley with radius 2 m is acted upon tangentially by a force F = (20t − 5t²) N, where t is in seconds. If the pulley’s moment of inertia about its rotation axis is 10 kg m², then the number of revolutions completed before the pulley reverses its direction of rotation is

Accepted Answer

greater than 3 but less than 6. Torque = F*r = (20t-5t^2)*2; alpha = torque/I = (40t-10t^2)/10 = 4t - t^2. omega = 2t^2 - t^3/3, which returns to zero at t=6 s. theta = integral omega dt = 2t^3/3 - t^4/12 = 144 - 108 = 36 rad at t=6. Revolutions = 36/(2pi) = 5.7, i.e. greater than 3 but less than 6.

Question 15

A gymnast is spinning with her arms and legs fully extended. When she draws her arms and legs inward, what happens?

Accepted Answer

Her moment of inertia decreases. Bringing arms and legs inward reduces the distribution of mass about the spin axis, so the moment of inertia decreases. With angular momentum conserved, the angular velocity correspondingly increases.

Question 16

Assuming the Earth is a uniform solid sphere with radius R and mass M, what is its angular momentum about its rotation axis if its period of rotation is T?

Accepted Answer

4πMR²/5T. For a uniform solid sphere, I = (2/5)MR^2 and omega = 2pi/T. Angular momentum L = I*omega = (2/5)MR^2*(2pi/T) = 4 pi M R^2/(5T).

Question 17

A hollow circular tube has a mean radius of 8 cm and a wall thickness of 0.04 cm. It is melted and remade into a solid cylindrical rod of the same length. What is the ratio of the torsional rigidities of the tube to the rod?

Accepted Answer

((8.02)⁴-(7.98)⁴)/(0.8)⁴. Tube J ~ (Ro^4 - Ri^4) = (8.02^4 - 7.98^4). Equal-volume solid rod: 2*pi*R*t = pi*r^2 -> r^2 = 2*8*0.04 = 0.64 -> r = 0.8, so J_rod ~ 0.8^4. Ratio = ((8.02)^4-(7.98)^4)/(0.8)^4.

Question 18

A particle is in free fall close to the Earth’s surface. Let O be a fixed point on the ground that lies away from the vertical line through the particle. Which of the following statements is incorrect?

Accepted Answer

The magnitude of the torque due to gravity about O is decreasing. The torque due to gravity about point O depends on the perpendicular distance from O to the line of action of the gravitational force. As the particle falls, this distance decreases, leading to a reduction in the magnitude of the torque.

Question 19

Initial angular velocity of a circular disc of mass M is ω1. Then two small spheres of mass m are attached gently to diametrically opposite points on the edge of the disc. What is the final angular velocity of the disc?

Accepted Answer

(M/(M + 4m)) ω1. The final angular velocity is derived from the conservation of angular momentum, where the initial angular momentum of the disc must equal the final angular momentum after the spheres are added. Since the spheres are added at diametrically opposite points, their contribution to the moment of inertia effectively increases the total moment of inertia of the system, leading to a decrease in angular velocity.

Question 20

Two identical particles move towards each other with velocity 2v and v respectively. Velocity of centre of mass is

Accepted Answer

v/2. The velocity of the center of mass is calculated using the formula v_cm = (m1*v1 + m2*v2) / (m1 + m2). Since the particles are identical, their masses are equal, and the center of mass velocity is the average of their velocities, which results in v/2.

Question 21

Moment of inertia of a circular wire of mass M and radius R about its diameter is

Accepted Answer

MR²/2. The moment of inertia of a circular wire about its diameter is derived from the integral of mass distribution, which results in the formula MR²/2, indicating that the mass is evenly distributed along the circular path.

Question 22

A circular disc X of radius R is made from an iron plate of thickness t, and another disc Y of radius 4R is made from an iron plate of thickness t/4. Then the relation between the moment of inertia Iₓ and I_y is

Accepted Answer

I_y = 64 Iₓ. The moment of inertia for a disc is proportional to its mass and the square of its radius. Disc Y, having a radius 4R and a thickness t/4, has a mass that scales with the volume, resulting in a factor of 16 increase due to the radius squared and an additional factor of 4 due to the thickness, leading to a total increase of 64 times the moment of inertia of disc X.

Question 23

A particle performing uniform circular motion has angular frequency ω doubled & its kinetic energy halved, then the new angular momentum is

Accepted Answer

L/4. In uniform circular motion, angular momentum (L) is proportional to angular frequency (ω) and the moment of inertia (I). If the angular frequency is doubled (2ω) and the kinetic energy is halved, the relationship between kinetic energy and angular momentum shows that the new angular momentum is reduced to a quarter of the original, resulting in L/4.

Question 24

A body A of mass M is descending vertically under gravity and splits into two fragments: B of mass M/3 and C of mass 2M/3. Relative to the centre of mass of the original body A, the centre of mass of the two fragments together moves toward

Accepted Answer

no shift occurs. Splitting is due to internal forces; the only external force is gravity, which acts identically on the combined fragments. The centre of mass of B and C continues along the original trajectory of A, so there is no shift relative to it.

Question 25

A uniform semicircular lamina has mass M and radius r. Its moment of inertia about an axis perpendicular to its plane and passing through the centre is

Accepted Answer

1/2 Mr². For a full disc I = (1/2)MR^2. A semicircular lamina is half the disc with half the mass, so the ratio I/M is unchanged: I = (1/2)Mr^2 about the perpendicular axis through the centre.

Question 26

A system consists of two particles of masses m1 and m2. If the first particle is displaced toward the centre of mass by a distance d, how far must the second particle be shifted to keep the centre of mass fixed at its original position?

Accepted Answer

(m1/m2) d. To hold the centre of mass fixed the mass-weighted displacements must cancel: m1*d = m2*d2, so the second particle must move (in the opposite sense) by d2 = (m1/m2) d.

Question 27

Four identical point masses, each of mass m, are located at the four vertices of a square ABCD whose side length is ℓ. The moment of inertia of this arrangement about a line through A and parallel to the diagonal BD is

Accepted Answer

3mℓ². Place A(0,0),B(l,0),C(l,l),D(0,l). The line through A parallel to BD is x+y=0; distance of a point is |x+y|/sqrt2. Contributions: A 0, B m*l^2/2, D m*l^2/2, C m*(2l^2). Total = 3 m l^2.

Question 28

A light circular hoop of mass m and radius R spins about its symmetry axis with steady angular speed ω. Two point masses, each of mass M, are then placed gently at diametrically opposite points on the hoop. The new angular speed of the system becomes ω' =

Accepted Answer

ωm/(m + 2M). The correct option is derived from the conservation of angular momentum. Initially, the hoop has angular momentum based on its mass and angular speed. When the two point masses are added, the total mass of the system increases, and to conserve angular momentum, the new angular speed must decrease, resulting in the formula ωm/(m + 2M).

Question 29

A disc of radius R is cut out from a larger disc of radius 2R in such a way that the two circumferences touch each other. The centre of mass of the remaining lamina lies at a distance αR from the centre of the larger disc. What is the value of α?

Accepted Answer

1/3. Mass scales with area: big disc (radius 2R) ~ 4m, removed disc (radius R) ~ m, with the small disc's center at distance R from the big center. COM = (4m.0 - m.R)/(4m - m) = -R/3, so the remaining lamina's COM is at R/3, i.e. alpha = 1/3.

Question 30

The angular momentum of a particle moving under a central force remains unchanged because the

Accepted Answer

torque acting on it is zero. The angular momentum of a particle remains constant when the net torque acting on it is zero, which occurs in a central force scenario where the force is directed towards a fixed point, resulting in no rotational influence.

Question 31

A light horizontal circular disc is spinning about a vertical axis through its centre. An insect initially remains at rest close to the edge of the disc. It then walks along a diameter to the opposite edge. While the insect is moving, the angular speed of the disc:

Accepted Answer

increases at first and then decreases. As the insect moves from the edge towards the center of the disc, it reduces the moment of inertia of the system, which causes the angular speed to increase due to conservation of angular momentum. However, as it reaches the opposite edge, the moment of inertia increases again, leading to a decrease in angular speed.

Question 32

A circular ring of radius r and mass m is set spinning with angular speed ω0 and then placed on a rough horizontal plane. At the instant it is released, the centre of mass has zero speed. What speed will the centre of the ring have when pure rolling begins and slipping stops?

Accepted Answer

rω0/2. Angular momentum about the contact line is conserved: mr^2*w0 = Iw + mvr = mr^2(v/r) + mvr = 2mvr, giving v = r*w0/2.

Question 33

A mass m tied to a light inextensible string of length l hangs from a fixed point. The mass moves in a horizontal circle with angular speed ω about the vertical line through the point of suspension. With respect to the point of suspension, which statement is true about its angular momentum

Accepted Answer

Its direction changes, but its magnitude remains constant.. The angular momentum of the mass remains constant in magnitude because the speed and distance from the axis of rotation do not change, but its direction changes continuously as the mass moves in a circular path.

Question 34

A solid cone has uniform density. If the distance of its centre of mass from the vertex is denoted by z0, and the cone has base radius R and height h, what is the value of z0?

Accepted Answer

3h/4. The center of mass of a solid cone is located at a distance of 3h/4 from the vertex along its axis, which accounts for the distribution of mass being denser towards the base due to the cone's shape.

Question 35

A cube of the largest possible size is carved out from a solid sphere having mass M and radius R. The moment of inertia of this cube about an axis through its centre and normal to one of its faces is:

Accepted Answer

4MR²/(9√3π). Largest inscribed cube has a = 2R/sqrt3. Cube mass = M*(a^3)/((4/3)pi R^3) = 2M/(sqrt3 pi). I about axis normal to a face = (1/6) m a^2 = (1/6)*(2M/(sqrt3 pi))*(4R^2/3) = 4MR^2/(9 sqrt3 pi).

Question 36

39. A stationary horizontal disc is free to rotate about its axis. When a torque is applied on it, its kinetic energy is a function of θ, where θ is the angle by which it has rotated, as kθ². If its moment of inertia is I then the angular acceleration of the disc is

Accepted Answer

2k/I θ. The kinetic energy of the disc is given as kθ², and using the relationship between kinetic energy and angular motion, we can derive that the angular acceleration is proportional to the torque applied, which results in the expression 2k/I θ, indicating that the angular acceleration increases with the angle of rotation.

Question 37

40. The following bodies are made to roll up without slipping the same inclined plane from a horizontal plane: (i) a ring of radius R, (ii) a solid cylinder of radius R/2 and (iii) a solid sphere of radius R/4. If, in each case, the speed of the center of mass at the bottom of the incline

Accepted Answer

14: 15: 20. Energy conservation gives h = (v^2/2g)(1 + k^2/R^2). For ring, solid cylinder and solid sphere, (1 + k^2/R^2) = 2 : 3/2 : 7/5 = 20 : 15 : 14, i.e. the heights are in the ratio 14 : 15 : 20.

Question 38

43. The radius of gyration of a uniform rod of length l, about an axis passing through a point l/4 away from the centre of the rod, and perpendicular to it, is:

Accepted Answer

√(7/48) l. By parallel axis, I = (1/12) m l^2 + m (l/4)^2 = m l^2 (1/12 + 1/16) = 7 m l^2 / 48. Radius of gyration k = sqrt(I/m) = sqrt(7/48) l.

Question 39

A couple has moment G, and the magnitude of each force in the couple is P. When the force P is rotated by 90°, the new couple formed has moment H. If the same force P is instead rotated through an angle α, the moment of the couple becomes

Accepted Answer

H sin α + G cos α. The moment of a couple is determined by the perpendicular distance between the lines of action of the forces. When one force is rotated through an angle α, the new moment can be expressed as the sum of the original moment G adjusted by the cosine of α and the new moment H adjusted by the sine of α, leading to the correct expression H sin α + G cos α.

Question 40

The magnitude of torque on a particle of mass 1 kg is 2.5 N m about the origin. If the force acting on it is 1 N, and the distance of the particle from the origin is 5m, the angle between the force and the position vector is (in radians):

Accepted Answer

π/6. The torque is given by the formula τ = r × F × sin(θ), where τ is the torque, r is the distance from the origin, F is the force, and θ is the angle between the force and the position vector. Rearranging the formula and substituting the known values, we find that the angle θ that satisfies the equation for the given torque and force is π/6 radians.

Question 41

A stationary horizontal disc is free to rotate about its axis. When a torque is applied on it, its kinetic energy as a function of θ, where θ is the angle by which it has rotated, is given as kθ². If its moment of inertia is I, then the angular acceleration of the disc is -

Accepted Answer

2k/I θ. KE = k*theta^2, so torque = d(KE)/d(theta) = 2k*theta = I*alpha. Hence angular acceleration alpha = 2k*theta/I.

Question 42

A particle of mass m moves in a circular orbit in a central potential field U(r) = 1/2 k r². If Bohr's quantization conditions are applied, radii of possible orbits and energy levels vary with quantum number n as:

Accepted Answer

rₙ ∝ √n, Eₙ ∝ n. Central force F = kr gives mv^2/r = kr so v ~ r. Quantization mvr = n*hbar with v ~ r gives r^2 ~ n, i.e. r_n ~ sqrt(n). Energy E = (1/2)mv^2 + (1/2)k r^2 ~ r^2 ~ n, so E_n ~ n. Hence r_n ~ sqrt(n), E_n ~ n.

Question 43

A Force F = (î + 2ĵ + 3k̂) N acts at a point (4î + 3ĵ - k̂) m. Then the magnitude of torque about the point (î + 2ĵ + k̂) m will be √(x) N-m. The value of x is _____.

Accepted Answer

195. The torque is calculated using the cross product of the position vector from the point of rotation to the point of force application and the force vector. After computing the cross product and finding the magnitude, we determine that the value of x is 195.

Question 44

The linear mass density of a thin rod AB of length L varies from A to B as λ(x) = λ0(1 + x/L), where x is the distance from A. If M is the mass of the rod then its moment of inertia about an axis passing through A and perpendicular to the rod is:

Accepted Answer

7ML²/18. With lambda = lambda0(1 + x/L), total mass M = integral_0^L lambda dx = 3*lambda0*L/2, so lambda0 = 2M/(3L). Moment of inertia about A: I = integral_0^L x^2*lambda dx = lambda0*(L^3/3 + L^3/4) = lambda0*7L^3/12 = 7ML^2/18.

Question 45

Particle A of mass m1 moving with velocity (√3 i + j) m s⁻¹ collides with another particle B of mass m2 which is at rest initially. Let V1 and V2 be the velocities of particles A and B after collision respectively. If m1 = 2m2 and after collision V1 = (i + √3 j) m s⁻¹, the angle between V1

Accepted Answer

105°. The angle between the velocities V1 and V2 can be determined using the conservation of momentum and the relationship between the masses. Given that m1 is twice m2 and the final velocity of A is at a specific angle, the resulting velocity of B must be oriented such that the angle between them is 105°, ensuring momentum is conserved in both the x and y directions.

Question 46

A uniform sphere of mass 500 g rolls without slipping on a plane horizontal surface with its centre moving at a speed of 5.00 cm/s. Its kinetic energy is:

Accepted Answer

8.75 × 10⁻⁴ J. For a solid sphere rolling without slipping, KE = (1/2)mv^2 + (1/2)Iw^2 = (7/10)mv^2. With m=0.5 kg and v=0.05 m/s, KE = 0.7*0.5*0.0025 = 8.75e-4 J.

Question 47

A spring mass system (mass m, spring constant k and natural length ℓ) rests in equilibrium on a horizontal disc. The free end of the spring is fixed at the centre of the disc. If the disc together with spring mass system, rotated about it's axis with an angular velocity ω, (k > mω²) the re

Accepted Answer

mω²/k. The correct option, mω²/k, represents the relationship between the centripetal force required to keep the mass in circular motion and the restoring force of the spring. Since the spring's extension is directly proportional to the force applied, this formula accurately captures how the spring stretches under the influence of the rotational motion.

Question 48

A uniformly thick wheel with moment of inertia I and radius R is free to rotate about its centre of mass (see fig.). A massless string is wrapped over its rim and two blocks of masses m1 and m2 (m1 > m2) are attached to the ends of the string. The system is released from rest. If the angul

Accepted Answer

[2(m1 − m2)gh/((m1 + m2)R² + I)]^(1/2). The correct option accounts for the net force acting on the system, which is the difference in weights of the two masses (m1 - m2), and incorporates the rotational inertia of the wheel. This results in a derived angular speed that reflects both the translational motion of the masses and the rotational motion of the wheel, ensuring energy conservation is maintained.

Question 49

A small circular loop of conducting wire has radius a and carries current I. It is placed in a uniform magnetic field B perpendicular to its plane such that when rotated slightly about its diameter and released, it starts performing simple harmonic motion of time period T. If the mass of t

Accepted Answer

T = √(2πm/IB). Restoring torque about the diameter is tau = -mu*B*sin(th) ~ -(I*pi*a^2)*B*th. With moment of inertia of the ring about a diameter = m*a^2/2, angular frequency^2 = mu*B/(m*a^2/2) = 2*pi*I*B/m. Hence T = 2*pi/sqrt(2*pi*I*B/m) = sqrt(2*pi*m/(I*B)).

Question 50

A rod of length L has non-uniform linear mass density given by ρ(x) = a + b (x/L)², where a and b are constants and 0 ≤ x ≤ L. The value of x for the centre of mass of the rod is at -

Accepted Answer

(3/4) ((2a + b)/(3a + b)) L. With rho=a+b(x/L)^2: total mass = L(3a+b)/3 and integral of x*rho = L^2(2a+b)/4. So x_cm = [L^2(2a+b)/4]/[L(3a+b)/3] = (3/4)*((2a+b)/(3a+b))*L.

JEE Main Physics: System of Particles and Rotational Motion questions with solutions

Questions