27. Refraction at Plane surfaces and Total internal reflection
28. Refraction through prism and Dispersion of Light
29. Refraction through Lenses
30. Chromatic abberation in Lenses, Optical instruments and Human eye
31. Velocity of Light
32. Photometry
33. Wave nature of Light
34
Electrostatics
34. Charge and Force
35. Electric Field and Potential
36. Capacitance
37
Electrodynamics
37. Electric current
38. Heating Effect of Current
39. Thermoelectricity
40. Chemical effect of Current
41. Meters
42
Electromagnetism
42. Properties of Magnets
43. Magnetic effects of Current
44. Electromagnetic induction
45. Alternating current
46
Modern Physics
46. Cathode rays, Positive rays and Electrons
47. Photoelectric effect
48. X-rays
49. Atomic structure and Spectrum
50. Radioactivity
51. Nuclear physics
52. Semiconductor and Semiconductor devices
53. Diode and Triode valves
54. Logic gates
55. Relativity and Universe
56. Particle physics
57
Mechanics
9. Rotational motion
1. A fan makes 10 revolutions in 3 second which is just switched on. Considering uniform acceleration the number of revolution made by fan in next 3 second is:
2. The spokes are used in bicycle wheel to [BP 201 1]
[BP 201 1]
Increases frictional force
Decreases frictional force
Increases moment of inertia
Increase angular momentum
(c) Spokes in a bicycle wheel are located away from the center, which increases the moment of inertia of the wheel, making it more resistant to changes in rotation and providing stability.
3. A small mass of 10 gm, lies in a hemispherical bowl of radius 0.4 m at a height of 0.2 m from the bottom of the bowl. The mass will be in equilibrium of the bowl rotates at an angular speed of
4. A thin uniform rod of mass 'm' moves translationally with acceleration 'a' due to two antiparallel force of lever arm '. One force is of magnitude F and acts at one extreme enThe length of the rod is [BP 2009]
[BP 2009]
2(F + ma)/ ma
ma F + ma
(F + ma)/ 2 ma
mal ma + F
(a) Net force Fnet = F - F' = ma => F' = F - mTorque τ = Iα. Since motion is translational, α = 0, so net torque about any point is zero. Taking torque about the end where force F is applied: F' * l = 0 => (F - ma)l = 0 => l = 0 (not possible). Taking torque about center of mass: F * l/2 - F' * l/2 = 0 => F = F'. This is wrong. Let the forces be F and F'. F - F' = mTorque about center of mass: F * l/2 - F' * l/2 = 0 => F = F'. So ma = 0 which implies a=0 or m=0. This is contradiction. Let lever arm be Torque τ = Fd - F'd = Iα = 0. So F = F'. Net force F - F' = ma => ma = 0. This is incorrect. Let distance between forces be Torque = FNet force F - F' = m F' = F - ml is length. Torque about center: F(l/2) - (F-ma)(l/2) = 0 => F = F - ma => ma = 0. This is wrong. The question is flaweIf lever arm is distance between forces, F - F' = ma, Torque = FIf translational motion only, torque should be zero if forces are applied along the line. Assuming 'lever arm l' refers to length of the roThen Torque = 0. Forces are F and F-m If forces are applied at ends and rod moves translationally, Torque = F(l/2) - (F-ma)(l/2) = 0 => F = F - ma => a = 0. This is contradiction.
5. A wire of length / and mass 'm' is bent in the form of a rectangle ABCD with 2. The moment of inertia of this wire frame about the side BC is :
[BP 2009]
203 ml
7 252 ml
165 ml 252 ml
136 ml 252 ml
() Let sides be x and y. 2(x+y) = l. x = l/7. y = 3l/14. MI of BC = 0. MI of AD = m/l * (l/7) * (3l/14)² + m/l * (l/7) * (3l/14)² + m/l * (3l/14) * (l/7)² = ml (2/7 * 9l²/196 + 3/14 * l²/49) = ml³ (18/1372 + 3/686) = ml³ (18/1372 + 6/1372) = 24ml³/1372 = 6ml³/343. Option b has ml.
6. A billiard ball is hit by a cue at a height " above the center. It acquires a linear velocity Vo. Mass of the ball is m and radius is r. The angular velocity acquired by the ball is: [BP 2009)
7. A ring, a dice, solid sphere, hollow sphere are dropped from the same inclined plane of same height then which one of the following reaches the ground first [MOE 2014)
[MOE 2014)]
ring
disc
solid sphere
hollow sphere.
(c) Acceleration of a rolling body down an inclined plane a = g sinθ / (1 + I/Mr²). For solid sphere I = (2/5)Mr², a = g sinθ / (1 + 2/5) = 5/7 g sinθ. For disc I = 1/2 Mr², a = g sinθ / (1 + 1/2) = 2/3 g sinθ. For ring I = Mr², a = g sinθ / (1 + 1) = 1/2 g sinθ. For hollow sphere I = (2/3)Mr², a = g sinθ / (1 + 2/3) = 3/5 g sinθ. Solid sphere has maximum acceleration.
8. The moment of inertia of a body of mass M about a given axis is I. What is the radius of gyration? [MOE 2014)
[MOE 2014)]
I/M
IM
√I/M
√IM
(c) Moment of inertia I = Mk², where k is the radius of gyration. So, k = √(I/M).
9. The torque due to gravitational force on body about its centre of mass is: [MOE 2014)
[MOE 2014)]
a, infinite
finite
Zero
can not be measured
(c) The gravitational force acts on every particle of the body. The torque due to the gravitational force about the center of mass is zero because the weight acts through the center of mass.
10. Two forces of 2N and 4N attached at the ends of a 0.5 meter rod act vertically downwards. A third force will keep the system in equilibrium if applied at a point between the ends of the ro magnitude, direction and position of the third force will be: [MOE 2011]
[MOE 2011]
6 N downwards at 33.33 cm from 2N force.
2N downwards at 33.33 cm from 2N force.
2N upward at 16.66 cm from 4N force.
6 N upward at 16.66 cm from 4N force.
(d) For equilibrium, net force must be zero and net torque about any point must be zero. Net downward force = 2 + 4 = 6 N. So, third force must be 6 N upwards. Let the distance of the third force from the 2N force be x. Taking torque about the point where the third force is applied: 2 * x - 4 * (0.5 - x) = 0 => 2x - 2 + 4x = 0 => 6x = 2 => x = 1/3 m = 33.33 cm. This is distance from 2N force. Distance from 4N force = 0.5 - 0.33 = 0.17 m = 16.67 cm.
11. 11. Two point masses of 1 kg and 2 separated by 0.5 m constitute a system The distance of the centre of mass of the system from 1 kg mass is: [IMOE 20131
[IMOE 20131]
0.15 m
0.25 m
. 0.33 m
0.4 m
(c) Distance of center of mass from m₁ (1 kg) = m₂(r₁ + r₂) / (m₁ + m₂) = 2 * 0.5 / (1 + 2) = 1 / 3 ≈ 0.33 m.
12. 12. A circular body of mass 2 kg of radius I then of inertia about diameter is? [MOE 2011
[MOE 2011]
0.5 kg m
4 kg m'
4 kg m'
I kg m?
() Assuming the circular body is a disMoment of inertia of a disc about its diameter is I = 1/4 MR² = 1/4 * 2 * (1)² = 0.5 kg m².
13. 13. Moment of inertia doesn't depend upon [MOE 2010
[MOE 2010]
Mass
Distribution of mass
Radius
Angular velocity
(d) Moment of inertia depends on mass, distribution of mass, and the axis of rotation (which often involves radius or distance from the axis) but not on angular velocity.
14. 14. If 'M' and 'r' are respectively the mass of electrons and radius of the orbit in which the electron revolves about the nucleus, the moment of inertia of electron will be: [MOE 2009]
[MOE 2009]
Mr
MY
(c) Assuming 'r' is the radius of the orbit, the moment of inertia of a point mass (electron) about an axis is given by I = Mr².
15. 15. When a body rolls downs an inclined plane. The total potential energy of the body changes into:
[IE 2011]
Rotational K.E.
Translational K.E
Both rotational and translational K.E.
None
(c) When a body rolls down an inclined plane without slipping, its initial potential energy is converted into both translational kinetic energy (due to the motion of the center of mass) and rotational kinetic energy (due to the rotation about the center of mass).
16. 16. If no internal force is applied in a body the velocity of the centre of mass: [IOM 2013]
[IOM 2013]
Zero
Increases
Decreases
Remains constant
(d) In the absence of external forces, the velocity of the center of mass of a system remains constant (according to the law of conservation of momentum). Internal forces can change the velocities of individual parts of the system but not the velocity of the center of mass.
17. 17. The product of moment of inertia and [IOM 2013] angular acceleration gives,
[IOM 2013]]
Linear momentum.
Angular momentum.
Torque
Force.
(c) Torque (τ) is the rotational equivalent of force and is given by the product of the moment of inertia (I) and the angular acceleration (α): τ = Iα.
18. 18, A cylinder has mass "M" a length T and Radius 'R' then M.I. about own axis is:
[IOM 2012]
MR
MR
MR
() Moment of inertia of a solid cylinder about its own axis is I = 1/2 MR².
19. Two bodies of masses m, and m; move in circles of radii r, and ra respectively. If they complete the circles in equal time, the ratio of their angular speed @ [KU 2014]
[KU 2014]
my/m_
() Since they complete the circles in equal time, their time periods are equal (T₁ = T₂). Angular speed ω = 2π/T. Therefore, their angular speeds are also equal (ω₁/ω₂ = 1).
20. . A uniform heavy disc is rotating with a constant angular velocity about a vertical axis through its center. Some wax is dropped gently on the disc near to the edge. The angular velocity of the disc [KU 2012]
[KU 2012]
Decreases
Becomes zero
Increases
d Does not change
(a) When wax is dropped gently on the disc, it increases the moment of inertia of the system (disc + wax) because mass is added at a distance from the axis of rotation. Since no external torque is applied, the angular momentum (L = Iω) remains constant. Therefore, if the moment of inertia (I) increases, the angular velocity (ω) must decrease to keep the angular momentum constant.
21. A uniform metal disc of radius R lies in XY - plane and rotates with uniform angular velocity w about the Z - axis, the total induced EMF between the center and the rim of the disc is equal to; [KU 2011]
[KU 2011]
- WBR
CBR
-QBR'
5 WB'R
(d) Induced EMF e = (1/2)ωBR², where ω is the angular velocity, B is the magnetic field perpendicular to the plane of the disc, and R is the radius of the disThe options seem to have a mix of variables (W, B, R, Q). Assuming W is ω, the correct formula is (1/2)ωBR².
22. Two masses of 1 kg and 2 kg are 9 m apart and make tw mass from 1 kg mass will be at [Bangladesh 09]
[Bangladesh 09]
6 m
4 m
3n
2 m
(a) Distance of center of mass from m₁ = m₂(r₁ + r₂) / (m₁ + m₂) = 2 * 9 / (1 + 2) = 18 / 3 = 6 m.
23. 3. A uniform disc is rotating at a constant speed about a vertical axis through its centre. Some wax is gently dropped on the disc, the angular velocity of the disc[KU 091
[KU 091]
doesn't change
increases
decreases
becomes zero
(c) As explained before, adding mass to the disc increases the moment of inertia, and to conserve angular momentum, the angular velocity must decrease.
24. . A circular disc of mass m and radius r is rotating about its axis with uniform speed of v. What is its kinetic energy? [TOM 04]
[TOM 04]
mv-
my
(a) Kinetic energy of a rotating disc = 1/2 Iω², where I is moment of inertia and ω is angular velocity. I = 1/2 mr², ω = v/r. KE = 1/2 * (1/2 mr²) * (v/r)² = 1/4 mv².
25. When the size of the earth is reduced to half, mass remaining same, the time period of the earth rotation will be: [IOM 031
[IOM 031]
6 hours
24 hours
12 hours
48 hours
(a) By conservation of angular momentum, Iω = constant. (2/5)MR²(2π/T) = constant. If R' = R/2, T' = T/4 = 6 hours.
26. A rotating disc has ...., kinetic energy, i mass is M & velocity is V [IOM 98
27. A fly-wheel of mass 10 kg and radius 50 cm is rotating with constant angular speed of @ with its kinetic energy 20 Joule. The angular speed of flywheel is [MOE 066]
28. The body applied with constant torque changes the angular momentum Io to final angular momentum 41, in 3 sethen find torque [MOE 2008]
[MOE 2008]
210
(b) Torque τ = rate of change of angular momentum = (Lf - Li) / t = (4I - I) / 3 = 3I / 3 = I.
29. Kinetic energy of a body is given by 1/2 mv. Which one of the following expression is correct for the kinetic energy of the rigid body where I andw represent the moment of intertia and angular velocity of the rigid body? [MOE 2065]
[MOE 2065]
(1/2) I w
(1/2) I w
(1/2) (1 w)
(1/2) (I @)
(b) Kinetic energy of a rotating rigid body is given by KE = 1/2 Iω².
30. If a body starts from rest with angular acceleration a= 6t. What is time taken to complete 10 revolution?
2.14
2.8
(a) ω = ω₀ + ∫α dt = 0 + ∫6t dt = 3t². θ = θ₀ + ∫ω dt = 0 + ∫3t² dt = t³. 10 revolution = 20π radians. 20π = t³ => t = (20π)^(1/3) ≈ (62.8)^(1/3) ≈ 3.97 s. None of the options are close.
31. If there is a change of angular momentum from 2 J to 4 J in 4 seThen the torque is [TE-04)
32. When torque acting upon a system is zer Which of the following will be constant? [TE-051
[TE-051]
Force
Linear momentum
Angular momentum
Linear impulse
(c) If the net external torque acting on a system is zero, the total angular momentum of the system remains constant (law of conservation of angular momentum).
33. A shell at rest explodes. The centre of mass of the fragments
[IE-08, BP 2017]
Moves along the parabolic path
Moves along the straight line
Move along an elliptical path
remains at rest
(d) In the absence of external forces (like gravity acting on the shell as a whole before explosion), the velocity of the center of mass of the system remains constant. Since the shell was initially at rest, the center of mass of the fragments will also remain at rest at the initial position of the shell.
34. The moment of inertia of a disc of mass M and radius R about an axis which is tangent to the circumference of the disc and parallel to its diameter is:
[BPKIHS-08]
MR
MR?
MR?
(a) Moment of inertia about diameter = 1/4 MR². Using parallel axis theorem, moment of inertia about tangent parallel to diameter = Idiameter + M(distance)² = 1/4 MR² + M(R)² = 5/4 MR².
35. A particle of mass m and radius of gyration k is rotating with an angular acceleration o. The torque acting on it is
mk a
mk /a
ma/k
1/4mk a
(a) Torque τ = Iα = (mk²)α.
36. The centre of gravity of a body [BPKTHS-94)
[BPKTHS-94)]
Lies always outside the body
May lie whether outside or on the surface the body
Lies inside the body
Lies on the surface of the body
(b) The center of gravity of a body may lie inside, outside, or on the surface of the body, depending on its shape and mass distribution (e.g., the center of gravity of a ring lies outside the body).
37. Radius of Gyration of an uniform rod about an axis through its middle is [BPKIHS-94]
[BPKIHS-94]]
L/(3)12
L/ (8)12
L/ (12) 12
.L/ (2) 12
(c) Moment of inertia of a uniform rod about an axis through its center and perpendicular to its length is I = ML²/12. Moment of inertia I = Mk². So Mk² = ML²/12 => k = L/√12.
38. Let I, and I be the moments of inertia of two bodies of identical geometrical shape, the first made of almunium and the second of iron [BPKIHS-95]
[BPKIHS-95]
1, <12
1/ = 12
1, > 12
Information is insufficient to derive the relation between I, and Iz
(a) Moment of inertia I = ∫r²dm = ∫r²ρdV. For identical geometrical shape, I ∝ ρ (density). Density of aluminium (2710 kg/m³) is less than density of iron (7870 kg/m³). So Ialuminium < Iiron.
39. Three point masses each of mass m are placed at the corners of an equilateral triangle of side /. The moment of inertia of system about an axis along one side of the triangle is [BPKIHS-96]
[BPKIHS-96]
3ml
m/'
3/4 m/
3/2 ml
(c) Consider the axis along one side (base). Moment of inertia of the two masses on this axis is 0. Moment of inertia of the third mass at the opposite vertex = m * (height)² = m * (√3/2 l)² = 3ml²/4.
40. Ratio of the angular velocity of the earth about its axis and the hour hand of a clock is [BPKIHS 1999]
[BPKIHS 1999]
12 : 11
1:2
2:1
(d) Angular velocity of Earth ωE = 2π / 24 rad/hr. Angular velocity of hour hand ωH = 2π / 12 rad/hr. Ratio ωE / ωH = (2π / 24) / (2π / 12) = 12 / 24 = 1/2.
41. If the radius of the earth's orbit is made one fourth, the duration of year will become [BPKIHS 2000]
[BPKIHS 2000]
1/2 times
1/4 times
1/8 times
1/16 times
(c) According to Kepler's third law, T² ∝ r³. If radius becomes r/4, then T'² ∝ (r/4)³ = r³/64 ∝ T²/64. So T' = T/8.
42. The moment of inertia of a circular ring of mass M and radius R about its diameter is
MR
MR 2
MR
MR?
(b) The moment of inertia of a circular ring of mass M and radius R about an axis perpendicular to its plane and passing through its center is MR². By perpendicular axis theorem, the moment of inertia about diameter is half of this, i.e., MR²/2.
43. The moment of inertia of a thin rod of mass M, length L, about an axis passing through a point from one end and perpendicular to length is
TO ML?
ML
ML
(c) Moment of inertia of a thin rod about an axis through its center and perpendicular to its length is ML²/12. Using parallel axis theorem, moment of inertia about an axis through one end and perpendicular to length is I = ML²/12 + M(L/2)² = ML²/12 + ML²/4 = ML²/3.
44. The moment of inertia of a solid sphere of mass M radius R about its diameter is
MR
2 MR2 5
2MR
MR
(b) The moment of inertia of a solid sphere of mass M and radius R about its diameter is (2/5)MR².
45. The M.I of a solid cylinder of length /, radius R about its geometrical axis is same as about equatorial axis, then the ratio of R and I will be
3
13
(b) Moment of inertia about geometrical axis = 1/2 MR². Moment of inertia about equatorial axis = M(R²/4 + l²/12). Given 1/2 MR² = M(R²/4 + l²/12) => R²/2 = R²/4 + l²/12 => R²/4 = l²/12 => R²/l² = 1/3 => R/l = 1/√3 = √3/3.
46. A uniform metallic disc of moment of inertia Io about its own axis is melted and a uniform ring of equal radius is then casted from it. Then, M.I of the ring about its diameter will be
Le 2
10
10 4
(a) Moment of inertia of disc about its axis I₀ = 1/2 MR². Moment of inertia of ring about its axis = MR². So Moment of inertia of ring is 2I₀. Moment of inertia of ring about its diameter = 1/2 MR² = 1/2 (2I₀) = I₀.
47. A uniform metallic disc has its M.I I. about its diameter. Then its M.I about an axis through its rim perpendicular to the plane will be
41₀
5I₀
(b) Moment of inertia about diameter I = 1/4 MR². Moment of inertia about axis through rim perpendicular to plane = Icenter + MR² = 1/2 MR² + MR² = 3/2 MR² = 3/2 (4I₀) = 6I₀. There might be error in options. Moment of inertia about axis through center perpendicular to plane is 2I₀. Then using parallel axis theorem, I' = 2I₀ + MR² = 2I₀ + 4I₀ = 6I₀.
48. The radii of two steel balls are R and 2R. Then, their moment of inertia about their diameters are in the ratio [KU 2009]
[KU 2009]
1:4
1 : 16
1: 32
(d) Moment of inertia of solid sphere about diameter I = (2/5)MR². Mass M ∝ R³. So I ∝ R⁵. Ratio I₁/I₂ = (R / 2R)⁵ = (1/2)⁵ = 1/32.
49. A circular portion of diameter R is cut out from the edge of a uniform disc of mass M and radius R. The M.I of the remaining portion of the disc about an axis passing through the centre O of the disc and perpendicular to its plane is
15 32 8 MR?
16 MR² 8
13 8 MR
(a) MI of complete disc I = 1/2 MR². MI of removed portion I' = 1/2 (M/4) (R/2)² = 1/32 MR². MI of remaining portion = I - I' = 1/2 MR² - 1/32 MR² = 15/32 MR².
50. A uniform rod of mass M and length L is rotating with angular speed ωo with two beads of mass m on either side of the axis passing through its centre and perpendicular to its length. The beads slide outward as it rotates. What will be the final angular speed when the beads reach the ends ?
M CDo M+2m
M C M+2m
M Wo M+6m
M Wo M+12m
(a) By conservation of angular momentum, Iᵢωᵢ = Ifωf. Initial MI Iᵢ = ML²/12 + 2m(0)². Final MI If = ML²/12 + 2m(L/2)² = ML²/12 + mL²/2 = (ML² + 6mL²) / 12 = (M+6m)L²/12. ML²/12 * ω₀ = (M+6m)L²/12 * ωf => ωf = Mω₀ / (M+6m). Option a has M+2m.
51. Three thin rods each of length L and mass M are placed along X, Y and Z-axis in such a way that one end of rod is at the origin. The moment of inertia of the system about Z-axis is
ML
ML'
ML?
ML2
(c) MI of rod along Z axis = 0. MI of rod along X axis about Z axis = M L²/3. MI of rod along Y axis about Z axis = M L²/3. Total MI about Z axis = 0 + ML²/3 + ML²/3 = 2ML²/3.
52. The M.I of two spheres of equal masses about their respective diameters are same. If one of them is solid and other is hollow, then the ratio of their radii (solid to hollow), will be
53. Two circular discs of same mass an thickness are made from metals having densities d, and dy respectively. The ratio of their moments of inertia about the central axis will be
di : d2
dz : di
did2 : 1
1 : didz
(b) Mass M is same. Thickness t is same. Density d = mass/volume = mass/(πr²t) => r² = M/(πtd). Moment of inertia I = 1/2 Mr² = 1/2 M * M/(πtd) = M² / (2πtd). I ∝ 1/So I₁/I₂ = d₂/d₁.
54. A wheel of moment of inertia 5x10 kg -m is making 20 rev/seThe torque required to stop it in 10 sec is
55. A thin hollow cylinder open at both ends, Slides without rotating rolls without slipping with the same ii. speeThe ratio of K.E in the two cases is:
1: 1
2:1
4 : 1
1. 1:2
(a) Kinetic energy for sliding KEsliding = 1/2 mv². Kinetic energy for rolling KErolling = 1/2 mv² + 1/2 Iω² = 1/2 mv² + 1/2 (mr²) (v/r)² = 1/2 mv² + 1/2 mv² = mv². Ratio KEsliding / KErolling = (1/2 mv²) / mv² = 1/2.
56. A solid sphere of mass M is rolling on a horizontal surface without sliding with velocity v. Its kinetic energy will be
my
7 1 my
To mv
(a) Total KE = Translational KE + Rotational KE = 1/2 mv² + 1/2 Iω² = 1/2 mv² + 1/2 (2/5 MR²) (v/R)² = 1/2 mv² + 1/5 mv² = 7/10 mv².
57. A solid sphere of mass M is rotating about its diameter and linear velocity of a point on its equator is v. Then its kinetic energy will be
58. A body rolling without sliding has its rotational kinetic energy equal to 40% of total energy. Then body should be
Circular disc
Hollow sphere
Solid sphere
Hollow cylinder
(c) Rotational KE = 1/2 Iω². Total KE = 1/2 mv² + 1/2 Iω² = 1/2 Iω² (1 + mr²/I). Rotational KE / Total KE = 1 / (1 + mr²/I) = 0.4 = 2/5. 1 + mr²/I = 5/2 => mr²/I = 3/2 => I = 2/3 mr². This is moment of inertia of hollow sphere. Error in question. Let's try Total KE = 1/2 mv² (1 + k²/r²) = Rotational KE * (1 + r²/k²). Rotational KE / Total KE = k² / (r² + k²) = 0.4 = 2/5. 5k² = 2r² + 2k² => 3k² = 2r² => k²/r² = 2/3. For solid sphere k²/r² = 2/5. For hollow sphere k²/r² = 2/3. So it should be hollow sphere.
59. A solid spherical ball rolls on a table. Ratio of rotational. K.E to the total K.E is
|179
(a) Rotational KE = 1/2 Iω² = 1/2 (2/5 MR²) (v/R)² = 1/5 Mv². Total KE = 1/2 Mv² + 1/5 Mv² = 7/10 Mv². Ratio = (1/5 Mv²) / (7/10 Mv²) = 2/7.
60. The least coefficient of friction for an inclined plane of inclination a with the horizontal in order that a solid cylinder will roll down without slipping is
tan o 2
- tan o
tan a
- tan o
(a) For rolling without slipping, μ ≥ tanα / (1 + MR²/I). For solid cylinder I = 1/2 MR². μ ≥ tanα / (1 + 1/2) = tanα / (3/2) = 2/3 tanα. Option a has tan o / 2 which might be typo.
61. A wheel of mass 10kg has a moment of inertia 160kg-m' about its own axis. The radius of gyration is:
10m
4m
om
(b) I = Mk² => k = √(I/M) = √(160 / 10) = √16 = 4 m.
62. The radius of gyration of a solid disc of mass 1kg and radius 50cm about an axis through centre of mass and perpendicular to its face is
25 cm
25 V2 cm
50 cm
25 16 cm
(b) I = 1/2 MR² = Mk². k = R/√2 = 50 / √2 = 25√2 cm.
63. A uniform circular disc, 20g is rotating about its own vertical axis at 30 rpm. When 20g sand falls on its surface at distance 5cm from the centre of the disc, the rate of rotation decreases to 24 rpm. Then the radius of the disc should be:
64. A particle performs uniform circular motion with an angular moment L. If the frequency of particle's motion is doubled and its kinetic energy is halved, the angular momentum becomes:
2L
4L
L/2
1/4
(b) KE = L²/2I. If KE' = KE/2, L'²/2I' = L²/4I. L = Iω = I(2πf). If frequency doubles, angular velocity doubles. ω' = 2ω. KE = 1/2 Iω². KE' = 1/2 I'(2ω)² = 2I'ω² = KE/2 = 1/4 Iω². 8I' = I. L' = I'ω' = (I/8)(2ω) = Iω/4 = L/4. Option d.
65. A constant torque acting on a uniform circular wheel changes its . angular omentum from Jo to 43, in 4 seconds. The magnitude of the torque is:
66. A flywheel of moment of inertia 0.5kgm i rotating 300 rpm initially comes to rest in 10 seconds under constant retarding torque. Then the number of revolutions made by the wheel until rest is:
68. Two particles A and B initially at rest move towards each other under a mutual force of attraction. At the instant when velocity of A is v and that of B is 2v, the velocity of centre of mass of the system is:
31
zer
(c) Initial momentum = 0. Momentum at any instant = mAvA + mBvB = 0 (by conservation of momentum). Velocity of center of mass Vcm = Total momentum / Total mass = 0 / (mA + mB) = 0.
69. Two particles of masses m, and my are at distance x. Then, their centre of mass lies at distance from my.
max
(mitmy X m2
mz (mi+mz
(d) Distance of center of mass from m₂ = m₁x / (m₁ + m₂).
70. Out of two particles of masses m, and m₂, the towards their centre of mass. What is the displacement of centre of mass?
mud mz
(m,+my/
(a) Displacement of center of mass Δxcm = (m₁Δx₁ + m₂Δx₂) / (m₁ + m₂) = (m₁d + m₂d) / (m₁ + m₂) (if both move towards each other) or if m₁ moves towards center of mass by d, then m₁d + m₂d' = 0 => d' = -m₁d/m₂. Displacement of center of mass = 0.
71. Two blocks of masses 5kg and 2kg ar placed on a frictionless surface and connected by a spring. An external kick gives a velocity of 14 m/s to the heavier block in the direction of lighter one Calculate the velocity gained by the centre of mass.
14 m/s
7 m/s
10 m/s
zero
(c) Velocity of center of mass Vcm = (m₁v₁ + m₂v₂) / (m₁ + m₂) = (5 * 14 + 2 * 0) / (5 + 2) = 70 / 7 = 10 m/s.
72. A shell is fixed a gun with a muzzle velocity u m/s at an angle 0 with the horizontal. At the top of the trajectory, the shell explodes into two fragments P and Q of equal mass. If the speed of fragment P immediately after explosion becomes zero where does the fragment Q hit the ground from the point of projection?
3 u sin e 2 g
u' sin28 2 g
u sin20 g
u sin'e g
(c) At the top, velocity of shell is ucosθ (horizontal). Mass of each fragment = M/2. Momentum before explosion = Mu cosθ. Momentum after explosion = 0 + (M/2)v => v = 2u cosθ (horizontal). Time to reach ground from top = √(2H/g) = √(2 * u²sin²θ / 2g) / g = usinθ/g. Horizontal distance travelled by Q from the point of explosion = v * t = (2u cosθ) * (usinθ/g) = 2u²sinθcosθ / g = u²sin2θ / g = Range.
73. . A circular plate of uniform thickness has diameter of 56cm. A circular portion of diameter 42cm is removed from one edge as shown in the fig. The centre of mass of remaining from the centre of plate will be
5 cm
7 cm
CA 9 cm
1 1 cm
(c) Let radius of complete disc be R = 28 cm, mass M. Radius of removed portion r = 21 cm, mass M' = M(r/R)² = M(21/28)² = 9M/16. Distance of center of mass of removed portion from center of complete disc = R - r = 28 - 21 = 7 cm. Center of mass of remaining part xcm = (M * 0 - M' * 7) / (M - M') = (-9M/16 * 7) / (7M/16) = -9 cm.
74. 83. Two masses of 1kg and 2kg are 9m apart and make a two body system. Their centre of mass from 1kg mass will be at [MOE]
[MOE]
6
4m
3m
2m
(a) Center of mass from 1 kg mass = (2 * 9) / (1 + 2) = 18 / 3 = 6 m.
75. Let F be a force acting on a particle having position vector r . Let + be the torque of this force about the origin, then [KU 2015]
[KU 2015]
() Torque τ = r × F. Therefore, r . τ = r . (r × F) = 0 (scalar triple product) and F . τ = F . (r × F) = 0.
76. If a gymnast on a rotating stool with his arms outstretched suddenly lower his arms
[TOM 2015]
The angular velocity decrease
The moment of inertia decrease
The angular velocity remains constant
The angular momentum increase
(b) When the gymnast lowers his arms, the mass distribution becomes closer to the axis of rotation, decreasing the moment of inertia (I). According to the law of conservation of angular momentum (L = Iω), if no external torque acts on the system, the angular momentum (L) remains constant. Therefore, if the moment of inertia (I) decreases, the angular velocity (ω) must increase.
