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
8. Circular motion
1. A body is moving with uniform velocity in circular path then, centripetal acceleration is: [BP 2014]
[BP 2014]
zero
directed towards radius
along tangent
along the axis perpendicular to plane.
(b) In uniform circular motion, the centripetal acceleration is always directed towards the center of the circular path, along the radius.
2. A four wheeler is moving in a banked track of radius 24 m with velocity 10 m/s. What is the minimum angle of banking so that it won't overturn?
230
240
250
20
(a) For no overturning, the angle of banking θ is given by tanθ = v²/rg = (10)² / (24 * 9.8) = 100 / 235.2 ≈ 0.425. θ = tan⁻¹(0.425) ≈ 23.04°. The closest option is 20°.
3. A stone of mass 'm' is tied to a string of length 'I' and rotated in a circle at constant speed 'v'. If the string is released the stone flies. [BP 2011]
[BP 2011]
Radially inward
Radially outward
Tangentially outward
With an acceleration mv-/1
(c) When the string is released, there is no longer a centripetal force acting on the stone. According to Newton's first law, the stone will continue to move in a straight line along the direction of its velocity at the moment of release, which is tangential to the circular path.
4. A road of 50 m radius is banked at correct angle for a given speed. If the speed is to be doubled, keeping the same banking angle, the radius of the road should be changed [BP 201 1]
[BP 2011]
25 m
100 m
150 m
200 m
(d) The correct angle of banking θ is given by tanθ = v²/rg. If θ and g are constant, then v² ∝ r. If the speed is doubled (v' = 2v), then (2v)² ∝ r' => 4v² ∝ r'. Since v² ∝ r, we have r' = 4r = 4 * 50 m = 200 m.
5. "In a death well", a motorcyclist performs his race on circular path of radius (r) then minimum velocity at lowest point is: [MOE2014]
[MOE2014]
√5 rg
√7 rg
3 rg
(b) For completing the circular path in a vertical death well, the minimum velocity at the lowest point is given by v = √(5rg).
6. A body of mass m is moving in a vertical circle with constant speed of V. The tension on the mass at the bottom of the. circle is [MOE 2012]
[MOE 2012]
mg - mrw
mg + mrw
mg x mro
mg/ mro
(b) At the bottom of the vertical circle, the tension T acts upwards and the weight mg acts downwards, providing the necessary centripetal force mv²/r (here speed is V). So, T - mg = mV²/r => T = mg + mV²/r (ω = V/r). Option b has mrw which should be mV²/r.
7. A can filled with water is revolved in a vertical circle of radius 4 m so that water does not spill down. The maximum period of revolution will be:
[MOE 2011]
2 sec
3 sec
4 se
5 sec
(c) For water not to spill at the top, the minimum velocity at the top is v = √(rg) = √(4 * 9.8) = √39.2 ≈ 6.26 m/s. Velocity v = 2πr/T => T = 2πr/v = 2π * 4 / 6.26 ≈ 4.03 s. Using g=10, v = √(4*10) = √40 = 2√10 ≈ 6.32. T = 2π*4 / (2√10) = 4π/√10 ≈ 12.56/3.16 ≈ 3.97 s ≈ 4 s.
8. A cyclist moving with velocity 10 m/s in radius 20 m than angle of inclination of cycle [MOE 2011]
9. A particle of mass m is moving along a circular path of radius R with a frequency and period T. It's centripetal acceleration can be expressed as: [MOE 2010]
[MOE 2010]
4π² R'n²
4π²Rn
4πRn²
4π²R/T²
(c) Centripetal acceleration ac = ω²R = (2πn)²R = 4π²n²R. Also, n = 1/T, so ac = 4π²(1/T)²R = 4π²R/T².
10. The radius of circular path is Im. A particle makes 10 revolutions in 6 sec. The linear velocity of the particle is: [MOE 20141
[MOE 20141]
0.5
10
2
(b) Frequency n = 10 rev / 6 sec = 5/3 Hz. Angular velocity ω = 2πn = 2π(5/3) = 10π/3 rad/s. Linear velocity v = ωR = (10π/3) * 1 = 10π/3 ≈ 10 * 3.14 / 3 ≈ 10.47 m/s. The closest option is 10.
11. A cyclist turns around a curve at 20 km/hr. If he turns at the double of this speed, the tendency to overturn is [KU 2014/ BP 2015/ 2017)
[KU 2014, BP 2015, BP 2017]
Double
Quadrupled
halved
Unchanged.
(b) Tendency to overturn is related to the angle of inclination θ, where tanθ = v²/rg. If speed v is doubled, tanθ becomes (2v)²/rg = 4v²/rg, which is 4 times the initial value. So, the tendency to overturn is quadrupled.
12. If a ball is attached to a string and whirled in a vertical circle then the velocity at the lowest point is :
[KU 2013]
Virg
Vgr
V2gr
er
(a) The minimum velocity required at the lowest point to complete a vertical circle is v = √(5gr). The options seem to have typos.
13. A particle moves in a circle of a radius 25 cm at 2 rev/sec. The acceleration of the particle in m/s' is
14. moving along convex bridge of radius of curvature 100m is speeding at 3m/s'. What is its acceleration when its velocity is 20m/s?
4m/s2
3m/s
5m/s
2m/s
(c) Total acceleration = √(atangential² + acentripetal²). Tangential acceleration is given as 3 m/s². Centripetal acceleration ac = v²/r = (20)² / 100 = 400 / 100 = 4 m/s². Total acceleration = √(3² + 4²) = √25 = 5 m/s².
15. The speed of revolution of a particle going around a circle is doubled and its angular speed is halved. What happens to the centripetal acceleration?
Unchanged
Doubled
Halved
Becomes four times
(a) Centripetal acceleration ac = v²/r = (ωr)²/r = ω²r. If speed v' = 2v and angular speed ω' = ω/2, then r' = v'/ω' = 2v / (ω/2) = 4v/ω = 4r. New centripetal acceleration ac' = v'²/r' = (2v)²/(4r) = 4v²/(4r) = v²/r = ac. Or ac' = ω'²r' = (ω/2)²(4r) = (ω²/4)(4r) = ω²r = ac. The centripetal acceleration remains unchanged. There might be error in my derivation. Let me retry. v' = 2v, ω' = ω/2. ac = ωv. ac' = ω'v' = (ω/2)(2v) = ωv = ac. Still unchanged. There's likely error in question or options. Let's try ac = v²/r. v' = 2v. ω' = ω/2 => v'/r' = ω/2 => 2v/r' = ω/2 => r' = 4v/ω = 4r. ac' = (2v)²/(4r) = 4v²/(4r) = v²/r = ac. Unchanged. Let's try ac = ω²r. ω' = ω/2. r' = 4r. ac' = (ω/2)²(4r) = ω²r = ac.
16. A 600g object is tied to a string Im long and it is rotated in a horizontal circle of radius 0.8m. Thus the tension produced on the string is [MOE Bangladesh]
[MOE Bangladesh]
g
g/2
3g/2
5g/2
(a) Let mass m = 0.6 kg, length l = 1 m, radius r = 0.8 m. Vertical angle θ = cos⁻¹(0.8/1) = cos⁻¹(0.8) ≈ 36.87°. Tangent of the angle tanθ = r/h = 0.8/√(1² - 0.8²) = 0.8/√0.36 = 0.8/0.6 = 4/3. Tension T. Vertical component Tcosθ = mg => T = mg/cosθ = 0.6 * 9.8 / 0.8 = 7.35 N. Horizontal component Tsinθ = mv²/r => 7.35 * (0.6/0.8) = 0.6 * v² / 0.8 => v² = 7.35 => v = √7.35 ≈ 2.71 m/s. The tension produced on the string is along the string. Vertical component balances weight. Horizontal component provides centripetal force. Tcosθ = mg => T = mg/cosθ = 0.6 * 9.8 / 0.8 = 7.35 N. tanθ = v²/rg => (0.6/0.8) = v² / (0.8 * 9.8) => v² = 0.6 * 9.8 = 5.88. T = 7.35 N. None of the options match.
17. A particle is moving in a circle of radius with a uniform speed. Its angular acceleration is,
v/r directed towards centre
\(\frac{v^2}{r}\) directed towards centre
\(\frac{v^2}{r}\) directed along the axis
Zero
(d) For uniform circular motion, the angular speed is constant, so the angular acceleration (rate of change of angular velocity) is zero. The options provided are directions, which is also zero.
18. A cyclist turns around a curve at 20 km/hr. If he turns at the double of this speed, the tendency to overturn is
double
quadrupled
halved
unchanged
(b) Tendency to overturn is proportional to v².
19. A stone tied to one end of string, 5 m long is whirled in horizontal circle at constant speed 5m/s. What is average acceleration of the stone in half revolution?
5π \(m/s^2\)
\(\frac{5}{π} m/s^2\)
\(\frac{10}{π} m/s^2\)
zero
(c) Initial velocity v₁ = 5 î. Final velocity v₂ = -5 î. Change in velocity Δv = v₂ - v₁ = -10 î. Time taken for half revolution t = πr/v = π * 5 / 5 = π s. Average acceleration = Δv / t = -10 î / π = -10/π m/s².
20. A stone tied to one end of light string 50 cm long, is whirled in horizontal circle with constant K.E. of 20J. Then, tension in the string will be,
80 N
40 N
20 N
10 N
(a) Radius r = 0.5 m. KE = 1/2 mv² = 20 J => mv² = 40. Tension T = mv²/r = 40 / 0.5 = 80 N.
21. Uniform circular motion is motion with
Constant velocity
Constant acceleration
Variable acceleration
Variable speed
(c) In uniform circular motion, the speed is constant, but the direction of velocity is continuously changing, so the acceleration (centripetal acceleration) is variable in direction.
22. A man swings a bucket of water in a vertical circle of diameter 2m. The minimum speed with which he must swing the bucket so that the water does not spill is
1.56 m/s
9.8 m/s
3.13 m/s
4.4 m/s
(c) Radius r = 1 m. Minimum speed at the top v = √(rg) = √(1 * 9.8) = √9.8 ≈ 3.13 m/s.
23. A stone tied to one end of a string is whirled in horizontal circle so that length increases gradually. If T, and T, are the tensions in the string when its length be / [BPKIHS]
[BPKIHS]
T1 = 2T2
T1 = 4T2
T1=8T2
T1 = T2/2
(c) Tension T = mv²/r = mω²r. If length increases, radius increases. Assuming constant angular velocity, T ∝ r. If r₂ = 2r₁, then T₂ = 2T₁ => T₁ = T₂/2. None of the options match.
24. The speed of revolution of a particle going around a circle is doubled and its angular speed is halved. What happens to the centripetal acceleration?
Becomes four times
Doubled
Halved
Remains unchanged
(d) ac = v²/r. v' = 2v, ω' = ω/2 => v'/r' = ω/2 => 2v/r' = ω/2 => r' = 4v/ω = 4r. ac' = (2v)²/(4r) = 4v²/(4r) = v²/r = ac. Unchanged. Error in my thought process. ac = ωv. ac' = ω'v' = (ω/2)(2v) = ωv = ac. Unchanged. Let's try ac = ω²r. ω' = ω/2. r' = 4r. ac' = (ω/2)²(4r) = ω²r. Unchanged. There is likely an error in the question or options. Let's reconsider a_c = v^2/r. If v doubles and ω halves, then r quadruples. a_c' = (2v)^2/(4r) = 4v^2/4r = v^2/r. Remains unchanged.
25. For a particle in circular motion, the centripetal acceleration is
less than its tangential acceleration
equal to its tangential acceleration
more than its tangential acceleration
may be more or less than its tangential acceleration
(d) If the speed is constant, tangential acceleration is zero, so centripetal acceleration is more. If speed is changing, it can be more or less.
26. What should be the angular velocity of earth so that a body on its equator is weightless
28. A block released at the top of a smooth hemispherical bowl of radius R slip tangentially with radius vector makes angle 0 with vertical, when its velocity is v. Then angle 0 is,
\(\theta = \tan^{-1}\left(\frac{v^2}{gR}\right)\)
\(\theta = \sin^{-1}\left(\frac{v^2}{gR}\right)\)
\(\theta = \cos^{-1}\left(\frac{v^2}{gR}\right)\)
\(\theta = \cot^{-1}\left(\frac{v^2}{gR}\right)\)
(c) Using conservation of energy: mgR = 1/2 mv² + mgRcosθ => v² = 2gR(1 - cosθ). For slipping tangentially, normal reaction becomes zero. Equation of motion along radius: mgcosθ - N = mv²/R => mgcosθ = m(2gR(1 - cosθ))/R => cosθ = 2(1 - cosθ) => cosθ = 2 - 2cosθ => 3cosθ = 2 => cosθ = 2/3 => θ = cos⁻¹(2/3). None of the options match.
29. A block released at the top of a smooth hemispherical bowl of radius R slips tangentially after falling a vertical height h. Then,
\(h = \tfrac{2R}{3}\)
\(h = \tfrac{R}{3}\)
\(h = \tfrac{3R}{4}\)
\(h = \tfrac{R}{2}\)
(b) cosθ = R-h / R. cosθ = 2/3 => R-h/R = 2/3 => 3R - 3h = 2R => R = 3h => h = R/3.
30. A block released at the top of a smooth hemispherical bowl of radius R slips tangentially after falling a vertical heighth where radius makes angle 0 with vertical Then 0 is
\(\theta = \cos^{-1}\left(\frac{1}{3}\right)\)
\(\theta = \cos^{-1}\left(\frac{3}{4}\right)\)
\(\theta = \cos^{-1}\left(\frac{1}{2}\right)\)
\(\theta = \cos^{-1}\left(\frac{2}{3}\right)\)
(d) cosθ = 2/3.
31. What minimum horizontal speed imparted to the block at the top of smooth hemisphere of radius R so that it slips instantly?
\(\sqrt{2gR}\)
\(\sqrt{5gr}\)
\(zero\)
\(\sqrt{gR}\)
(d) At the top, θ = 0, cosθ = 1. Normal reaction N + mg = mv²/R. For slipping instantly, N = 0. mg = mv²/R => v² = gR => v = √(gR).
32. A stone tied to one end of a string of length 5/6m is whirled in a vertical circle so that maximum tension in the string is 4 times the minimum tension in it. The speed of the stone at highest point is,
5m/s
4m/s
6m/s
3m/s
(a) Tmax = mv²/r + mg = 4(Tmin) = 4(mV'²/r - mg). v' is speed at highest point, v is speed at lowest. v² = v'² + 4gr. (v'² + 4gr)/r + g = 4(v'²/r - g) => v'² + 4gr + gr = 4v'² - 4gr => 9gr = 3v'² => v'² = 3gr = 3 * g * 5/6 = 2.5g = 2.5 * 10 = 25. v' = 5 m/s.
33. For what minimum height h, a body is released on a smooth inclined plane so that it revolves in vertical circle of radius R after reaching the ground
\(h = \frac{5}{2}R\)
\(h = \frac{3}{2}R\)
\(h = 2R\)
h = 3R
(a) Height of inclined plane h should be equal to the height of the highest point in vertical circle for just completing it, which is 2R from the center or 2R + R = 3R from the lowest point if released from rest. If released from some height to gain speed at the bottom, then height should be 5R/2 for just completing vertical circle.
34. A horizontal turn table is rotating steadily at angular speed c with a coin on its distance 4cm from the centre o the table. If angular speed of the tu is made 2w then distance of the coin on the table in equilibrium condition is
35. The radius of the circular path of a particle is doubled but its frequency of rotation remains constant. If the initial centripetal force be F, then final centripetal force will be
F
2F
F/2
4F
(b) Centripetal force F = mω²r = m(2πf)²r = 4π²mf²r. If radius r' = 2r and frequency f remains constant, then final centripetal force F' = 4π²mf²(2r) = 2(4π²mf²r) = 2F.
36. A body of mass 1kg is rotating in a vertical circle of radius 1m. What will be th difference in its kinetic energy at the top and bottom of the circle?
10J
20J
30J
50J
(b) Difference in potential energy between bottom and top = mg(2r) = 1 * 10 * 2 * 1 = 20 J. By conservation of energy, loss in PE = gain in KE. So difference in KE = 20 J.
37. A bucket of filled with water is revolved in a vertical circle of radius 4cm. If g = 10m/s', the time period of revolution will be nearest
10s
8s
5s
4s
(d) Radius r = 0.04 m. Minimum speed at top v = √(rg) = √(0.04 * 10) = √0.4 ≈ 0.63 m/s. Time period T = 2πr/v = 2π * 0.04 / 0.63 ≈ 0.25 s. None of the options match. Let's check if radius was 4m. v = √(4*10) = √40 = 6.32. T = 2π*4 / 6.32 ≈ 4 s. Option c is closest if g=9.8, v=6.26, T=4.03.
38. An object is tied to string and rotated in a vertical circle of radius r. Constant speed is maintained along the trajectory. If Tmax/Tmin = 2, then v/(rg) is
39. A particle is making uniform circular motion with angular momentum L. If its kinetic energy is made half and angular frequency be doubled, its new angular momentum will be
L/4
L/2
2L
4L
(a) KE = L²/2I. If KE' = KE/2, then L'²/2I' = L²/4I. L = Iω. KE = 1/2 Iω². If ω' = 2ω, KE' = 1/2 I'(2ω)² = 2I'ω². KE' = KE/2 => 2I'ω² = 1/2 (1/2 Iω²) => 8I' = I. L' = I'ω' = (I/8)(2ω) = Iω/4 = L/4. Option a.
40. If an object of mass 'm' moving in circular path-with uniform speed 'v' which of th following change occurs in half circle [IOM 2009]
[IOM 2009]
kinetic energy changes by \(\frac{mv^2}{2}\)
kinetic energy changes by \(\frac{mv^2}{4}\)
momentum changes by 2mv
momentum doesn't change
(c) Kinetic energy remains constant as speed is uniform. Momentum changes in direction. Initial momentum = mv î. Final momentum = -mv î. Change in momentum = -2mv î. Magnitude = 2mv.
41. The distance between two rails is 1.5m. The centre of gravity of the train is at height of 2m from the ground. The maximum speed of the train on a circular path of radius 120m can be:
42. A bob of a simple pendulum of length L is at its lowest point. What minimum velocity is imparted to the bob so that the string makes a maximum angle of 60 with vertical.
\(V=\sqrt{gl}\)
\(V=\sqrt{2gl}\)
\(V=\sqrt{3gl}\)
\(V=\sqrt{5gl}\)
(a) Using conservation of energy. 1/2 mv² = mgl(1 - cos60) = mgl(1 - 1/2) = mgl/2 => v² = gl => v = √(gl).
43. A stone tied to a string of length L is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is in the lowest position and has a speed u. The magnitude of change in velocity as it reaches a point where the string is horizontal is
a. Vu -2gL
b. V2gl
c. Vu -gl
V2(u - gL)
(d) Velocity at lowest point = u. Velocity at horizontal position v' = √(u² - 2gL). Change in velocity = √(u² + v'²) = √(u² + u² - 2gL) = √(2u² - 2gL) = √2(u² - gL).
44. A particle of mass m is tied to one end of a long in extensible string whose other end is fixed to a rigid support. The particle is taken one side so that the angular displacement from vertical is of The particle is released so that the angular displacement of any instant is 0 (< <). Then the tension in the string must [IOM 2008]
[IOM 2008]
a. mgcose
mgcosa
c. greater than mgcose
d. less than mgcose
(c) T - mgcosθ = mv²/l. v² = 2gl(cosθ - cosα). T = mgcosθ + 2mg(cosθ - cosα) = mg(3cosθ - 2cosα). Since θ < α, cosθ > cosα. So T > mgcosθ.
45. 48 Two particles of equal masses are revolving in circular paths of radii r, and Tr respectively with same speed. What will IKU 2009 be the ratio of their forces
[KU 2009]
() Centripetal force F = mv²/r. For same mass and speed, F ∝ 1/r. F₁/F₂ = r₂/r₁.
46. A particle of mass m is moving along a circular path of radius R with a frequency n and period T. Its centripetal acceleration can be expressed as [MOE 2010)
[MOE 2010]
4π²R'n²
4π RT
C. 4πRn
(a) ac = ω²R = (2πn)²R = 4π²n²R.
47. A body of mass 0.1kg tied by a string is rotating round a vertical circle with speed of 10ms" of radius Im. What is the tension experienced by the string at the highest point: [MOE 2055]
[MOE 2055]
a. 8N
11N
C. 10N
d. ON
(a) At the highest point, tension T + mg = mv²/r => T = mv²/r - mg = 0.1 * (10)² / 1 - 0.1 * 10 = 10 - 1 = 9 N. Option a is 8N.
48. A body of mass 2kg is attached to one end of string of length Im & rotated in vertical circle such that velocity at highest point is 4m/s. Here tension is [IE-06]
[IE-06]
a. 18N
12N
c. 45N
0
(a) At the highest point, tension T + mg = mv²/r => T = mv²/r - mg = 2 * (4)² / 1 - 2 * 9.8 = 32 - 19.6 = 12.4 N. Closest option is 12N or 18N. Let's use g=10. T = 32 - 20 = 12N.
49. If an object of mass 10kg is whirled round a horizontal circle of radius 4m by revolving string inclined at 30" to vertical If the uniform speed of the object is 5ms", the tension of the string (approx) is [ IE-08]
50. When a pendulum is displaced through 90%, the length / and mass m of the pendulum. The minimum strength of string that can withstand tension at the mean position is [IOM 2015]
[IOM 2015]
a. Img
b. 2mg
3mg
d. 4mg
(c) At extreme position (90°), velocity is zero. At mean position, by conservation of energy, 1/2 mv² = mgl => v² = 2gl. Tension at mean position T = mv²/l = m(2gl)/l = 2mg. Net tension = T + mg = 3mg.
51. A cyclist turned over at 15 miles/hr. It doubled the speed of cycle then what is the chance of overturning? [IOM 2016)
[IOM 2016]
Doubled
0. Halved
Quadrupled
Constant
(c) Tendency to overturn proportional to v².
52. Centrifugal force is
[IOM 2016]
Centripetal force
Gravitational force
Viscous force
Apparent force on rotating frame
(d) Centrifugal force is a fictitious force that appears to act outwards on an object moving in a circular path when viewed from a rotating frame of reference.
53. A circular object of radius 25 cm, is revolving with a speed of 2 rev/sec. Then what is the acceleration produced? [IOM 2016]
[IOM 2016]
a.
b. 4π
d. π²
(c) Radius R = 0.25 m. Frequency f = 2 Hz. Acceleration ac = ω²R = (2πf)²R = 4π²f²R = 4π²(2)²(0.25) = 4π².
54. For vertical circular motion, radius is Im then its critical velocity is :
[IOM 2017]
a. 3.13m/s
b. 6.36 m/s
c. 9.48m/s
d. 12 m/s
(a) Critical velocity at the lowest point to complete vertical circle is √(5gr) = √(5 * 9.8 * 1) ≈ √49 = 7 m/s. Critical velocity at the highest point is √(gr) = √(9.8 * 1) ≈ 3.13 m/s. Option a refers to velocity at highest point.
55. The mas of particle is 'm' is revolving around a circle having radius r & angular momentum L. Then its centripetal force will be:
[IOM 2017]
L'
b. mr^3
L mr
d. mr
(a) Angular momentum L = Iω = mr²ω => ω = L/(mr²). Centripetal force F = mrω² = mr(L/(mr²))² = mrL²/(m²r⁴) = L²/(mr³). Option a has L'
