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
4. Projectile Motion
1. When air resistance is taken into account
while dealing with the motion of the
projectile. Of the following properties of
the projectile, the one which shows an
increase is:
[BP 2014)
[BP 2014]
Range.
Max" height
Speed at which it strikes the ground.
The angle at which the projectile strikes
the ground.
(d) When air resistance is considered, it opposes the motion of the projectile, leading to a decrease in range, maximum height, and the speed at which it strikes the ground. The angle at which it strikes the ground will increase compared to the case without air resistance.
2. A body is thrown with velocity 50 m/s. The
maximum horizontal distance it can cover
[BP 2010]
[BP 2010]
200 m
250 m
300 m
400 m
(b) Maximum horizontal range is achieved at a projection angle of 45°. Rmax = u²/g = (50)² / 10 = 2500 / 10 = 250 m (assuming g = 10 m/s²).
3. The range of a projectile fired at an angle
of 15" is 50 m. If it is fired with the same
speed at an angle of 45", its range will be;
100m
25 m
50 m
37 M
(a) Range R = u²sin2θ/g. R₁/R₂ = sin2θ₁/sin2θ₂. 50/R₂ = sin(30°)/sin(90°) = (1/2)/1 = 1/2. R₂ = 100 m.
4. In a projectile throw at the same angles the
velocity of A is two times of velocity of B
their ranges are related as
[IOM 2014]
[IOM 2014]
R. 2R
R1 = 2R.
Ru = 4R
(b) Range R = u²sin2θ/g. R ∝ u². If velocity of A is 2 times velocity of B, then Range of A will be (2)² = 4 times the range of B. RA = 4RB.
5. A body is projected at an angle 45" with
a velocity 200 m/s. The maximum height
attained by the projectile is
[MOE 2012,11]
[MOE 2012, MOE 2011]
200 m
400 m
800 m
1000 m
(d) Maximum height H = u²sin²θ / 2g = (200)² * sin²(45°) / (2 * 10) = 40000 * (1/√2)² / 20 = 40000 * (1/2) / 20 = 20000 / 20 = 1000 m (assuming g = 10 m/s²).
6. A body projected at angle of 60" attains the
same height at two points a and b at two
different times. If air resistance is
neglected the ratio of mechanical energies
at with respect to that at b will be:
(MOE 2011]
[MOE 2011]
More than one.
Less than one.
Equal to one.
Dependent on their directions of
Velocity
(c) In the absence of air resistance, the mechanical energy (sum of kinetic and potential energy) of a projectile remains constant throughout its flight. Therefore, the ratio of mechanical energies at points a and b will be equal to one.
7. A fielder can throw a cricket ball to a
maximum horizontal distance of 100 m.
How high the fielder can throw the same
ball?
[IOM 2013)
[IOM 2013]
25 m
50 m
40 m
100 m
(b) Maximum range Rmax = u²/g = 100 m (at θ = 45°). Maximum height Hmax = u²/2g = Rmax / 2 = 100 / 2 = 50 m. There is a mistake. Hmax = u²sin²(90)/2g = u²/2g. Rmax = u²sin(90)/g = u²/g = 100. Hmax = 100/2 = 50 m. There is another mistake in applying formula for max height. For max height throw angle is 90 degree. H = u²/2g. For max range throw angle is 45 degree. R = u²/g = 100. So H = 100/2 = 50 m.
8. A stone is projected horizontally at 30m/s
from the top of a tower 45m high. At what
angle with horizontal, the stone strikes the
horizontal ground at the same level of the
foot of the tower?
30
45
2 60
90
(b) Time of flight t = √(2h/g) = √(2*45/10) = √9 = 3 s. Vertical velocity at ground vy = gt = 10 * 3 = 30 m/s. Horizontal velocity vx = 30 m/s. Angle with horizontal tanθ = vy / vx = 30 / 30 = 1 => θ = 45°.
9. If maximum height of a projectile is
increased by 10% keeping 0 same, then the
time of flight increases by
5%
10%
15%
20%
(a) Maximum height H = u²sin²θ / 2g. Time of flight T = 2usinθ / g. H ∝ u², T ∝ u. If H increases by 10%, u increases by √1.1 = 1.05 (5%). So time of flight increases by 5%.
10. 10. A stone is thrown horizontally forward at
30m/s from the top of a tower 45m high.
Then displacement of the stone from the
foot of the tower after one second is
10m
20m
40m
50m
(d) Horizontal distance x = ut = 30 * 1 = 30 m. Vertical distance y = 1/2 gt² = 1/2 * 10 * 1² = 5 m. Displacement from foot of tower = √(x² + y²) = √(30² + 5²) = √925 ≈ 30.4 m. There seems to be a mistake in options. Let's calculate displacement from the point of projection. Displacement vector = 30 î - 5 ĵ. Magnitude = √925.
11. 11. A projectile is thrown at angle 0 with
vertical with initial kinetic energy Eo- If air
resistance is neglected the K.E. at highest
point will be
[MOE 2013]
[MOE 2013]
Eccos'8
E.cose
Eosin 0
zero
(c) Angle with vertical θ, so angle with horizontal is 90 - θ. Initial velocity u. E₀ = 1/2 mu². Velocity at highest point = u cos(90 - θ) = u sinθ. KE at highest point = 1/2 m(u sinθ)² = 1/2 mu² sin²θ = E₀ sin²θ. There seems to be a mistake in the options. If θ is angle with horizontal, velocity at highest point = u cosθ. KE = 1/2 m(u cosθ)² = 1/2 mu² cos²θ = E₀ cos²θ.
12. 12. It is possible to project a particle with a
given velocity in two possible ways so as to
make it pass through a point at a distance
r from the point of projection. The product
of times taken to reach this point in the
two possible ways is then proportional to
r
\(r^2\)
1/r
1/(r^2)
(a) Let the two angles of projection be θ and 90 - θ for the same range. Time of flight T = 2usinθ/g. Product of times T₁T₂ = (2usinθ/g) * (2usin(90-θ)/g) = (4u²sinθcosθ)/g² = (2u²sin2θ)/g² = 2R/g. So, the product of times is proportional to the range R.
13. 13. From the top of a tower of height 40m a
ball is projected upwards with a speed of
20m/sec at an angle of elevation of 30.
Then the ratio of the total time taken by
the ball to hit the ground to its time of
flight is (g = 10m/s')
2:1
3:1
3:2
(a) Time of flight T = 2usinθ / g = 2 * 20 * sin30 / 10 = 2 * 20 * (1/2) / 10 = 2 s. Total time to hit ground: s = utsinθ - 1/2 gt². -40 = 20 * (1/2) * t - 1/2 * 10 * t² => -40 = 10t - 5t² => t² - 2t - 8 = 0 => (t - 4)(t + 2) = 0. t = 4 s. Ratio of total time to time of flight = 4/2 = 2:1. Option a.
14. 14. A gun fires two bullets at 60' and 30' with
horizontal. The bullets strike at some
horizontal distance. The ratio of maximum
height for the two bullets is in the ratio
2:1
4:1
(b) Maximum height H = u²sin²θ / 2g. H₁/H₂ = sin²(60) / sin²(30) = (√3/2)² / (1/2)² = (3/4) / (1/4) = 3:1. There might be an error in the options.
15. 15. The maximum range of a gun on
horizontal train is 16km. If g is 10m/s', the
muzzle velocity of the shell will be
400m/s
200m/s
800m/s
256m/s
(a) Maximum range R = u²/g = 16000 m. u² = R * g = 16000 * 10 = 160000. u = √160000 = 400 m/s.
16. 16. A javelin thrown into air at an angle with
the horizontal has a range of 200m. If the
time of flight is 5 second, then the
horizontal component of velocity of the
projectile at the highest point of the
trajectory is
40m/'s
Om/
c. 98m/s
(a) Range R = ux * T. 200 = ux * 5. ux = 200 / 5 = 40 m/s. Horizontal component of velocity remains constant throughout the trajectory. So, at the highest point, the horizontal component of velocity is 40 m/s.
17. 17. A ball as projected upwards from the top
of tower with vertical velocity 50m/s
making an angle 30" with the horizontal.
The height of the tower is 70m. After how
many seconds from the instant of throwing
will the ball reach the ground?
18. 18. The equation of trajectory from a
projectile thrown horizontally from
certain height above earth is y = 45 - 20
where x and y are in metre. Then initial
horizontal velocity is
30m/s
15m/s
10m/s
(d) The equation of trajectory for a projectile thrown horizontally is y = h - (g/(2u²))x². Comparing with y = 45 - 2x², we have h = 45 m and g/(2u²) = 2. Taking g = 10 m/s², 10/(2u²) = 2 => 10 = 4u² => u² = 10/4 = 2.5 => u = √2.5 m/s. There is a mistake in the question. Let's assume y = 45 - 2x². Then 2 = g/(2u²) => 4u² = g => u = √(g/4) = √(10/4) = √2.5. There must be typo in the question's equation. If y = 45 - 2x²/45. Then 2/45 = 10/(2u²) => 4u² = 450 => u² = 112.5 => u = √112.5.
19. 19. From the above question, find the
horizontal range?
a. 20m
45m
30m
(b) Vertical displacement y = 0 at ground. 0 = 45 - 2x² => 2x² = 45 => x² = 22.5 => x = √22.5. Using equation y = 45 - (g/(2u²))x². 0 = 45 - (10/(2*2.5))x² => 0 = 45 - 2x² => x² = 22.5 => x = √22.5.
20. 20. A projectile thrown with speed u at angle 0
with horizontal is moving at right angle to
its initial direction when its velocity is
utane
usine
c. ucote
usece
c
(c) Initial velocity vector v₀ = ucosθ î + usinθ ĵ. Velocity vector at time t, v = ucosθ î + (usinθ - gt) ĵ. v₀ . v = 0 (since they are perpendicular). (ucosθ)(ucosθ) + (usinθ)(usinθ - gt) = 0 => u²cos²θ + u²sin²θ - ugt sinθ = 0 => u² - ugt sinθ = 0 => t = u / (g sinθ). At this time, vertical component of velocity vy = usinθ - g(u/gsinθ) = usinθ - u/sinθ = (usin²θ - u) / sinθ = -u cos²θ / sinθ = -u cosθ cotθ. Velocity vector v = ucosθ î - ucosθ cotθ ĵ. Speed = √(u²cos²θ + u²cos²θ cot²θ) = ucosθ √(1 + cot²θ) = ucosθ cosecθ = ucosθ (1/sinθ) = u cotθ.
21. 21. For what angle of projection with
horizontal, the horizontal range of a
projectile is equal to its maximum vertical
height?
tan (1)
tan (2)
tan (3)
tan (4
(d) Range R = u²sin2θ / g = 2u²sinθcosθ / g. Max height H = u²sin²θ / 2g. Given R = H => 2u²sinθcosθ / g = u²sin²θ / 2g => 2cosθ = sinθ / 2 => tanθ = 4.
22. 22. A projectile thrown at certain angle with
horizontal from horizontal ground is
moving horizontally after 2 seconds and is
moving at 45" after one second. Then the
angle of projection for the projectile is
tan (1)
tan (2)
-tan (3)
tan (4)
(b) Moving horizontally after 2 seconds means vertical velocity is zero at t=2s. vy = u sinθ - gt = 0 => u sinθ = 2g. Moving at 45° after 1 second means vy = vx at t=1s. vx = u cosθ. vy = u sinθ - g. So, u sinθ - g = u cosθ. 2g - g = u cosθ => g = u cosθ. tanθ = u sinθ / u cosθ = 2g / g = 2. θ = tan⁻¹(2). There is a mistake. Moving horizontally after 2 sec means v_y = 0. u sinθ = 2g. Moving at 45 degree after one second means v_y = -v_x. u sinθ - g = -u cosθ. 2g - g = -u cosθ => g = -u cosθ. tanθ = u sinθ / u cosθ = 2g / -g = -2.
23. 23. A ball is thrown vertically upward at
initial velocity 20m/s from roof of a bus
travelling along constant velocity 30m/s
along a straight road. The ball returns to
thrower's hand after the bus has travelled
90cm
120m
150m
180n
(b) Time of flight T = 2u/g = 2 * 20 / 10 = 4 s. Horizontal distance travelled by bus in 4 s = velocity * time = 30 m/s * 4 s = 120 m.
24. 24. A body is projected with velocity v1 from
the point A as shown in the figure. At the
same time, another body is projected
vertically upwards from B with velocity v2.
The point B lies vertically below the
highest point. For both the bodies to
collide, - should be
0.5
(b) For collision, horizontal velocities should be same and vertical positions should be same at same time. v1cosθ = 0 => θ = 90 (vertical projection). But v1 has horizontal component. This question is likely flawed or missing information.
25. 25. The equation of the trajectory of an
oblique projectile is y = V3x - -gx
Here x and y are in metre and g is in m/s'.
The angle of projectile is
90
450
tan V3
(d) Equation of trajectory y = x tanθ - gx² / (2u²cos²θ) = √3x - gx²/45. Comparing, tanθ = √3 => θ = 60°.
26. 26. A projectile is fired with velocity u making
angle 0 with the horizontal. What is the
angular momentum of the projectile at the
highest point about the starting point?
Given the mass of the projectile is m
mu'sin 0cost
mcose
2g
mu coste
(a) At highest point, velocity = u cosθ î. Position vector r = (R/2) î + H ĵ = (u²sin2θ/2g) î + (u²sin²θ/2g) ĵ. Angular momentum L = r × p = r × (mu cosθ î) = [ (u²sin2θ/2g) î + (u²sin²θ/2g) ĵ ] × (mu cosθ î) = (u²sin²θ/2g) mu cosθ (-k̂) = -m u³ sin²θ cosθ / 2g (-k̂). Magnitude = m u³ sin²θ cosθ / 2g = m u (u sinθ) (u cosθ) / 2g = m u (u sinθ) (R g / 2u sinθ) / 2g = m u R / 4.
27. 27. Two projectile are fixed from the same
point with the same speed at angles of
projection 60"and 30 respectively. Which
one of the following is true?
their range will be same
b. their maximum height will be the same
c. their landing velocity will be the same
d. their time of flight will be same
(a) For complementary angles of projection (θ and 90 - θ), the range of a projectile remains the same. Here, 60° and 30° are complementary angles (60 + 30 = 90).
28. 28. The range of a projectile when projected at
an angle 0 degree is R. If the angle of
projection is 20 degree but the range
remains the same the angle will be
[MOE 2009]
100
20
30
40
(c) For the same range, the two angles of projection are θ and 90 - θ. If one angle is 20°, the other angle will be 90° - 20° = 70°. There is a typo in the question, initial angle is θ, and then 20 degree gives same range. So θ = 70 degree.
29. 29.
A projectile's time of flight 'T' is related to
horizontal range by equation gT = 2R
The angle of projection in degrees is:
[TOM 063]
45
90
30
60
(a) T = 2usinθ/g. R = u²sin2θ/g = 2u²sinθcosθ/g. gT = g(2usinθ/g) = 2usinθ. Given gT = 2R => 2usinθ = 4u²sinθcosθ/g => 1 = 2ucotθ/g => tanθ = 2u/g. Also, T = 2usinθ/g => u = gT/(2sinθ). tanθ = 2(gT/(2sinθ))/g = T/sinθ. Using gT = 2R = 2(u²sin2θ/g) = 4u²sinθcosθ/g. gT² = 4u²sinθcosθ. Substitute u = gT/(2sinθ). gT² = 4(gT/2sinθ)² sinθcosθ = 4(g²T²/4sin²θ) sinθcosθ = g²T²cosθ/sinθ = g²T²cotθ. tanθ = g.
30. 30. A ball is thrown horizontally from the
top of a tower. Horizontal component of
its velocity will
[MOE 066]
[MOE 066]
increase
decrease
remain unchanged
first increase and then decrease
(c) In the absence of air resistance, the horizontal component of the velocity of a projectile remains constant throughout its motion.
31. 31. Two bullets A and B are fired horizontally
at the same instant of time with velocity
V. and V, from the same height. If V. is
greater than V, which will reach the
ground first?
[MOE 2065]
[MOE 2065]
A
B
Both A and B simultaneously
Depends upon their masses
(c) The vertical motion of both bullets is independent of their horizontal velocities. Since they are fired from the same height and experience the same acceleration due to gravity, they will both reach the ground simultaneously.
32. 32.
A cricket ball is struck with a K.E of'K' at
an angle of 45' with
the horizontal. The
at the maximum height will be.
[MOE 20621
K
K/2
2K
(b) Initial kinetic energy K = 1/2 mu². At the maximum height, the vertical component of velocity is zero, and the horizontal component is u cos 45° = u/√2. Kinetic energy at maximum height = 1/2 m(u/√2)² = 1/2 m(u²/2) = 1/2 (1/2 mu²) = K/2.
33. K is the horizontal range of projectile for
an angle of projection of 150. For the same
range an another angle of projection will
be
[MOE 2000]
[MOE 2000]
30
750
60
(b) For the same range, two angles of projection are θ and 90° - θ. If one angle is 15°, the other angle will be 90° - 15° = 75°.
34. 34. For a projectile fired at equal inclination to
horizontal and vertical line with velocity u,
the horizontal distance travelled is: [MOE]
u-/2g
u sine/g
none of the above
(a) Equal inclination to horizontal and vertical means the angle of projection is 45°. Horizontal range R = u²sin2θ/g = u²sin(90)/g = u²/g.
35. 35. Two projectile A & B are thrown at a sam
angle with velocity us = Zu, then what is
the relation between their range [TE-02]
RA-RB
2RA RE
RB = 4R
2Ra RA
(c) Range R = u²sin2θ/g. Range is proportional to the square of the velocity. If velocity of B is twice that of A (uB = 2uA), then RB = (2uA)²sin2θ/g = 4uA²sin2θ/g = 4RA.
36. 36. If mass & velocity of a body in a projectile
motion is doubled then linear momentum
becomes
LIE-05
2 times
4 times
1/2 times
constant
(b) Linear momentum p = mv. If mass becomes 2m and velocity becomes 2v, then new momentum p' = (2m)(2v) = 4mv = 4p. There seems to be a mistake in the options. The momentum becomes 4 times.
37. 37. Ball A is dropped vertically B is thrown
horizontally from the same height and a
the same moment, then:
[IE-071
Ball. A reaches the ground first
Ball B reaches the ground first
Ball A reaches the ground with more
velocity
Ball B reaches the ground with more
velocity
(d) Both balls A and B will reach the ground simultaneously because their vertical motion is the same (initial vertical velocity is zero, and they experience the same acceleration due to gravity). However, ball B will have a greater velocity upon striking the ground because it has a horizontal component of velocity in addition to the vertical component.
38. 38. The greatest height to which man can
throw a stone is h. The greatest distance to
which he can throw it will be [BPKIHS-09]
[BPKIHS-09]
h
h
2h
4h
(c) Greatest height h = u²/2g (at 90° projection). Greatest horizontal distance R = u²/g (at 45° projection). Therefore, R = 2h.
39. 39. If the maximum range of a projectile is
four times of its maximum height, then the
angle of projection is equal to [ BPKIHS-07]
[BPKIHS-07]
30
45
60
750
(b) Maximum range R = u²/g. Maximum height H = u²/2g. Given R = 4H => u²/g = 4(u²/2g) => u²/g = 2u²/g which is not possible. Range R = u²sin2θ/g. Max height H = u²sin²θ/2g. R = 4H => u²sin2θ/g = 4(u²sin²θ/2g) => 2sinθcosθ = 2sin²θ => tanθ = 1 => θ = 45°.
40. 40. A plane is moving with 300ms" when it is
just above the observer at a height of 2km.
At what angle with the vertical must he
fire the gun with velocity 400 m/sec
directly to hit the aeroplane? [BPKIHS 02]
[BPKIHS 02]
30
60
450
90
(a) Horizontal velocity of aeroplane = 300 m/s. Let angle with vertical be θ. Horizontal velocity of bullet = 400 sinθ. For direct hit, 400 sinθ = 300 => sinθ = 3/4. θ = sin⁻¹(3/4) ≈ 48.6°. None of the options match. If angle is with horizontal, 400 cosθ = 300 => cosθ = 3/4. θ ≈ 41.4°.
41. 41. A man in a train moving with a constant
velocity drops a ball on the platform. The
path of the ball as seen by an observer
[KU 2015]
standing on the platform is
[KU 2015]
straight line
a circle
a parabola
a helix
(c) As seen by a stationary observer on the platform, the ball has an initial horizontal velocity equal to the train's velocity and an initial vertical velocity of zero. Under the influence of gravity, it will follow a parabolic path.
42. 42. The range of projectile fired at an angle of
15 is 50 m. If it is fired with same speed at
an angle of 450 then, range will be:
[KU 2017]
[KU 2017]
100m
15m
50m
37m
(a) Range R = u²sin2θ/g. R ∝ sin2θ. R₁/R₂ = sin(30°)/sin(90°) = (1/2)/1 = 1/2. 50/R₂ = 1/2 => R₂ = 100 m.
43. 43. A stone is released from a top of tower.
When a constant force is applied by high
speed wind on that stone, then the path it
[IOM 2017]
follows is:
[IOM 2017]
Parabolic
Straight
Hyperbolic
circular
(b) The stone has an initial vertical velocity of zero and is accelerated downwards by gravity. The constant horizontal force due to wind will provide a constant horizontal acceleration. The combination of constant vertical acceleration and constant horizontal acceleration results in a parabolic trajectory.
