NEET PRACTICE TEST (180 Question) For the student who is giving NEET exam.

th
11 NEET

,SKC SIR
PHYSICS
1. A particle starts oscillating simple harmonically from 8. The mass M shown in the figure oscillates in simple
its equilibrium position with time period T. The ratio harmonic motion with amplitude A. The amplitude of
of kinetic energy and potential energy of the particle point P is
at time t = T/12 is
(1) 1 : 2 (2) 2 : 1 (3) 3 : 1 (4) 4 : 1
2. If the displacement x and velocity  of a particle
executing simple harmonic motion are related (1) k1A (2) k 2 A
through the expression 42 = 25 – x2, then its time k2 k1
period is
(3) k1A (4) k 2 A
(1)  (2) 2  (3) 4  ()   k1 + k 2 k1 + k 2
3. The kinetic energy of a particle executing SHM will
be equal to (1/8)th of its potential energy when its 9. Two springs of spring constants k1 and k2 have equal
displacement from the mean position is(where A is maximum velocities when executing simple
the amplitude) harmonic motion. The ratio of their amplitudes
A 2 2
(masses are equal will be)
(1) A 2 (2) (3) 2
A (4) A




R
1/2 1/2
k  k 
2 3 3 (1) k1 (2)  1  (3) k 2 (4)  2 
4. Two particles A and B execute simple harmonic
k2  k2  k1  k1 
5T
motion with periods of T and respectively. They 10. The bob of a simple pendulum executes SHM in




(1) 0 (2)

(3)

4
start simultaneously from mean position. The phase
difference between them when A completes one
oscillation will be-
(4)
2
SI water with a period T, while the period of oscillation
of the bob is T0 in air. Neglecting frictional force of
water and given that the density of the bob is  4  ×

1000 kgm–3. The relationship between T and T0 is
3  

2 4 5 (1) T = T0 (2) T = 4 T0
C
5. The maximum velocity of a particle executing simple T
(3) T = 2 T0 (4) T = 0
harmonic motion with an amplitude 7 mm is 4.4 m/s. 2
The time period of oscillation is
(1) 100 s (2) 0.01 s (3) 10 s (4) 0.1 s 11. A sound wave travelling through a medium of bulk
modulus B is represented as
SK

6. The equation of motion of two particles executing y(x, t) = Asin(kx – t) where symbols have their usual
S.H.M. are meanings. Then, the corresponding pressure
y1 = 10sin 10t +   m, y2 =10cos  8t +   amplitude is
 3   4  (1) BAk (2) B(A/k)1/2
The phase difference between these particles at (3) B (4) B(Ak)1/2
t = 0.5 s is
12. A string of linear density 0.2 kg/m is stretched with a
7 13 25  force of 500 N. A transverse wave
(1) (2) (3) (4)
12 12 12 12 of length 4.0 m is set up along it. The speed of wave
7. Two simple harmonic motions are represented by the is
(1) 50 m/s (2) 75 m/s
equations x1 = 10sin  3t +   and (3) 150 m/s (4) 200 m/s
 4
13. A uniform rope of mass 0.1 kg and length 2.45 m
x2 = 5(sin3t + 3 cos3t). Their amplitudes are in hangs from a ceiling. The time taken by a transverse
the ratio of wave to travel the full length of the rope is (g = 9.8
(1) 1: 2 (2) 1 : 1 (3) 2 : 3 (4) 2 : 1 m/s2)
(1) 1 s (2) 2 s (3) 3 s (4) 4 s




SKC SIR

,14. Two sound wave travel in the same direction in a 21. String A has length L, radius of cross-section r,
medium. The amplitude of each wave is A and the density of material  and is under tension T. String B
phase difference between the two waves is 120°. The has all these quantities double those of string A. If A
resultant amplitude will be and B are the corresponding fundamental
(1) 2 A (2) 2A (3) 3A (4) A frequencies of the vibrating string, then
(1) A = 2B (2) A = 4 B
15. Assertion: The percentage change in time period is (3) B = 4 A (4) A = B
1.5%, if the length of simple pendulum increases by
3%. 22. The shape of a string on which standing waves are
Reason : Time period is directly proportional to produce at t = 0 is shown here. The suitable equation
length of pendulum.
of standing wave can be
(1) If both assertion and reason are true and the reason
is the correct explanation of the assertion.
(2) If both assertion and reason are true but reason is
not the correct explanation of the assertion.
(3) If assertion is true but reason is false.
(4) If the assertion and reason both are false. (1) y = A sin kx cos t
(2) y = A sin kx sin t
16. The stationary wave y = 2a sinkx cost in a closed (3) y = A cos kx sin t
organ pipe is the result of the superposition of y1 = (4) y = A cos kx cos t
asin(t – kx) and
(1) y2 = – a cos(t – kx) 23. A wave pulse is generated in string that lies along x –
(2) y2 = – a sin(t + kx) axis. At the points A and B, as shown in figure, if RA
(3) y2 = a sin(t – kx) and RB are ratio of wave speed to the particle speed
(4) y2 = a cos(t + kx) respectively then

17. A closed organ pipe and an open pipe of same length
produce 4 beats when they are set into vibrations
simultaneously. If the length of each of them were
twice their initial lengths, the number of beats
produced will be [Assume same mode of vibration in
both cases] (1) RA > RB
(1) 2 (2) 4 (3) 1 (4) 8 (2) RB > RA
(3) RA = RB
18. For a wave displacement amplitude is 10–8 m, density
(4) Information is not sufficient to decide
of air 1.3 kg m–3, velocity in air 340 ms–1 and
frequency is 2000 Hz. The intensity of wave is 24. The potential energy of a particle of mass 1 kg in
(1) 5.3 × 10–4 Wm–2 (2) 5.3 × 10–6 Wm–2 motion along the x − axis is given by: U = 4(1 −
(3) 3.5 × 10–8 Wm–2 (4) 3.5 × 10–6 Wm–2 cos 2x)J, where x is in metres. The period of small
oscillations (in sec) is:
19. If the length of a stretched string is shortened by 40%
and the tension is increased by 44%, then the ratio of (1) 2π (2) π
π
the final and initial fundamental frequencies is (3) (4) √2π
2
(1) 3 : 4 (2) 4 : 3 (3) 1 : 3 (4) 2 : 1
25. The period of small oscillation of a simple pendulum
20. A wall clock uses a vertical spring-mass system to is T. The ratio of density of liquid to the density of
measure the time. Each time the mass reaches an material of the bob is ρ(ρ < 1). When immersed in the
extreme position, the clock advances by a second. The liquid, the time period of small oscillation will now
clock gives correct time at the equator. If the clock is be
taken to the poles it will (1) T (2) T(1 − ρ)
(1) run slow (2) run fast T
(3) stop working (4) give correct time (3) (4) T√1 − ρ
√1−ρ

, 26. Motion of an oscillating liquid column in a 2πt
(1) x(t) = B sin ( 30 )
U-tube is πt
(1) Periodic but not simple harmonic (2) x(t) = B cos (15)
πt π
(2) Non-periodic (3) x(t) = B sin (15 + 2 )
(3) Simple harmonic and time period is independent πt π
(4) x(t) = B cos (15 + 2 )
of the density of the liquid
(4) Simple harmonic and time period is directly
31. In this question, a statement of assertion (A) is
proportional to the density of the liquid
followed by a statement of reason (R). Mark the
27. The speed (v) of a particle moving along
correct choice as:
a straight line, when it is at a distance (x)
Assertion: If a pendulum is suspended in a lift and
from a fixed point on the lines is given by
lift is falling freely, then its time period becomes
v 2 = 144 − 9x 2 . Which of the following infinite.
statement(s) is/are correct? Reason: Free falling body has acceleration equal to
(1) The magnitude of acceleration at a distance 3 acceleration due to gravity.
units from the fixed point is 27 units. (1) If both assertion and reason are true and reason is
2π
(2) The motion is simple harmonic with T = 3 units. the correct explanation of assertion.
(3) The maximum displacement from the fixed point (2) If both assertion and reason are true but reason is
is 4 units. not the correct explanation of assertion.
(4) All of the above (3) If assertion is true but reason is false.
28. A solid ball of mass m is made to fall from a height H (4) If both assertion and reason are false.
on a pan suspended through a spring of spring
constant K as shown in figure. If the ball does not 32. The displacement of a linear harmonic oscillator is
rebound and the pan is massless, then amplitude of given by x = A cos t. The curves showing the
oscillation is variation of the potential energy with t and x (see
figure) are displayed respectively by:



mg mg 2HK 1/2
(1) K
(2) k
(1 + mg
)
mg 2HK 1/2 mg 2HK 1/2
(3) +( ) (4) [1 + (1 + ) ]
K mg K mg
29. A point mass is subjected to two simultaneous
sinusoidal displacements in x-direction, x1 (t) = A sin
2π
t and x2 (t) = A sin(ωt + 3 ). Adding a third
sinusoidal displacement x3 (t) = B sin (t + ɸ) brings
the mass to a complete rest. The values of B and ɸ are
3π 4π
(1) √2A, (2) A,
4 3
5π π
(3) √3A, 6 (4) A, 3 (1) I and III (2) I and IV
30. Figure shows the circular motion of a particle. The (3) II and III (4) II and IV
radius of the circle, the period, sense of revolution and 33. A particle of mass m is allowed to oscillate near the
the initial position are indicated on the figure. The point of minima of a vertical parabolic path having
simple harmonic motion of the x-projection of the the equation x2 = 4ay, where x - axis is along the
radius vector of the rotating particle P is: horizontal direction and y - axis is along the vertical
direction. The angular frequency of small oscillations
of the particle is
8g 2g g g
(1) √ (2) √ (3) √ (4) √
a a a 2a