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Physics NEET MCQ
Quiz 10
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Q.1
A source of sound S emitting waves of frequency 100 Hz and an observer O are located at some distance from each other. The source is moving with a speed of 19.4 ms-1 at an angle of 60° with the source observer line as shown in the figure. The observer is at rest. The apparent frequency observed by the observer (velocity of sound in air 330 ms-1)is :-
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a) 97 Hz
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b) 100 Hz
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c) 103 Hz
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d) 106 Hz
Explanation
Component of Vs along observer = Vscosθ =19.4 × cos60 =9.7 Using formula for source moving to wards observer at rest Answer:c)
Q.2
A string is stretched between fixed points separated by 75.0 cm. It is observed to have resonant frequencies of 420 Hz and 315 Hz. There are no other resonant frequencies between these two. The lowest resonant frequencies for this string is :
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a) 105 Hz
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b) 155 Hz
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c) 205 Hz
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d) 10.5 Hz
Explanation
f1 is the minimum resonant frequency nf1 = (n-1)f1 nf1 = 420 …(i) (n-1)f1 = 315 …(ii) From (ii) and (i) we get f1 =420 -315 = 105 Hz Answer:(a)
Q.3
A siren emitting a sound of frequency 800 Hz moves away from an observer towards a cliff at a speed of 15ms–Then, the frequency of sound that the observer hears in the echo reflected from the cliff is : (Take velocity of sound in air = 330 ms–1) [AIPMT 2016]
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a) 765 Hz
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b) 800 Hz
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c) 838 Hz
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d) 885 Hz
Explanation
Asked for echo, that is the frequency heard by a observer on cliff General equation V0 =0 as observer is stationary Vs = -ve as source is moving towards the observer Answer:(c)
Q.4
A uniform rope of length L and mass m1 hangs vertically from a rigid support. A block of mass m2 is attached to the free end of the rope .A transverse pulse of wavelength λ1 is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope is 位The ratio λ2/λ1 is : ..[ AIMPT 2015]
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a)
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b)
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c)
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d)
Explanation
Here ρ is mass per unit length and T is tension Now Tension at both end is different T1 = m2 g and T2 = (m2 +m1)g Answer:(b)
Q.5
An air column, closed at one end and open at the other, resonates with a tuning fork when the smallest length of the column is 50 cm. The next larger length of the column resonating with the same tuning fork is :
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a) 66.7 cm
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b) 100 cm
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c) 150 cm
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d) 200 cm
Explanation
When pipe is closed at bottom during resonance antinode is form at the open end. Such antinode can be formed if length of pipe is increased for three times as shown in figure. Thus next length = 3L = 150 cm Answer:(c)
Q.6
A body of mass m is attached to the lower end of a spring whose upper end is fixed. The spring has negligible mass. When the mass m is slightly pulled down and released, it oscillates with a time period of 3s. When the mass m is increased by 1 kg, the time period of oscillations becomes 5 s. The value of m in kg is … [ NEET II – 2016]
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a) 16/9
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b) 9/16
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c) 3/4
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d) 4/3
Explanation
9m + 9=25m 16m = 9 ∴ m =9/16 Answer:(b)
Q.7
The second overtone of an open organ pipe has the same frequency as the first overtone of a closed pipe L metre long. The length of the open pipe will be …[ NEET II -2016]
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a) L/2
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b) 4 L
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c) L
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d) 2 L
Explanation
Frequency of Second overtone open pipe = Frequency First overtone over tone of close pipe L' = 2L Answer:(d)
Q.8
Three sound waves of equal amplitudes have frequencies (n – 1), n, (n + 1). They superimpose to give beats. The number of beats produced per second will be .. [NEET II 2016]
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a) 3
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b) 2
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c) 1
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d) 4
Explanation
Number of beats produced per second is equal to the reciprocal of time interval between two successive minima = f1-f2 We can get three combinations (n-1) and (n) thus number of beats = n-(n-1) = 1Hz (n) and (n+1) thus number of beats = (n+1)-(n) = 1Hz (n-1) and (n+1) thus number of beats = (n+1)-(n-1) = 2Hz Net beat frequency = LCM of individual beat frequencies = 2 Hz So, no. of beats per second = 2 Answer:(b)
Q.9
A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is …[ NEET 2016]
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a)
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b)
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c)
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d)
Explanation
a = -ω2y a = -ω2× 2 Answer:(c)
Q.10
Two cars moving in opposite directions approach each other with speed of 22 m/s and 16.5 m/s respectively. The driver of the first car blows a horn having a frequency 400 Hz. The frequency heard by the driver of the second car is [velocity of sound340 m/s] …[ NEET 2017]
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a) 350 Hz
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b) 361 Hz
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c) 411 Hz
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d) 448 Hz
Explanation
Observer moving towards source Vo is +Ve = 16.5 m/s , Source is moving towards or approaching observer Vs is –Ve = -22 m/s f'=400×1.12=448 Hz Answer:(d)
Q.11
A spring of force constant k is cut into lengths of ratio1 : 2 :They are connected in series and the new force constant is k’. Then they are connected in parallel and force constant is k”. Then k’ : k” is …[NEET 2017]
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a) 1 : 6
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b) 1 : 9
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c) 1 : 11
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d) 1 : 14
Explanation
If total length is 6l and have spring constant k k1 ∝6k,k2 ∝ 3k; k3 ∝ 2k For Series combination k’= k as Parallel combination:| k” = 6k+3k+2k = 11k ∴ k’:k” = k:11k = 1:11 Answer:(c)
Q.12
The two nearest harmonics of a tube closed at one end and open at other end are 220 Hz and 260 Hz.What is the fundamental frequency of the system?
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a) 10 Hz
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b) 20 Hz
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c) 30 Hz
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d) 40 Hz
Explanation
nf1 = 220 (n+2) f1 = 260 nf1 +2f1 = 260 220 + 2f1 = 260 f1 = 20 Answer:(a)
Q.13
The amplitude of a damped oscillator decreases to 0.9 times its original magnitude in 5s. In another 10s it will decrease to α times its original magnitude, where α equals …[ IIT Mains 2017]
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a) 0.7
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b) 0.81
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c) 0.729
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d) 0.6
Explanation
A = Ao e−kt 0.9A0=A0 e(-5k) 0.9 =e(-5k) ln0.9 =-5k …(i) αA0 = A0 e(-15k) α=e(-15k) lnα = -15k …(ii) Multiply (i) by 3 3ln0.9 = -15k …(iii) from (ii) and (iv); 3ln0.9 = lnα ln(0.9)3 = lnα ∴ α = (0.9)3 = 0.729 Answer:(c)
Q.14
A sonometer wire of length 1.5 m is made of steel. The tension in it produce an elastic strain of 1%.What is the fundamental frequency of steel if density and elasticity of steel are 7.7 × 103 kg / m3 and 2.2 × 1011 N/m2 respectively ?
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a) 188.5 Hz
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b) 178.2 Hz
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c) 200.5 Hz
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d) 770 Hz
Explanation
Elastic strain = 2.2 × 1011 N/m2 Tension = strain of 1% = 2.2 × 1011 ×(1/100) = 2.2× 109 Answer:(b)
Q.15
An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass M. The piston and the cylinder have equal cross sectional area A. When the piston is in equilibrium, the volume of the gas is V0 and its pressure is PThe piston is slightly displaced from the equilibrium position and released. Assuming that the system is completely isolated from its surrounding, the piston executes a simple harmonic motion with frequency …[ IIT mains 2013]
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a)
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b)
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c)
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d)
Explanation
Process is adiabatic If small displacement of piston is x the ΔV = Ax, A is constant Δp = change in pressure = F/A =Ma/A Comparing above equation with standard equation for SHM we get Answer:(c)
Q.16
than one correct answer Q291) A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation, y(x, t) = (0.01 m) sin [(62.8 m−1)x]. cos[(628 s−1)t]. Assuming π = 3.14,the correct statement(s) is (are)
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a) The number of nodes is 5.
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b) The length of the string is 0.25 m.
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c) The maximum displacement of the midpoint of the string, from its equilibrium position is 0.01 m.
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d) The fundamental frequency is 100 Hz.
Explanation
Option (a):Fifth over tone = 5f1 hus number of nodes =6 option “a” wrong Standard equation for standing wave y=(2a)cos(kx)sin(ωx)…(i) Comparing (i) with given equation K =62.8 and 𝜔 =628 thus λ = 10 and f = 100 Option (b) For One harmonic L = λ/2 Thus fifth harmonic L =5λ/2= 5/2 10=25 m option “b” wrong Option (c) : Amplitude is maximum when cos(kx)=1 Thus 0.01 m is maximum amplitude option “c” correct Option (d): f = 100 option “d” correct Answer:(c, d )
Q.17
than one correct answer Q292) Two vehicles, each moving with speed u on the same horizontal straight road, are approaching each other. Wind blows along the road with velocity w. One of these vehicles blows a whistle of frequency fAn observer in the other vehicle hears the frequency of the whistle to be fThe speed of sound in still air is V. The correct statements(s) is (are)
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a) If the wind blows from the observer to the source, f2 > f1.
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b) If the wind blows from the source to the observer, f2> f1.
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c) If the wind blows from observer to the source, f2 < f1.
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d) If the wind blows from the source to the observer, f2 < f1.
Explanation
Source and observer both moves towards each other If wind blows in the direction of source v-w is velocity of sound w.r.t air u+w velocity of observer w.r.t air u-w velocity of source w.r.t air If wind blows in the opposite to direction of source v+w is velocity of sound w.r.t air u-w velocity of observer w.r.t air u+w velocity of source w.r.t air Thus f2 is independent of direction of wind f2 > f1 irrespective of direction of wind Answer:(a, b)
Q.18
A particle of mass m is attached to one end of a mass-less spring of force constant k, lying on a frictionless horizontal plane. The other end of the spring is fixed. The particle starts moving horizontally from its equilibrium position at time t = 0 with an initial velocity u0.When the speed of the particle is 0.5 u0, it collides elastically with a rigid wall. After this collision,
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a) the speed of the particle when it returns to its equilibrium position is u0.
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b) the time at which the particle passes through the equilibrium position for the first time is
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c) the time at which the maximum compression of the spring occurs is
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d) the time at which the particle passes through the equilibrium position for the second time is
Explanation
The impulse of the wall merely reverses the direction of motion is thought changing particle speed. Hence option “a” correct From the formula of velocity of oscillator When v is max x = 0 thus u0= ωA Oscillator started its motion from equilibrium thus equation velocity by taking derivative of x Given that oscillator collides elastically when velocity was u0/2 “t” is the time when collision takes place and will take same time to pass through equilibrium from point of collision = 2t option “b” is wrong Oscillator have periodic time thus oscillator when reach another extreme end from equilibrium=T/4, it will take time t|1 Thus when for maximum compression total time is t+t1 Option c is wrong When oscillator passes through equilibrium from compression time t4 Time taken to pass through equilibrium second time Option “d” correct Answer:(a, d)
Q.19
One end of a taut string of length 3m along the x axis is fixed at x =The speed of the waves in the string is 100 ms−The other end of the string is vibrating in the y direction so that stationary waves are set up in the string. The possible waveform(s) of these stationary waves is (are) .. [ IIT Advance 2014]
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a)
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b)
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c)
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d)
Explanation
We have to verify from the options v=ω/k Option a , b, c , d have velocity = 100 m/s Node is point where the values of Amplitude is zero If by substituting x =3 in above sin values we can find node creates at X=3 Option a Option b Option c Option d Answer:(b)
Q.20
Two loudspeakers M and N are located 20 m apart and emit sound at frequencies 118 Hz and 121 Hz, respectively. A car is initially at a point P, 1800 m away from the midpoint Q of the line MN and moves towards Q constantly at 60 km/hr along the perpendicular bisector of MN. It crosses Q and eventually reaches a point R, 1800 m away from Q. Let v(t) represent the beat frequency measured by a person sitting in the car at time t. Let vP, vQ and vR be the beat frequencies measured at locations P, Q and R, respectively. The speed of sound in air is 330 m/s. Which of the following statement(s) is(are) true regarding the sound heard by the person?
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a) The plot below represents schematically the variation of beat frequency with time
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b) vP + vR = 2vQ
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c) The plot below represents schematically the variation of beat frequency with time
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d) The rate of change in beat frequency is maximum when the car passes through Q.
Explanation
When listener is at point P Source is stationary and listener is moving. Velocity of listener to wards the source is V0cosθ Number of beats n= f” – f’ As listener moves to wards the point Q angle θ increases thus number of beats starts to reduce with time At point Q angle θ =90° , n = 3 Frequency at point R n=f"-f’ As listener moves away from Q towards point R , θ starts to decrease, cosθ increases and beets will decrease Option “a” correct Now from (i) and (ii) Option “b” is correct Taking derivative of (i) with θ Rate of change of beats is maximum when θ =90°, it is at point Q Answer:(a, b)
Q.21
A block with mass M is connected by a massless spring with stiffness constant k to a rigid wall and moves without friction on a horizontal surface. The block oscillates with small amplitude A about an equilibrium position xConsider two cases : (i) when the block is at x0; and (ii) when the block is at x = x0 + A. In both the cases, a particle with mass m (< M) is softly placed on the block after which they stick to each other. Which of the following statement(s) is (are) true about the motion after the mass m is placed on the mass M? [IIT Advance 2016]
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a) The amplitude of oscillation in the first case changes by a factor of where as in the second case it remains unchanged.
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b) The final time period of oscillation in both the cases is same
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c) The total energy decreases in both the cases
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d) The instantaneous speed at x0 of the combined masses decreases in both the cases
Explanation
From law of conservation of momentum Mv= (M+m) v’ For case I maximum velocity at x0 ( equilibrium) v= ωA Case II Mass m is kept at extreme position when velocity is zero. Thus momentum does not change and it oscillate with same amplitude. (Option a is correct) Option b Angular velocity ω in both case is same thus period is same. Option b correct Option c In first case amplitude is less hence total energy is less than case II Option d Instantaneous speed in case I And in case II v=ω’A Answer:(a, b, d )
Q.22
A block M hangs vertically at the bottom end of a uniform rope of constant mass per unit length. The top end of the rope is attached to a fixed rigid support at O. A transverse wave pulse (Pulse 1) of wavelength λ0 is produced at point O on the rope. The pulse takes time TOA to reach point A. If the wave pulse of wavelength 位0 is produced at point A (Pulse 2) without disturbing the position of M it takes time TAO to reach point O. Which of the following options is/are correct ? [ IIT Advance 2017]
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a) The time TAO = TOA
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b) The velocities of the two pulses (Pulse 1 and Pulse 2) are the same at the midpoint of rope
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c) The wavelength of Pulse 1 becomes longer when it reaches point A
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d) The velocity of any pulse along the rope is independent of its frequency and wavelength.
Explanation
Speed of the wave is property of medium thus time will be same (Option a correct) Speed is maximum at O and minimum at A thus will have same Speed, but velocity is a vector, directions are different (Option b is wrong) since tension is maximum at O velocity is maximum. .Frequency is property of source and will remain same irrespective of velocity. Thus. Wave length is maximum at O and minimum at A ( Option c wrong) Velocity of wave depends on the tension and mass per-unit length, thus independent of wavelength and frequency (Option d correct) Answer:(a, d)
Q.23
The maximum velocity of a particle executing simple harmonic motion with an amplitude 7mm is 4.4m/s. The period of oscillation is ..[ AIEEE 2006]
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a) 0.01 s
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b) 10 s
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c) 0.1s
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d) 100s
Explanation
Maximum velocity vmax=aω vmax=a(2π /T) T=2πa / vmax T=2×3.14×7×10⁻³ / 4.4=0.01s Answer: (a)
Q.24
The displacement of an object attached to a spring and executing simple harmonic motion is given by x=2×10⁻²cosπt meter. the time at which the maximum speed first occurs is .. [ AIEEE 2007]
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a) 0.25s
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b) 0.5s
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c) 0.75s
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d) 0.125s
Explanation
Here, x=2×10⁻²cosπt speed is given by v=dx/dt=2×10⁻²πsinπt For the first time, the speed to be maximum sinπt=1 or sinπt=sinπ/2 ∴ πt=π/2 or t=1/2=0.5sec Answer: (b)
Q.25
A vibrating tuning fork of frequency n is placed near the open end of long cylindrical tube. the tube has a side opening and is fitted with a movable reflecting piston. As the piston is moved through 8.75cm, the intensity of sound changes from a maximum to minimum. If the speed of sound is 350 metre per second, then n is
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a) 500Hz
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b) 600Hz
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c)1000Hz
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d)2000Hz
Explanation
Stationary waves are formed in the cylinder. then distance between a node and antinode is λ/4=8.75 cm or λ=35.0 cm n=v / λn=350 / (35×10⁻²) =1000 HzAnswer: (c)
Q.26
A thin plane membrane separates hydrogen at 27° from hydrogen at 127°C, both being at the same pressure. A plane sound wave enters from the colder to the hotter side. If the angle of incidence on the membrane is 30°, then the angle of refraction is ..
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a) sin⁻¹ (1/√3)
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b) sin⁻¹(2/3)
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c)sin⁻¹(3/8)
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d)sin⁻¹(2/8)
Explanation
We know that v1 /v2=√T1 / √T2 And refractive index sini/ sinr=v1 /v2∴sini/ sinr=√T1 / √T2 sin30/sinr=√300 / √400 Or sinr=sin30 × (√400 /√300) sinr=(1/2)× ( √4 / √3)sinr=1/√3r=sin⁻¹ (1 / √3) Answer:(a)
Q.27
A stretched string of one metre length, fixed at both ends, having a mass of 5×10⁻⁴ kg is under a tension of 20 newtons. It is plucked at a point situated at 25cm from one end. The stretched string would vibrate with a frequency of [ CPMT 1988]
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a) 400 Hz
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b) 100 Hz
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c)200 Hz
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d)256 Hz
Explanation
velocity of wave in string v=√ (T/m)=√ (20/(5×10⁻⁴)=200 m/secNow λ/4=25 λ=100 cm=1 m Now n=v/λ=200 /1=200 Hz Answer:(c)
Q.28
A student is performing an experiment using a resonance column and a tuning fork of frequency 244 s−He is told that the air in the tube has been replaced by another gas(assume that the column remains filled with the gas). If the minimum height at which resonance occurs is (0.350 ± 0.005) m, the gas in the tube is…[ IIT Advance 2014] molar masses M in grams are given in the options. Take the values of √(10/M)for each gasas given there)
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a)
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b)
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c)
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d)
Explanation
λ=4×(0.035±0.005) = 0.14±0.02 v=λf = (0.14±0.02) 244 = 341.6 ± 4.88 m/s Velocity of sound in gas Option a: Neon M = 20×10⁻³ kg/mole and γ = 5/3 Option “a” wrong Option b: Nitrogen M =28×10⁻³ kg/mole and γ = 7/5 ( as rigid rotator) Option “b” wrong Option c: oxygen M =32×10⁻³ kg/mole and γ = 7/5 ( as rigid rotator) Option “c” wrong Option d : Argon M =36×10⁻³ kg/mole and γ = 5/3 Option “d” is correct Answer:(d)
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