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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 401. |
Explain red shift and blue shift in Doppler Effect. |
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Answer» If the spectral lines of the star are found to shift towards red end of the spectrum (called as red shift) then the star is receding away from the Earth. Similarly, if the spectral lines of the star are found to shift towards the blue end of the spectrum (called as blue shift) then the star is approaching Earth. |
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| 402. |
Two cars, one in the front of the other, are traveling down the highway at 25 m/s. The car behind sounds it's own horn, which has a frequency of 500Hz. What is the frequency heard by the driver of the lead car? (Sound = 340 m/s) |
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Answer» According to doppler effect, f' \(=(\frac{\nu+\nu_0}{\nu-\nu_0})f\) where f = real frequency = 500 Hz f' = apparent frequency \(\nu\) = 340 m/s \(\nu_0\) = 25 m/s f' \(=(\frac{340+25}{340-25})\times500\) f' = 365/315 x 500 f' = 579.365 Hz |
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| 403. |
A stationary source is emitting sound at a fixed frequency `f_(0)`, which is reflected by two cars approaching the source. The difference between the frequencies of sound reflected from the cars is `1.2%` of `f_(0)`. What is the difference in the speeds of the cars (in km per hour) to the nearest integer ? The cars are moving at constant speeds much smaller than the speed of sound which is `330 ms^(-1)`.A. 2B. 3C. 5D. 7 |
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Answer» Correct Answer - D `f_(app)=f_(0)""(c+v )/(c-v)` , where `c` is speed of sound `rArr" "df=(2f_(0)c)/((c-v)^(2))dv~~(2f_(0)c)/((c-v)^(2))dv=(2f_(0))/(c)dv" "`...(i) Given : `df=(1.2)/(100)f_(0) " "`...(ii) From eqs. (i) and (ii), `dv ~~2m//s=7km//h` |
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| 404. |
What is the distance between a node and an antinode in a stationary wave ? |
| Answer» The distance between node and antinode is `( lambda)/(4)` | |
| 405. |
In a stationary wave pattern that forms as a result of reflection pf waves from an obstacle the ratio of the amplitude at an antinode and a node is `beta = 1.5.` What percentage of the energy passes across the obstacle ? |
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Answer» We know that, `(A_(max))/(A_("min"))=(A_(i)+A_(r))/(A_(i)-A_(r))=(3)/(2)` (given) `2A_(i)+2A_(r)=3A_(i)-3A_(r)` `2A_(r)+3A_(r)=3A_(i)-2A_(i)` `5A_(r)=A_(i)` `(A_(r))/(A_(i))=(1)/(5)` `(l_(r))/(l_(i))=((A_(r))/(A_(i)))^(2)=((1)/(5))^(2)=(1)/(25)` `l_(r)=0.04l_(i)` `because4%` of the incidennt energy is reflected or 96% energy passes across the obstacle. |
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| 406. |
A wave `y = a sin (omegat - kx)` on a string meets with another wave producing a node at `x = 0`. Then the equation of the unknown wave isA. (a) `y = a sin (omegat + kx)`B. (b) `y = -a sin (omegat + kx)`C. ( c ) `y = a sin (omegat - kx)`D. (d) `y = -a sin (omegat - kx)` |
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Answer» Correct Answer - B (b) To form a node there should be superposition of this wwave with the reflected wave. The reflected wave should travel in opposite direction with a phase change of `pi`. The equation of the reflected wave will be `y = a sin (omegat + kx +pi)` rArr `y = -a sin (omegat + kx)` |
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| 407. |
A stretcded sting resomates with tuning fork of frequency 512 Hz. When length o fthe string is 0.5 m. the length of the string required to vibrate resonantly with a tuning fork of frequeny 256 Hz would beA. 0.25 mB. 0. 5 mC. 1 mD. 2 m |
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Answer» Correct Answer - c |
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| 408. |
A train , standing at the outer signal of a railway station blows a whistle of frequency 400 Hz in still air, (i) What is thefrequency of the whistle for a platform observer when the train (a) approaches the platform with a speed of `10ms^(-1)`, (b) recedes from the platform with a speed of `10 ms^(-1)` ? (ii) What is the speed of sound in still air can be taken as `340 ms^(-1)` . |
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Answer» (i) Here, `v = 400 Hz, v = 340 m//s ` (a) Train approaches the platform `v _(s) 10 m//s ` `v^(1)= ( v)/( v-v_(s)) = ( 340 xx 400)/( 340- 10) = 412.12 Hz` (b) Train receds from the platform `v_(s) = 10 m //s ` `v^(1)= ( v)/( v-v_(s)) = ( 340 xx 400)/( 340+ 10) ` `= 388.6 Hz` (ii) The speed of sound in each case is the same =` 340 m //s ` |
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| 409. |
A wave travelling along the x-axis is described by the equation y (x, t) = 0.005 sin (`alphax - betat`). If the wavelength and time period of the wave are 0.08 m and 2 s respectively, then `alpha, beta` in appropriate units areA. `alpha=25pi,beta=pi`B. `alpha=0.08/pi,beta=2/pi`C. `alpha=0.04/pi,beta=1/pi`D. `alpha=12.5pi,beta=pi/2` |
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Answer» Correct Answer - A Here, `lamda=0.08m,T=2 s` `y(x,t)=asin(kx-omegat)` We get,`alpha=k=(2pi)/lamda=(2pi)/0.08=25pi` `beta=omega=(2pi)/T=(2pi)/2=pi` |
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| 410. |
A travelling harmonic wave on a string is described by y(x, t) = 7.5 sin (0.005 x +12t + π /4) 1. what are the displacement and velocity of oscillation of a point at x = 1 cm, and t = 1 s? Is this velocity equal to the velocity of wave propagation? 2. Locate the points of the string which have the same transverse displacements and velocity as the x a 1 cm point at t = 2 s, 5 s, and 11s. |
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Answer» Given: y (x, t) = 7.5 sin (0.005 x + 12t + π /4) 1. At x=1 cm and t = 1s y (1, 1)= 7.5 Sin(0.005 +12 + π /4) = 7.5 sin (12.005 + π /4) = 1.67 cm Velocity of oscillation : v = \(\frac{d(Y (x,t))}{dt}\) = d/dt (7.5 sin (0.005 x + 12t + π/4) dt = 7.5 × 12 cos (0.005 x+ 12t + π/4) At x = 1 cm and t = 1 s v = 7.5 × 12 cos(0.005 + 12 + π/4) = 87.75 cm s-1 We know that velocity of wave propagation = w/k Here w = 12 s-1 and k = 0.005 cm-1 ∴ Velocity of wave propagation = \(\frac{12s^{-1}}{0.005cm^{-1}}\) = 24 ms-1 ∴ At x = 1 cm and t = 1 s velocity of oscillation is not equal to velocity of wave propagation. 2. In a wave, all the points which are separated by a distance ± λ, ±2λ ……..from x = 1 cm will have same transverse displacements and velocity. For the given wave , λ= \(\frac{2π}{0.005}\) = ±12.56 cm, +25.12 m….From x = 1 cm will have the same displacements and velocity as at x = 1 cm, t = 2s, 5s and 11 s. |
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| 411. |
A wave travelling along a string is described by, y(x,t) = 0.005 sin ( 80.0x - 3.0 x). In which the numerical constants are in SI units ( 0.005m, 80.0 rad `m^(-1)` , and 3.0 rad `s^(-1)`) . Calculate (a) the amplitude , ( b) the wavelength , and (c ) the period and frequency of the wave. Also,calculate the displacement of the wave. Also, calaculate the displacement y of the wave at a distance x = 30.0 cm and time t =20s ? |
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Answer» On comparing this displacement equation with Eq. y ( x,t) n =a sin (kx`- omegat+phi) y(x,t)=a sin (kx-omegat)` We find (a) the amplitude of the wave is 0.005m=5mm. (b) the angulare wave number k and angular frequency `omega `are k `= 80.0m^(-1)` and `omega=3.0 s^(-1)` We then relate the wavelength `lambda` to k throughEq. `lambda= ( 2pi )/( K )` `= ( 2pi)/( 80.0 m^(-1))= 7.85cm` (c ) Now we relate T to `omega ` by the relation `T = ( 2pi)/( omega)` `= ( 2pi)/( 3.0s^(-1))` `=2.09 s` and frequency ,`v= ( 1)/(T ) = 0.48 Hz` The displacement y at x= 30.0 cmand time t = 20s is given by `y = ( 0.005m ) sin (80.0 xx0.3-3.0 xx20)` `= (0.005m) sin ( - 36+ 12pi)` `=( 0.005m ) sin (1.699)` `= ( 0.005m ) sin( 97^(@)) ~= 5mm` |
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| 412. |
An open pipe and a closed pipe have the same fundamental frequency. Explain how their lengths are related? |
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Answer» Fundamental Frequency of an open pipe f1 = v/2t1 Fundamental Frequency of an closed pipe f2 = v/4l2 Given f1 = f2 ∵ 2l1 = 4l2 l1/l2 = 2/1 ⇒ l1:l2 = 2:1. |
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| 413. |
An open organ pipe is excited to vibrate in the thid overtone. It is observed that there areA. Three nodes and three antinodesB. Three nodes and four antinodesC. Four nodes and five antinodesD. Four nodes and four antinodes |
| Answer» Four nodes and five antiondes | |
| 414. |
What are nodes and antinomies in a stationary wave? |
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Answer» The points in a stationary wave where the amplitude of vibration of the particles zero are called nodes The points in a stationary wave where the amplitude of vibration of particles is maximum are called antinodes. |
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| 415. |
The distance between any two adjacent nodes in a stationary wave is 15 cm. if the speed of the wave is 294 ms/, what is its frequency? |
| Answer» Correct Answer - 980 Hz | |
| 416. |
Statement-1: In stationary waves nodes are permanently at rest, so no energy is transmitted acroo them. And Statement-2: Nodes are points of maximum pressure.A. Statement-1 is true, statement-2 is true, statement-2 is a correct explanation for statement-1B. Statement-1 is true, statement-2 is true, statement-2 is not a correct explanation for statement-1C. statement-1 is true, statement-2 is falseD. statement-1 is false, statement-2 is true |
| Answer» Correct Answer - B | |
| 417. |
Explain why (or how) `:` (a) In a sound wave, a displacement node is a pressure antinode and vice-versa, (b) Bats can ascertain distances, directions, nature and size of obstacles without any eyes, (c) a violin note and sitar note may have the same frequency, yet we can distinguish between the two notes, (d) Solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases, (e) The shape of pulse gets distorted during propagation in a dispersive medium. |
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Answer» a) A node is a point where the amplitude of vibration is the minimum and pressure is the maximum. On the other hand, an antinode is a point where the amplitude of vibration is the maximum and pressure is the minimum. Therefore, a displacement node is nothing but a pressure antinode, and vice versa. (b) Bats emit very high-frequency ultrasonic sound waves. These waves get reflected back toward them by obstacles. A bat receives a reflected wave (frequency) and estimates the distance, direction, nature, and size of an obstacle with the help of its brain senses. (c) The overtones produced by a sitar and a violin, and the strengths of these overtones, are different. Hence, one can distinguish between the notes produced by a sitar and a violin even if they have the same frequency of vibration. (d) Solids have shear modulus. They can sustain shearing stress. Since fluids do not have any definite shape, they yield to shearing stress. The propagation of a transverse wave is such that it produces shearing stress in a medium. The propagation of such a wave is possible only in solids, and not in gases. Both solids and fluids have their respective bulk moduli. They can sustain compressive stress. Hence, longitudinal waves can propagate through solids and fluids. (e) A pulse is actually is a combination of waves having different wavelengths. These waves travel in a dispersive medium with different velocities, depending on the nature of the medium. This results in the distortion of the shape of a wave pulse. |
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| 418. |
Statement-1: When a transverse wave is reflected from a denser medium, the trough is reflected as crest and vice-versa. Statement-2: in case of reflection from a denser medium, there is inversion of the reflected wave.A. Statement-1 is true, statement-2 is true, statement-2 is a correct explanation for statement-1B. Statement-1 is true, statement-2 is true, statement-2 is not a correct explanation for statement-1C. statement-1 is true, statement-2 is falseD. statement-1 is false, statement-2 is true |
| Answer» Correct Answer - A | |
| 419. |
A police car moving at `22m//s`, chases motorcyclist. The police man sounds his horn at `176 Hz`, while both of them move towards a ststionary siren of frequency `165 Hz`. Calculate the speed of the motorcycle, if it is given that he does not observes any beats. A. (a) `33m//s`B. (b) `22m//s`C. ( c)`zero`D. (d) `11m//s` |
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Answer» Correct Answer - B (b) `f_(1) = frequency of the police car heard by motorcyclist`, `f_(2) = frequency of the siren heard by motorcyclist`. `f_(1) = (330-v)/(330 - 22) xx 176` , `f_(2) = (330 + v)/(330) xx 165` , `because` `f_(1) - f_(2) = 0 rArr v = 22 m//s` |
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| 420. |
A string of mass per unit length `mu` is clamped at both ends such that one end of the string is at x = 0 and the other is at `x =l`. When string vibrates in fundamental mode, amplitude of the midpoint O of the string is a, tension in a, tension in the string is T and amplitude of vibration is A. Find the total oscillation energy stored in the string.A. `y=3sin((pix)/(L))sin((pi)/(L)vt)+2sin((2pix)/(L))sin((2pi)/(L)vt)`B. `Y=3sin((pi)/(L))cos((pi)/(L)vt)+2sin((2pix)/(L))cos((2pi)/(L)vt)`C. `y=3sin((pix)/(L))sin((2pi)/(L)vt)+2sin((2pix)/(L))sin((pi)/(L)vt)`D. `y=3sin((pix)/(L))cos((2pi)/(L)vt)+2sin((2pix)/(L))cos((pi)/(L)vt)` |
| Answer» Correct Answer - B | |
| 421. |
A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a source of 1237.5 Hz ?(sound velocity in air = 330 m s-1) |
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Answer» 3rd harmonic [since no=v/4 l=412.5 with v = 330 m/s] |
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| 422. |
A train standing at the outer signal of a railway station blows a whistle of frequency 400 Hz still air.The train begins to move with a speed of 10 m s-1 towards the platform. What is the frequency of the sound for an observer standing on the platform?(sound velocity in air = 330 m s-1) |
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Answer» Frequency of the sound for an observer standing on the platform 412.5Hz. |
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| 423. |
A sound wave of wavelength `0.40 m` enters the tube at `S`. The smallest radius `r` of the circular segment to hear minimum at detector `D` must be |
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Answer» The path difference between the waves moving along the straight path and the semicircular path is given by `Deltaxpir-2r=(pi-2)r` For minima this path difference should be at least `(lamda)/(2)` i.e., 20, hence we have, for the minimum value of r. `(pi-2)r=20cm` or `r=(20)/(pi-2)=17.54cm` |
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| 424. |
Calculate the velocity off sound in a gas in which two wavelength 204 cm and 208 cm produce 20 betas in 6 sec. |
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Answer» In the given case. Wavelength of first wave `lamda_(1)=204cm` Wavelength of the second wave `lamda_(1)=208cm` let velocity of sound in the gas is v cm/s then frequency of first wave `f_(1)=(v)/(lamda_(1))=(v)/(204)` and frequency of second wave `f_(2)=(v)/(lamda_(2))=(v)/(208)` Beat frequency `f=(20)/(6)=Hz` or `f_(1)-f_(2)=(20)/(6)` or `(v)/(204)-(v)/(208)=(20)/(6)` or `(v[208-204])/(204xx208)=(20)/(6)` or `v=(20)/(6)xx(204xx208)/(4)cm//s` `=35360cm//s=354.6m//s` |
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| 425. |
What will be the speed of sound in a perfectly rigid rod?A. 332 m/sB. les than 332 m/sC. InfiniteD. Zero |
| Answer» Correct Answer - C | |
| 426. |
Velocity of sound in a gaseous medium at constant temperature is independent ofA. pressureB. elasticityC. humidityD. density |
| Answer» Correct Answer - A | |
| 427. |
Two turning forks of frequencies `n_(1)` and `n_(2)` produces n beats per second. If `n_(2)` and n are known, `n_(1)` may be given byA. `n_(2)/n+n_(2)`B. `n_(2)n`C. `n_(2)pmn`D. `n_(2)/n-n_(2)` |
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Answer» Correct Answer - C Beat frequency number of beats/sec. `n=n_(2)-n_(1)orn_(1)-n_(2)thereforen_(1)=n_(2)pmn` |
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