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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.
| 351. |
At a distance `20 m` from a point source of sound the loudness level is `30 dB`. Neglecting the damping, find (a) the loundness at `10 m` from the source (b) the distance from the source at which sound is not heard. |
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Answer» `(a)` As there is no damping, so `L_(r_(0))=10 log ((I)/( I_(0))) =10 log .(1//2rho a^(2)omega^(2) v// r_(0)^(2))/(1//2 rho a^(2) omega^(2) v )=-20 log r _(0)` Similarly `L_(r)=-20 log r ` So, `L_(r)-L_(r_(0))=20 log ( r_(0)//r)` or `L_(r)=L_(r_(0))+20 log ((r_(0))/( r))=30 ++20xxlog ((20)/(10))=36 dB` `(b)` Let `r` be the sought distance at which the sound is not heard. So, `L_(r)=L_(r_(0))+20 log ((r_(0))/(r))=0` or, `L_(r_(0))=20 log ((r)/(r_(0)))` or `30=20 log ((r)/( 20))` So, `log _(10)((r)/(20))=3//2 `or ` 10-^((3//2))=r//20` Thus `r=200sqrt(10)=0.63 km.` Thus for `r gt 0.63km` no sound will be heard. |
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| 352. |
`4.0 g` of a gas occupies `22.4` litres at NTP. The specific heat capacity of the gas at constant volume is `5.0 JK^(-1)mol^(-1)`. If the speed of sound in this gas at NTP is `952 ms^(-1)`. Then the heat capacity at constant pressure isA. `8.5 JK^(-1)mol^(-1)`B. `8.0 JK^(-1)mol^(-1)`C. `7.5 JK^(-1)mol^(-1)`D. `7.0 JK^(-1)mol^(-1)` |
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Answer» Correct Answer - A Here, `C_(upsilon)=5.0JK^(-1)mol^(-1), upsilon=952ms^(-1)` As `4.0g` of gas occupies `22.4` litres at `N.T.P.,` hence molar mass of gas, `m=4g//mol` Velocity of sound, `upsilon=sqrt((gamma P)/(rho))=sqrt((gammaP)/(m//V))=sqrt((gammaPV)/(m))=sqrt((gammaRT)/(m))` ,brgt or `upsilon^(2)=(gammaRT)/(m)` or `gamma=(upsilon^(2)m)/(RT)=((952)^(2)xx(4xx10^(-3)))/(8.3xx273)=1.6` `gamma=(C_(p))/(C_(v))` or `C_(p)=gammaC_(v)=1.6xx5` `=8.0JK^(-1)mol^(-1)` |
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| 353. |
The kinetic energy of a particle executing SHM is 16J. When it is in its mean position. If the amplitude of oscillation is 25cm and the mass of the particle is 5.12 kg, the itme period of its oscillation in second isA. `pi//5`B. `2pi`C. `5pi`D. `20pi` |
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Answer» Correct Answer - A `pi//5` |
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| 354. |
Is the damping force constant on a system executing SHM? |
| Answer» No, because damping force depends upon velocity and is more when the system moves fast and is less when system moves slow. | |
| 355. |
A plane wave of frquency `omega` propagates so that a certain phase of oscillation moves along the `x,y,z` axes with velocities `upsilon_(1), upsilon_(2), upsilon_(3)`. Respectively. Find the wave vector`k`, assuming the unit vetors `e_(x), e_(y), e_(z)` of the coordinate axes to be assigned. |
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Answer» The phase of the oscillation can be written as `Phi=omegat-vec(k). vec(r)` When the wave moves alont the `x-` axis `Phi=omegat-k_(x)x(` On putting `k_(y)=k_(z)=0).` Since the velocity associated with this wave is `v_(1)` We have `k_(x)=(omega)/( v_(1))` Similarly `k_(y)=(omega)/( v_(2))` and `k_(z)=(omega)/(v_(3))` Thus `vec(k)=(omega)/(v_(1))hat(e)_(x)+(omega)/(v_(2))hat(e)_(y)+(omega)/(v_(3))hat(e)_(3)` |
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| 356. |
A wave travelling along a string is given by `y(x,t)=0.05 sin (40x-5t)`, in which numerical constants are in SI units. Calculate the displacement at distance 35 cm from origin, and time 10 sec. |
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Answer» The given equation is `y(x,t)=0.05sin(40x-5t)` Compare it with standard form `y=rsin((2pi)/(lambda)x-(2pit)/(T))` `r=0.05m,, (2pi)/(lambda)=40radm^(-1),(2pi)/(T)=5rad//s` `lambda=(2pi)/(40)=(pi)/(20)m=15.7cm` `T=(2pi)/(5)s=1.26s` `v=(1)/(T)=(5)/(2pi)=0.8Hz` At `x=35cm=0.35m and t=10s,` `y=0.05 sin (40xx0.35-5xx10)` `=0.05sin(-36)=-0.05sin36metre` |
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| 357. |
A displacement wave is represented by `xi=0.25xx10^(-3)sin(500t=0.025x)` Deduce (i) amplitude (ii) period (iii) angular frequency (iv)wavelength (v) amplitude of particle velocity (vi) amplitude of particle acceleration . `xi`, t and x are in cm, sec, and metre respectively. |
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Answer» We know that the displacement of a particle in a simple harmonic wave is `xi=rsin((2pit)/(T)-(2pix)/(lambda))` Comparing this equation with the equatino `xi=0.25xx10^(-3)sin(500t-025x)`, We have (i) amplitude ` r=0.25xx10^(-3)cm` (ii) `(2pi)/(T)=500 or T=(2pi)/(500)=(pi)/(250)sec` (iii) Angular frequency, `omega=(2pi)/(T)=500rad.//sec` (iv) `(2pi)/(lambda)=-025 or lambda=(2pi)/(0.025) =80pimetres` (v) Amplitude of particle velocity `=` max. velocity `=omega r=(2pi)/(T).r` `=500xx0.25xx10^(-3)=0.125cm//s` (vi) Amplitude of particle acceleratino `=` max. acceleration `=omega^(2)r` `=(500)^(2)xx0.25xx10^(-3)=62.5cm//sec^(2)` |
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| 358. |
A wave travelling along a strong is described by `y(x,t)=0.005 sin (80.0x-3.0t)` 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 (c) the period and frequency of the wave. Also , calculate the displacement y of the wave at a distance `x=30.0` cm and time t=20 s? |
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Answer» The given equation is `y(x,t)=0.005sin[80.0x-3.0t]` Comparing with the standard eqn. `y(x,t)=rsin[(2pix)/(lambda)-(2pit)/(T)]`, we get (i) `r=0.005m=5mm.` This is the amplitude. (ii) `(2pi)/(lambda)=80.0, lambda=(2pi)/(80.0)=(pi)/(400)metre` `=(pi)/(40)xx100cm=7.85cm` (iii) `(2pi)/(T)=3.0,T=(2pi)/(3.0)=(2xx22)/(3xx7)=2.09s,` `v=(1)/(T)=(1)/(2.09)=0.48Hz` (iv) At, `x=30.0cm and t=20s,` `y=0.005sin(80.0xx(30)/(100)-3.0xx20)` `=0.005sin(24-60)=0.005sin(-36)` `=0.005sin(-36+12pi)` `=0.005sin(-36+37.71)` `=0.005sin(1.71rad)` `y=0.005sin(98.03^(@))=0.005xx1m` `=5mm` |
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| 359. |
What is the distance between a compression and its nearest rarefaction in a longitudinal wave? |
| Answer» Distance between a compression and adjoining rarefaction is `lambda//2`. | |
| 360. |
Ocean waves hitting a beach are always found to be nearly normal to the shore. |
| Answer» Ocean waves are transverse waves travelling in concentric circles of ever increasing radii . When they hit the shore, their radius of curvature is so large that they can be treated as plane waves. Hence, they hit the shore nearly normal to the shore. | |
| 361. |
A ball of radius `R=50 cm` is located in a non`-` magnetic medium with permittivity ` epsilon=4.0 `. In that medium a plane electromagnetic wave propagates, the strength amplitude of whose electri component is equal to `E_(m) = 200 V //m`. What amount of energy reaches the ball during a time interval `t=1.0 `min ? |
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Answer» For the Poynting vector we can derive as `lt S gt = (1)/(2) sqrt((epsilon epsilon_(0))/(mu_(0))) E_(m)^(2)` along the direction of propagation. Hence in time `t`(which is much longer than the time period `I` of the wave, the enegry reaching the ball is `piR^(2) xx (1)/(2) sqrt((epsilon epsilon_(0))/(mu_(0))) E_(m)^(2)xxt = 5 kJ`. |
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| 362. |
If a simple harmonic motion is given by `y=(sin omegat+cosomegat)m` Which of the following statements are true:A. Time is considered from `y=0m`B. Time is considered from `y=1m`C. The amplitude is `sqrt(2)m`D. The amplitude is `1m` |
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Answer» Correct Answer - B::C `y=sin omegat+cos omegat` `=2pi[(1)/(sqrt(2))sin omegat+(1)/(sqrt(2))cos omegat]` `=sqrt(2)[cos((pi)/(4))sin omegat+sin ((pi)/(4))cos omegat]` `=sqrt(2)sin[omegat+(pi)/(4)]` Amplitude, `a=sqrt(2)m, ` when` t=0` `y=sqrt(2)sin((pi)/(4))=sqrt(2)xx(1)/(sqrt(2))=1m` |
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| 363. |
A standing electromagnetic wave `E=E_(m) cos kx. Cosomegat` is sustained along the `x` axis in vacuum. Find the projection of the Poyntind vector on the `x` axis `S_(x)(x,t)` and the mean value of that projection averaged over an oscillation period. |
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Answer» `oversetrarr(E) = oversetrarr(E_(m)) cos kx omega t` `oversetrarr(H) = (oversetrarr(k) xx oversetrarr(E_(m)))/(mu_(0)omega) sin kx sin omegat` `oversetrarr(S) = oversetrarr(E) xx oversetrarr(H) = (oversetrarr(E_(m)) xx (oversetrarr(k)xx oversetrarr(E)_(m)))/(mu_(0)omega) (1)/(4) sin 2kx sin 2 omegat` Thus `S_(x) = (1)/(4) epsilon_(0) cE_(m)^(2) sin 2kx sin 2 omegat` (as `(1)/(mu_(0)c) = epsilon_(0)c)` |
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| 364. |
A plane electromagnetic wave of frequency `v=10 MHz` propagates in a poorly conducting medium with conductivity `sigma=10 mS//m` and permittivity `epsilon=9`. Find the ratio of amplitudes of conduction and displacement current densities. |
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Answer» Conduction current density `= sigmavec(E)` Displacement current density `=(deltavec(D))/( deltat)=epsilonepsilon_(0)(deltavec(E))/( delta t ) = i omega epsilon epsilon_(0)vec(E)` Ration of magnitudes `=( sigma)/( omega epsilon epsilon_(0))=(j_(c))/( j_(dis))=2`, on putting the values. |
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| 365. |
A plane harmonic electromagnetic wave with plane polarization propagates in vacuum. The elctric component of the wave has a strength amplitude `E_(m)= 50 m V//m`, the frequency is `v=100MHz`. Find `:` `(a)` the efficient value of the displacement current density , `(b)` the mean energy flow density averaged ove an oscillation period. |
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Answer» `E = E_(m)cos (2pi vt - kx)` (a) `j_(dis) = (delD)/(delt) = - 2pi epsilon_(0)vE_(m)sin (omegat - kx)` Thus `(j_(dis))_(rms) = lt j_(dis)^(2)gt^(1//2)` `= sqrt(2) pi epsilon_(0)vE_(m) = 0.20 mA//m^(2)`. (b) `lt S_(x) gt = (1)/(2) sqrt((epsilon_(0))/(mu_(0))) E_(m)^(2)` as in. Thus `lt S_(x) gt = 3.3 muW//m^(2)` |
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| 366. |
Find the mean Plynting vector `( : S: )` of a plane electromagnetice wave `E=E_(m) cos ( omegat - kr )` if the wave propagates in vacuum. |
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Answer» `overset rarr(E) = overset rarr(E_(m)) cos (omegat - oversetrarr(k.)oversetrarr(r ))` then as before `oversetrarr(H) = sqrt((epsilon_(0))/(mu_(0))) (oversetrarr(k)xxoverset rarr(E_(m)))/(k)cos(omegat-overset rarr(k).oversetrarr(r ))` so `overset rarr(S) = overset rarr(E) xx overset rarr(H) = sqrt((epsilon_(0))/(mu_(0))) oversetrarr(E_(m)) xx (oversetrarr(k)xxoversetrarr(E_(m)))(1)/(k)cos^(2) (omegat- oversetrarr(k).oversetrarr(r))` `= sqrt((epsilon_(0))/(mu_(0)) oversetrarr(E_(m)^(2)) (oversetrarr(k))/(k)cos^(2) (omega t - overset rarr(k).oversetrarr(r ))` `lt oversetrarr(S) gt = (1)/(2) sqrt((epsilon_(0))/(mu_(0))) oversetrarr(E_(m)^(2)) (oversetrarr(k))/(k)` |
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| 367. |
Amplitude of a wave is represented by `A=(c)/(a+b-c)`. Under what conditions, the resonance will occur. |
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Answer» The resonance will occur when amplitude A becmes infinite. It will be so if `a+b-c=0`. Which will be possible if `(i) a=c-b (ii) b=0` and `a=c.` |
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| 368. |
Infinite springs with force constants k,2k, 4k and 8k … respectively are connected in series. What will be the effective force constant of springs, when `k=10N//m` |
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Answer» When springs are in series, the effective force constant K is given by `(1)/(K)=(1)/(k)=+(1)/(2k)+(1)/(4k)+(1)/(8k)+.........` `=(1)/(k)(1+(1)/(2)+(1)/(2^(2))+(1)/(2^(3))+........)` `(1)/(k)((1)/(1-(1)/(2)))=(2)/(k)` or `K=(k)/(2)=(10)/(2)=5N//m` |
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| 369. |
Three springs are connected to a mass m as shown in figure, When mass oscillates, what is the effective spring constant and time period of vibration? Given `k=2Nm^(-1)` and m=80 gram. |
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Answer» The equivalent arrangement of spring system is shown in figure. This shows that all springs are connected in parallel. Therefore effective spring constant `K=k+2k+k=4k=4xx2` `8Nm^(-1)` Time period, `T=2pisqrt(m//4k)=2xx(22)/(7)sqrt(0.08//8)=0.628` |
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| 370. |
A spring force constant k is cut into two parts such that one piece is double the length of the other. Then the long piece will have a force constatnt ofA. `(2//3)k`B. `(3//2)k`C. 3kD. 6k |
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Answer» Correct Answer - B `(3//2)k` |
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| 371. |
Estimate the speed of sound in air at standard temperature and pressure. The mass of 1 mole of air is `29.0xx10^(-3)kg`, `lambda` for air `=7//5`. |
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Answer» At S.T.P, `P=1` atmosphere `=1.01xx10^(5)N//m^(2)` As mass of 1 mole of air `=29.0xx10^(-3)kg` and its volume is 22.4litre `=22.4xx10^(-3)m^(3)` `:.` density of air, `rho=(M)/(V)=(29.0xx10^(-3))/(22.4xx10^(-3))=1.29kg//m^(3)` As , `v=sqrt((lambdaP)/(rho))` `:. v=sqrt((7)/(5)xx(1.01xx10^(5))/(1.29))=33.1m//s` |
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| 372. |
At what temperature will the speed of sound be double its value at 273K? |
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Answer» `T_(2)=?T_(2)=273K, v_(2)=2v_(1)` As `(v_(2))/(v_(1))=sqrt((T_(2))/(T_(1)))=2, T_(2)=4T_(1)=4xx273` `T_(2)=1092K` |
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| 373. |
A well with vertical lsides and water at the bottom resonates at 7Hz and at no other lower freqency. The air in the well has density `1.10kgm^(-3)` and bulk modulus of water is `1.33xx10^(5)N//m^(2)`. How deep is the well ? |
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Answer» Here, `v=7Hz, rho=1.10kgm^(-3)`, `K=1.33xx10^(5)Nm^(-2)` A well behaves as closed organ pipe of length `=` depth of water in the well `=l` `upsilon=sqrt((K)/(rho)) =sqrt((1.33xx10^(5))/(1.1))=347.7m//s` From `v=(upsilon)/(4l), l=(upsilon)/(4v)=(34.7)/(4xx7)=12.41metre` |
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| 374. |
You have a light spring, a metre scale and a known mass. How will you find the time period of oscillation of mass without the use of a clock ? |
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Answer» Suspended the known mass from one end of spring whose another end is connected to a rigid ceiling. Note the extension l in the spring with the help of metre scale. If k is the spring constant of the spring, then in equlibrium position, `kl=mg or (m)/(k)=(l)/(g)` Tiem period of the loaded spring, `T=2pisqrt((m)/(k))=2pisqrt((l)/(g))` Knowling, the value of l, time period of oscillation T can be determined. |
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| 375. |
The pendulum clock runs ……………….. Due to decrease in its time period and runs …………………… due to increase in its period.. |
| Answer» Correct Answer - fast,slow | |
| 376. |
A particle executing SHM possesses two types of energy. Potential energy is on account of displacement of particle from the mean positoin, anad kinetic energy is ono account of velocity of the particle. At any time t `P.E. =U=(1)/(2)momega^(2)y^(2)=(1)/(2)momega^(2)a^(2)sin^(2)omegat, ` and `K.E..=K=(1)/(2)momega^(2)(a^(2)-y^(2))=(1)/(2)momega^(2)a^(2)cos^(2)omegat` `T.E. =E=U+K=(1)/(2)momega^(2)a^(2)=` constant Read the above passage and answer the following questions: (i) At what distance from the mean position will K.E. of a particle be twice its P.E.? (ii) What are the implications of this study in day life? |
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Answer» (i) As K.E. `=2P.E`. `:. (1)/(2)momega^(2)(a^(2)-y^(2))=2xx(1)/(2)momega^(2)y^(2) or a^(2)-y^(2)=2y^(2) or 3y^(2)=a^(2), y=(a)/(sqrt(3))` (ii) This study shows that sum total of P.E. and K.E. of a particle in SHM stays constant at all position and at all times. However, P.E. and K.E. both keep on changing with position or time. The same is true in daya to day life. We can acquire one form of energy by spending some other form of energy, and vice-versa. For example, when we work too har, we forget about rest and relaxation. And when we relax, we store energy to work again. This is what goes on in life. |
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