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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.
| 1. |
A particle executed SHM iwht a time period of 3s and amplitude 6cm. Find (i) displacement (ii) velocity and (iii) acceleration, after `1//2` second, starting from the mean position. |
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Answer» Here `T=3s, A=6cm,t=1//2s` (i) As particle is starting from mean position, the displacement of particle at time t is given by `x=Asinomegat=Asin((2pi)/(T))t=6sin((2pi)/(3))xx(1)/(2)` `=6sin((pi)/(2))=6xx(sqrt(3))/(2)=5.196cm ` (ii) Velocity, `v=(dx)/(dt)=(2piA)/(T)cos((2pit)/(T))` `=(2pi)/(3)xx6cos((2pi)/(3))xx(1)/(2)` `=4picos((pi)/(3))=(4pixx1)/(2)=2xx(22)/(7)=6.28cm//s` (iii) Acceleration, `A=(dv)/(dt)=(4pi^(2))/(T^(2))Asin((2pit)/(T))` `=-(4pi^(2)xx6)/(3^(2))sin((2pi)/(3))xx(1)/(2)` `=-(8pi^(2))/(3)sin((pi)/(3))=-(8pi^(2))/(3)xx(sqrt(3))/(2)=-22.8cm^(-2)` In magnitude, `A=22.8cm^(2)`, Here `-ve` sign indicates that acceleration is directed towards mean position whereas displacement is directed away from mean position. |
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| 2. |
A particle performs harmonic oscillations along a straight line with a period of 6s and amplitude 4 cm. The mean velocity of the particle averaged over the time interval during which it travels a distance of 2 cm starting from the extreme position isA. `1cms^(-1)`B. `2cms^(-1)`C. `4cms^(-1)`D. `8cms^(-1)` |
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Answer» Correct Answer - B Here, `T=6s, r=4cm,` Displacement from the mean position when the particle travels 2 cm from the extreme position is,`x=4-2=2.0cm. ` If t is the time taken then `x=r cos omega t =r cos ((2pi)/(T))t` `:. 2=4cos ((2pi)/(6))t or cos ((pit)/(3))=(2)/(4)=(1)/(2)=cos((pi)/(3))` so `:. ` Average velocity `=` `(dist ance)/(time )=(2)/(1)=2cms^(-1)` |
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| 3. |
A body executed SHM of time period 6s. If its mass be 0.1kg, its velocity, 1s after it passes throuh mean position be `3ms^(-1)`, find its (i) KE (ii) PE and (iii) total energy. |
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Answer» Here, `T=6s,m=0.1kg, t=1s,` `v=3ms^(-1)` `v=Aomegacosomegat=A(2pi)/(T)cos(2pi)/(T)t` `3=Axx(2pi)/(6)cos((2pi)/(6)xx1)=Axx(2pi)/(6)xx(1)/(2)` `A=(18)/(pi)m` KE`=(1)/(2)mv^(2)=(1)/(2) xx0.1xx(3)^(2)=0*45J` `TE=(1)/(2)momega^(2)A^(2)=(1)/(2)m(4pi^(2))/(T^(2))xx(18^(2))/(pi^(2))` `=(1)/(2)xx0.1xx(4)/(6^(2))xx18^(2)=1*8J` `PE=TE-KE=1.8-0.45=1.35J` |
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| 4. |
Two air columns of resonance appatus, 100cm and 101 cm long give 17 beats in 20 second, when each is sounding its fundamental mode. Calculate the velocity of sound. |
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Answer» Here, `l_(1)=100cm, l_(2)=101cm, m=(17)/(20)` As `m=v_(1)-v_(2)` `(17)/(20)=(upsilon)/(4l_(1))=(upsilon)/(4l_(2))=(upsilon)/(4)[(1)/(100)-(1)/(101)]=(upsilon)/(400xx101)` `upsilon=(17xx400xx101)/(20)cm//s=34340cm//s` `=343.4m//s` |
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| 5. |
Two factories are sounding their sirens at 800 Hz. A man goes from one factory to the other at a speed of 2 m/s. The velocity of sound is 320 m/s. The number of beats heard by the person is 1 s will be |
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Answer» Here, `v=800Hz, v_(L)=2m//s, upsilon=320m//s.` Number of beats `//` sec. `m=?` The man moves from one factory to the other. The apparent frequency of siren heard from the factory from which he is moving away `v_(1)=((upsilon-upsilon_(L))v)/(upsilon)=((320-2)800)/(320)=(318)/(320)xx800Hz` The apparent frequency of siren heard from the factory to which the man is moving. `v_(2)=((upsilon+upsilon_(L))v)/(upsilon)=((320+2)800)/(320)=(322)/(320)xx800Hz` `:. ` Number of beats heard per sec., `m=v_(2)-v_(1)` `=(322)/(320)xx800-(318)/(320)xx800` `=(800)/(320)(322-318)` `m=(800xx4)/(320)=10` |
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| 6. |
If tension of a sire is increased to four times, how is the wave speed changed? |
| Answer» As `vprop sqrt(T)` , therefore, wave speed becomes twice. | |
| 7. |
A string is vibrating in n loops. The numbers of nodes and anitnodes respectively areA. `n,n`B. `(n+1),n`C. `n,(n-1)`D. `(n-1),n` |
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Answer» Correct Answer - B `(n+1),n` |
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| 8. |
An open organ pipe is closed at one end. How will the frequency of fundamental note of pipe change? |
| Answer» The fundamental frequency shall reduce to half the original value. | |
| 9. |
The temperature coefficient of velocity of sound in air isA. `1m//s//^(@)C`B. `0.61m//s//^(@)F`C. `1m//s//K`D. `0.61m//s//K` |
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Answer» Correct Answer - D `0.61m//s//K` |
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| 10. |
The changes in pressure an volume of air, when sound wave passes through air areA. isothermalB. isobaricC. isovolumicD. adiabatic |
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Answer» Correct Answer - D adiabatic |
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| 11. |
An organ pipe emits a fundamental note of 100Hz. On blowing into it more strongly, the note produced has frequency 300Hz. Is the pipe open or closed? |
| Answer» The organ pipe mus be closed as the frequency of first overtone is three times the fundamental frequency. | |
| 12. |
Statement-1 `:` The velocity of sound in a gas is not affected by changes in pressure provided temperature of gas remains constant. Statement-2 `:` This is because velocity of sound is inversely proportional to square root of density of the gas.A. Statement -1 is true, Statement-2, Statement -2 is correct explanation of statement-1.B. Statement-1 is true, Statement-2 is true, Statement-2 is not a correct explanantion of Statement-1.C. Statement-1 is correct, Statement-2 is false.D. Statement-1 is false, Statement-2 is true. |
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Answer» Correct Answer - d Infact, `upsilon=sqrt((gammaP)/(rho))` In temp, remains constant `sqrt(P//rho)` remains constant. Therefore, velocity of sound `(upsilon)` is not affected. Statement-2 is not a correct explanantion of statement-1. |
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| 13. |
The fundamental frequency of a vibrating organ pipe is `200Hz`A. The first overtone may be `400Hz`B. The first overtone is `400Hz`C. The first overtone may be `600Hz`D. The second overtone may be `600Hz` |
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Answer» Correct Answer - A::C::D If the pipe is closed, the first overtone may have frequency`=3v_(1)=3xx200=600Hz. ` If the pipe is open, the first overtone may have frequency `=2v_(1)=2xx200=400Hz.` The second overtone may have frequency , `v=3v_(1)=3xx200=600Hz.` |
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| 14. |
When a sound wave passes through a gas, the changes in pressure and vollume of gas are …………………… and……………………… of heat……………………… . |
| Answer» adiabatic, no exchange, takes place. | |
| 15. |
For aluminium , the modulus of rigidity is `2.1xx10^(10)Nm^(-2` and density is `2.7xx10^(3)kg//m^(3)`.Find the speed of transverse waves in the medium. |
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Answer» Here, `eta=2.1xx10^(10)N//m^(2)`, `rho=2.7xx10^(3)kg//m^(3),v=?` `v=sqrt((eta)/(rho))=sqrt((2.1xx10^(10))/(2.7xx10^(3)))=2.79xx10^(3)m//s` |
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| 16. |
The fundamental frequency of vibration of a stretched string varies ………………….. As…………………… of the string, provided …………………. Are constant. |
| Answer» inversely, diameter, length, tension and density of material. | |
| 17. |
The transverse displacement of a string (clamped at its two ends ) is given by `y(x,t)=0.06 sin (2pi)/(3)x cos 120pit,` where x, y are in m and t is in s. Do all the points on the string oscillate with theh same (a) frequency (b) phase (c) amplitude Explain your answer. |
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Answer» In the given equation, terms containing x and t are independent. Therefore, the equation t are independent. Therefore, the equation represents a stationary wave. Hence, (i) All the points on the string do not oscillate with the same frequency. For example, frequency at nodes is zero. (ii) Phase diff. between particle in between successive pair of nodes is `180^(@)`. (iii) The amplitude of vibration of different points on the string is different. At nodes, the amplitude of vibration is zero. |
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| 18. |
A pipe 20cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430Hz source? Will the same source be in resonance with the pipe if both ends are open? Take speed of sound in air `340m//s`. |
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Answer» Here, `L=20cm=0.2m, v_(n)=430Hz, upsilon=340ms^(-1)` The frequency of nth normal mode of vibration of closed pipe is `v_(n)=(2n-1)(upsilon)/(4L):. 430=(2n-1)(340)/(4xx0.2)` `2n-1=(430xx4xx0.2)/(340)=1.02` `2n=2.02, n=1.01` Hence it will be the 1st normal mode of vibration. In a pipe, open at both ends, we have `v_(n)=nxx(upsilon)/(2L)=(nxx340)/(2xx0.2)=430:. n=(430xx2xx0.2)/(340)=0.5` As n has to be an integer, therefore, open organ pipe cnnot be in resonance with the source . |
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| 19. |
A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45Hz. The mass of the wire is `3.5xx10^(-2)`kg and its linear mass density is `4.0xx10^(-2)kgm^(-1)`. What is (a) the speed of a transverse wave on the string , and (b) the tension in the string? |
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Answer» Here,`v=45Hz, M=3.5xx10^(-2)kg,` mass`//` length `=m=4.0xx10^(-2)kgm^(-1)` `:. l=(M)/(m)=(3.5xx10^(-2))/(4.0xx10^(-2))=(7)/(8)m` As, `(lambda)/(2)=l=(7)/(8) :. lambda=(7)/(4)m =1.75m. ` The speed of the transverse wave `upsilon=vlambda=45xx1.75=78.75m//s` (b) As, `upsilon=sqrt((T)/(m)):. T=upsilon^(2)xxm=(78.75)^(2)xx4.0xx10^(-2)=248.06N`. |
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| 20. |
Consider a pair of identical pendulums, which oscillate with equal amplitude independently such that when one pendulum is at its extreme position making an angle os `2^(@)` to the right with the vertical , the other pendulum makes an angle of `1^(@)` to the left of the vertical. What is the phase difference between the pendulums? |
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Answer» Let `phi_(1)` and `phi_(2)` be the angular displacement of first and second pendulums respectively at an instant. `phi_(0)` be the maximum angular displacement of each pendulum. Let `theta_(1)` and `theta_(2)` be the initial phases of two pendulums, then `phi_(1)=phi_(0)sin(omegat-theta_(1))` ...(i) and `phi_(2)=phi_(0)sin(omegat+theta_(2))` ...(ii) For first pendulum , `phi_(1)=2^(@)=phi_(0)` , From (i), we have `:. 2=2sin(omegat+theta_(1)) or sin (omegat+theta_(1))=1 or omegat +theta_(1)=90^(@)` ...(iii) For second pendulum, `phi_(2)=-1^(@),` From (ii), we have `-1=2sin(omegat +theta_(2))or sin(omegat+theta_(2))=-(1)/(2)=sin(180+30^(@))=sin210^(@)` `:. omegat+theta_(2)210^(@)` ...(iv) Subtracting (iii) from (iv), we have `(omegat+theta_(2))-(omegat+theta_(1))=210-90=120^(@) or theta_(2)-theta_(1)=120^(@)` |
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| 21. |
Two pendulums of length 90 cm and 100 cm start oscillating in phase. After how many osxillations will they b again in the same phase ? |
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Answer» The two pendulums will be in same phase again when pendulum of large length completes v oscillations in time t and small pendulum completes `(v+1)` oscillations in time t. For pendulum of large length, `(v)/(t)=(1)/(2pi)sqrt((g)/(l))=(1)/(2pi)sqrt((g)/(100))` and `(v+1)/(t)=2pisqrt((g)/(90))` `:. (v+1)/(v)=sqrt((100)/(90))=(sqrt(10))/(3)=(3.162)/(3)=1.054` `1+(1)/(v)=1.054 or (1)/(v)=1.054-1=0.054` or `v=(1)/(0.054)=18.5` Thus the two pendulums will be in same phase when the pendulum of larger length completes 18.5 oscillations and pendulum of smaller length completes `18.5+1=19.5` oscillations. |
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| 22. |
Two pendulums of time period 3 s and 8 s respectively starts oscillating simultaneously from two oppositee extreme positions. After how much time they will be in the same phase ?A. `(24)/(5)s`B. `(12)/(5)s`C. `(24)/(11)s`D. `(12)/(11)s` |
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Answer» Correct Answer - B `y_(1)=Asin (omega_(1)t+pi//2)` and `y_(2)=A sin (omega^(2)t-pi//2)` When, `omega_(1)t+pi//2=omega_(2)t-pi//2` then `t=(pi)/(omega_(2)-omega_(1))=(pi)/((2pi//T_(2))-(2pi//T_(1)))=(T_(1)T_(2))/(2(T_(1)-T_(2)))` `=(8xx3)/(2(8-3))=(24)/(10)=(12)/(5)s` |
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| 23. |
A particle starts oscillating from half the amplitude position. What is its initial phase? |
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Answer» `y=asin(omegat+phi_(0))`. At `t=0, y=a//2` `:. (a)/(2)=asin(omegaxx0+phi_(0))asinphi_(0)` or `sinphi_(0)=(1)/(2)=sin((pi)/(6)) or phi_(0)=(pi)/(6)` |
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| 24. |
A coil with active resistance `R` and inductance `L` was connected at the moment `t=0` to a source of voltage `V=V_(m)cos omegat.` Find the current in the coil as a function of time `t`. |
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Answer» The equatio of the circuit is `(I` is the current `)` `L(dI)/(dt)=RI=V_(m) cos omegat ` From the theory of different equations. `I=I_(p)+I_(C)` where `I_(p)` is a particular integral and `I_(C)` is the complementary function `(` Solution of the differential equation with the RHS `=0). N`ow `I_(C)=I_(CO)e^(-tR//L)` and for `I_(p)` we write `I_(p)=I_(m) cos ( omegat-varphi)` Substituting we get `I_(m)=(V_(m))/( sqrt(R^(2)+omega^(2)L^(2))), varphi=tan ^(-1)(omegaL)/(R)` Then `I_(m)=(V_(m))/( sqrt(R^(2)+omega^(2)L^(2)))cos ( omegat- varphi)+I_(CO) e^(-tR//L` Now in an inductive circuit `I=0` at `t=0` because a current cannot change suddently. Thus `I_(CO)=-(V_(m))/(sqrt(R^(2)=omega^(2)L^(2))) cos varphi` and so ` I=(V_(m))/(sqrt(R^(2)+ omega^(2)L^(2)))[cos ( omegat - varphi)- cos varphie^(-tR//L)]` |
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| 25. |
At the moment `t=0` a point starts oscillating along the `x` axis according to the lasw` x=a sin omegat t`. Find: `(a)` the mean value of its velocity vector projection `( : v_(x) : )`, `(b)` the modulus of the mean velocity vector `|( : v : )|`, (c) the mean value of the velocity modulus `( : v : )` averaged over `3//8` of the period after the start. |
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Answer» As `x=a sin omega t` so,` v_(x)=a omega cos omega t` Thus, `lt v_(x) gt = int v_(x)dt//int dt=(int_(0)^((3)/(8)T)a omegacos(2pi//T)t dt)/((3)/(8)T)=(2sqrt(2)a omega)/(3pi)` (using `T=(2pi)/(omega))` `(b)` In accordance with the problem `vec(v)=v_(x)vec(i)`, so,`|lt vec(v) gt|=|lt v_(x) gt|` Hence, usning part (a) `|lt vec(v) gt|=|(2sqrt(2)a omega)/(3pi)|=(2 sqrt(2)a omega)/(3pi)` We have got, `v_(x)=a omega cos omegat` So, `{:(v=|v_(x)|=a omega cos omega t, "for", t leT//4),(= -aomega cos omega t,"for", T//4 le t3/8 T):}]` Hence, `lt v gt =(int f dt )/(int dt)(int _(0)^(T//4)a omega cos omegat dt+int_(T//4)^(3T//4)-a omega cos t dt )/(3T//8)` Using `omega=2pi//T`, an on evaluating the integral we get `lt v gt = (2(4-sqrt(2))a omega)/(3pi)` |
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| 26. |
The function `sin^(4)(omegat)` representsA. a simple harmonic motion with a period `pi//omega`B. a simple harmonic motion with a period` 2pi//omega`C. a period , but not simple harmonic motion with a period `pi//omega`D. a periodic but not simple harmonic motion with a period `2pi//omega` |
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Answer» Correct Answer - C `y=rsin^(4)(omegat)=rsinomegatxxsin^(3)omegat` `=rsin omegat[(1)/(4)(3sin omegat-sin 3 omegat)]` `=r[(3)/(4)sin^(2)omegat-(1)/(4)sinomegat sin 3 omegat ]` `=(3r)/(4)(1-cos2omegat)` `=-(r)/(4)xx(1)/(2)[cos(3omegat -omegat)-cos (3omegat+omegat)]` `=(3r)/(4)(1-cos2omegat)-(r)/(8)[cos 2 omegat -cos 4 omegat]` It does not represent a SHM but represents a periodic motion with time period `T=(2pi)/(2omega)=(pi)/(omega)` |
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| 27. |
A tolley of mass 3.0kg is connected to two identical spring each of force constant `600Nm^(-1)` as shown in figure. If the trolley is displaced from its equilibrium position by 5.0 cm and released, what is (a) the period of ensuing oscillation? (b) the maximum speed of trolley? (c) How much is the total energy dissipated as heat by the time the trolley comes to rest due to damping forces? |
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Answer» Here, `m=3.0kg. ` `k_(1)=k_(2)=600Nm^(-1), a=5.0cm =0.05m` When the mass is displaced a little to one side, one spring gets compressed and another is elongated. Due to which the combination of spring works as parallel combination of springs. Here, effective spring factore K will be given by `K=k_(1)+k_(2)=600+600=1200Nm^(-1)` (a) `T=2pisqrt((m)/(K))=2xx3.14sqrt((m)/(1200))=3.14s` (b) `V_(max)=aomega =asqrt((K)/(m))` `=0.05sqrt((1200)/(3))=1ms^(-1)` (c) Total energy, `E=(1)/(2)Ka^(2)` `=(1)/(2)xx1200xx(0.05)^(2)=1.5J` |
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| 28. |
A trolley of mass 3.0 kg is connected to two identical rubber bands each of force constant `600Nm^(-1)` as shown in figure. If the trolley is displaced from an equilibrium position by 5.0 cm and released, what is the period of ensuing oscillations? |
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Answer» When the trolley is displaced a little to one side, one rubber band gets stretched and provides restoring force whereas othe rband becomes slag and provides no restoring force. Thus the effective restoring force in this case is due to one rubber band. Hence, force constant, `k=600Nm^(-1) :. T=2pisqrt((m)/(k))=2pisqrt((3)/(600))=0.45s` |
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| 29. |
A small trolley of mass 2.0kg resting on a horizontal turn table is connected by a light spring to the centre of the table. When the turn table is set into rotation a speed of 600rpm, the length of stretched spring is 50cm. If the original length of the spring is 42cm, find the force constant of the spring. |
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Answer» Here, mass of trolley, `m=2.0kg,` Frequency of rotation, `v=(600)/(60)=10rps` Extension produced in string, `x=50-42=8cm =8xx10^(-2)m` When the turn table is set into rotation, the tension (restoring force), in the spring is equal to the centrifugal force. Then length of stretched spring, `r=50cm=0.50m.` `:. F=kx=mr(2piv)^(2)=4pi^(2)mrv^(2)` or `k=(4pi^(2)mrv^(2))/(x)` `=(4xx(22//7)^(2)xx2.0xx0.50xx(10)^(2))/(8xx10^(-2))` `=49387.7Nm^(-1)` |
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| 30. |
A small trolley of mass 2.0kg resting on a horizontal turn table is connected by a light spring to the centre of the table. When the turn table is set into rotation at a speed of 360 rpm,, the length of the stretched spring is 43cm. If the original length of the spring is 36cm, determing the force constant of the spring. |
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Answer» Here, `m=2.0kg, v=(360)/(60)=6rps` Extension produced in spring , `y=43-36=7cm=7xx10^(-2)m` When turn table is set into rotation, the tension in the spring will provide the required centripetal force i.e., `ky=mr(2piv)^(2)=4pi^(2)v^(2)mr` [where r is the length of stretched spring`=43cm`] or `k=(4pi^(2)v^(2)mr)/(y)` `=(4xx(22//7)^(2)xx6^(2)xx(2.0)xx(43xx10^(-2)))/((7xx10^(-2))` `=17475Nm^(-1)` |
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| 31. |
A string of mass 2.50kg is under a tension os 200N. The length of the stretched string is 20.0m. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?A. one secondB. 0.5 secondC. 2 secondD. data given is insufficient |
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Answer» Correct Answer - B Here, `m=(2.5)/(20)kg//m, T=200N` `upsilon=sqrt((T)/(m))=sqrt((200)/(2.5//20))=sqrt((200xx200)/(25))=(200)/(5)=40m//s` Time taken, `t=(l)/(upsilon)=(20)/(40)=0.5s` |
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| 32. |
A string of mass 2.50kg is under a tension os 200N. The length of the stretched string is 20.0m. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end? |
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Answer» Here, `M=2.50kg, T=200N,l=20.0m` Mass per unit length, `m=(M)/(l)=(2.5)/(20.0)=0.125kgm^(-1)` velocity, `v=sqrt((T)/(m))=sqrt((200)/(0.125))=40ms^(-1)`. Time taken by disturbance to reach the other end, `t=(1)/(v)=(20)/(40)=0.5s` |
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| 33. |
An external harmonic force `F` whose frequency can be varied, with amplitude maintained constant, acts in a vertical direction on a ball suspended by a weightless spring. The damping coefficient is `eta` times less than the natural oscillation frequency `omega_(0)` of the ball. How much, in per cent, does the mean power `( :P: )` developed differ from the maximum mean power `( :P: )_(max)`? Averaging is performed over one oscillation period. |
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Answer» Given ` beta = omega_(0)//eta` . Then from the previous problem `lt Pgt =(F_(0)^(2)omega_(0))/( etam). (1)/( ((omega_(0)^(2))/(omega)-omega)^(2)+ 4 ( omega_(0)^(2))/( eta^(2)))` At displacement resonance ` omega=sqrt(omega_(0)^(2)-2 beta^(2))` `lt P gt_(res)=(F_(0)^(2)omega_(0))/( etam)(1)/((4 beta^(4))/( omega_(0)^(2)- 2 beta^(2))+(4 omega_(0)^(2))/( eta^(2)))=(F_(0)^(2)omega_(0))/( eta m)(1)/((4 omega_(0)^(4)//eta^(4))/( omega_(0)^(2)(1-(2)/( eta^(2))))+4( omega_(0)^(2))/( eta^(2)))` `=(F_(0)^(2))/( 4 eta m omega_(0))(n^(2))/((1)/( n^(2)-2)+1)=(F_(0)^(2)eta)/( 4 m omega_(0))(n^(2)-2)/( n^(2)-1)` while `lt P gt_(max)=(F_(0)^(2)eta)/(4 m omega_(0))`. Thus `(lt P gt_(max)- lt P gt _(res))/(lt P gt_(max))=(100)/(eta^(2)-1)%` |
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| 34. |
The density `rho` of a liquid varies with depth h from the free surface as `rho=kh`. A small body of density `rho_(1)` is released from the surface of liquid. The body willA. come to a momentary rest at a depth `2rho_(1)//k` from the free surfaceB. execute S.H.M. about a point at a depth `rho_(1)//k` from the surfaceC. execute S.H.M. of amplitude `rho_(1)//k`D. all of these |
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Answer» Correct Answer - D Let V be the volume of body which is released on the surface of liquic. When body reaches at depth x from the surface of liquid, then upward thrust on body due to liquid `=V(kx)g` Weight of body `=Vrho_(1)g` At equilibrium , `V(kx)g=Vrho_(1)g or x=rho_(1)//k` If the body comes to rest at depth h, then work done on body by the upward thrust dur to liquid `=` change in potential energy of body `:. -int_(0)^(h)V(kx)g dx=-Vrho_(1) gh or h =(2rho_(1))/(k)` Now, the body will move upwards towards equilibrium position due to displacemtn x from equilibrium position . Hence the body will execute S.H.M. Choice (d) is right. |
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| 35. |
A mass m is performing linear simple harmonic motion, then which of the following graph represents correctly the variation of acceleration a corresponding to linear velocity `upsilon` ?A. B. C. D. |
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Answer» Correct Answer - D In SHM, `y=rsin omegat` `upsilon=romega cos omegat and a=(dupsilon)/(dt)=-omega^(2)rsinomegat` `:. cos omegat =(upsilon)/(romega) and sin omegat =-(a)/(omega^(2)r)` Hence, `cos^(2)omegat + sin^(2)omegat=((upsilon)/(romega))^(2)+[-(a)/(omega^(2)r)]^(2)` or `1=(upsilon^(2))/(r^(2)omega^(2))+(a^(2))/(omega^(4)r^(2))` or ` upsilon^(2)=-(a^(2))/(omega^(2))+r^(2)omega^(2)` ...(i0 Comparing (i), with general equation of st. line, `y=mx+c` , we note theat, the equation (i) is a st. line betweent `upsilon^(2) and a^(2)` with negative slope `(1//omega^(2))`. Hence, optiond (d) is correct. |
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| 36. |
A particle is performing simple harmonic motion along `x-`axis with amplitude `4 cm` and time period `1.2 sec` .The minimum time taken by the particle to move from `x = 2 cm to x = +4 cm` and back again is given byA. `0.3s`B. `0.4s`C. `0.5s`D. `0.2s` |
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Answer» Correct Answer - B Here, `a=4cm, T=1.2s` As `x=asin omegat =asin ((2pi)/(T))t` `:. T=(T)/(2pi) sin ^(-1) ((x)/(a))` `a=4c, At x=2, t=(T)/(2pi)sin^(-1)((2)/(4))` `t=(T)/(2pi)xx(pi)/(6)=(T)/(12)=(1.2)/(12)=(1)/(10)s` At `x=4, t=(T)/(2pi)sin^(-1)((4)/(4)) or t=(T)/(2pi)xx(pi)/(2)=(T)/(4)`j `t=(1.2)/(4)=(3)/(10)s` Time taken in going from `x=+2cm` to `x=+4cm` will be `=(3)/(10)-(1)/(10)=0.2s` and same time will be taken for coming back. So total time taken `=0.2+0.2=0.4s` |
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| 37. |
A tuning fork has a frequency of 256. Distance travelled by sound emitted during the time it makes 16 vibrations is `(v=332m//s)`A. 256mB. 332mC. 20.7mD. `256xx16m` |
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Answer» Correct Answer - C 20.7m |
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| 38. |
A particle moving with SHM in a straight line has a speed of `6m//s` with when 4m from the centre of oscillation and a speed of `8m//s` when 3m from the centre of oscillation . Find the amplitude of oscillation and the shortest time taken by the particle in moving from the extremen position to a point mid way between the extreme position and the centre. |
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Answer» `V^(2)=omega^(2)(r^(2)-y^(2))` (i) `6^(2)=omega^(2)(r^(2)--4^(2))` (ii) `omega^(2)(r^(2)-3^(2))` or `(64)/(36)=(r^(2)-9)/(r^(2)-16)` On solving, `r=+-5m and omega =2s^(-1)` For the given displacement, `x=(r)/(2),t=?` Asx `x=r cos omega t` `:. (r)/(2)=r cos 2t or cos 2t=(1)/(2)=cos((pi)/(3))` or `2t=(pi)/(3)or t=(pi)/(6)s` |
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| 39. |
In a sonometer experiment, the graph between frequency `n` and reciprocaal of reonance length `(1//l)` is given by figure.A. B. C. D. |
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Answer» Correct Answer - C As, `n prop (1)/(l)`, therefore, choise (c) is the correct graph. |
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| 40. |
In a sonometer experiment, the density of material of the wire used in `7.5xx10^(3)kg//m^(3)`. If the stress of the wire is `3.0xx10^(8)N//m^(2)`, find out the speed of transvers waves in the wire. |
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Answer» Here, `rho=7.5xx10^(3)kg//m^(3)`, Stress `=3.0xx10^(8)N//m^(2)` If A is area of cross section of the wire, then tension in wire, `T=stressxxarea=3.0xx10^(8)A N` Mass per unit length of wire, `m=Axx1xxrho=Axx7.5xx10^(3)kg//m` `v=sqrt((T)/(m))=sqrt((3.0xx10^(8)A)/(Axx7.5xx10^(3)))=sqrt(4xx10^(4))=200m//s` |
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| 41. |
The density of oxygejn is 16 times the density of hydrogen. What is the ration of speeds of sound in them? |
| Answer» `(upsilon_(o x y))/(upsilon_(hy))=sqrt((rho_(hy))/(rho_(o xy)))=(1)/(16)=(1)/(4)` | |
| 42. |
The length of a sonometer wire is `0.75m`, and density `9xx10^(3)m`. It can bear a stress of `8.1xx10^(8)N//m^(2)` without exceeding the elastice limit. What is the fundamental frequency that can be produced in the wire? |
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Answer» Here, `l=0.75m, rho=9xx10^(3)kg//m^(3)`. stress`=8.1xx10^(8)N//m^(2),v=?` Let a be the area of cross section of the wire. If m is mass of unit length of wire, then `m=axx1xxrho` `v=(1)/(2l)sqrt((T)/(m))=(1)/(2l)sqrt((stressxxa)/(axx1xxrho))` `=(1)/(2xx0.75)sqrt((8.1xx10^(8))/(9xx10^(3)))` `v=(1)/(1.5)xx3xx10^(2)=200Hz` |
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| 43. |
Calculate the ration of speed of sound in neon to that in water vapours at any temperature. Molecular weight on neon `=2.02xx10^(-2)kg//mol e` and for water vapours, molecular weight is `1.8xx10^(-2)kg//mol e`. |
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Answer» Here, `M_(n)=2.02xx10^(-2)kg//mol e` `M_(w)=1.8xx10^(-2)kg//m ol e` As neon is monoatomic, therefore `gamma_(n)=5//3` and water vapours are non-linear triatomic. `:. Gamma_(w)=4//3` As `upsilon=sqrt((gammarho)/(rho))=sqrt((gamma RT)/(M))` and T is same, therefore, `(upsilon_(n))/(upsilon_(w))=sqrt((gamma_(n))/(M_(n))xxM_(w)/(gamma_(w)))=(5//3xx1.8xx10^(-2))/(2.02xx10^(-2)xx4//3)` `=sqrt((5xx18)/(2.02xx4))=sqrt((9)/(8.08))=sqrt(1.1138)` `(upsilon_(n))/(upsilon_(w))=1.055` |
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| 44. |
A uniform string of length `20m` is suspended from a rigid support. A short wave pulse is introduced at its lowest end. It starts moving up the string. The time taken to reach the support is : (take `g = 10ms^(-2)`)A. `2pisqrt(2)s`B. `2 s`C. `2sqrt(2)s`D. `sqrt(2)s` |
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Answer» Correct Answer - C As `mu` is mass per unit length of the rop, then `mu=m//L` As `upsilon=sqrt((T)/(mu))` `:. (dx)/(dt)=sqrt((mgx//L)/(m//L))=sqrt(g)sqrt(x)` or `(dx)/(sqrt(x))=sqrt(g)dt` Integrating both sides, we get `int_(0)^(L)x^(-1//2)dx=sqrt(g)int_(0)^(L)dt` `[(x^(1//2))/(1//2)]_(0)^(L)=sqrt(g)t` or` (sqrt(L)-0)/(1//2)=sqrt(g)t` `t=(2sqrt(L))/(sqrt(g))=2xxsqrt((20)/(10))=2sqrt(2)s` |
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| 45. |
A sonometer wire of length `1.5m` is made of steel. The tension in it produces an elastic strain of `1%`. What is the fundamental frequency of steel if density and elasticity of steel are `7.7 xx 10^(3) kg//m^(3)` and `2.2 xx 10^(11) N//m^(2)` respectively ?A. `188.5Hz`B. `178.2Hz`C. `200.5Hz`D. `770Hz` |
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Answer» Correct Answer - B Here,`l=1.5m, (DeltaL)/(L)=1%, n=?` `rho=7.7xx10^(3)kg//m^(3)` and `Y=2.2xx10^(11)N//m^(2)` From `Y=(F)/(A(DeltaL//L))` `F=YA((DeltaL)/(L))=2.2xx1^(11)Axx(1)/(100)=2.2Axx10^(9)` From `v=(1)/(2l)sqrt((T)/(m))` `v=(1)/(2l)sqrt((F)/(Axx1xxrho))=(1)/(2xx1.5)sqrt((2.2xx10^(9)A)/(Axx7.7xx10^(3)))` `v=(10^(3))/(3)sqrt((2)/(7))=178.Hz` |
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| 46. |
A parallel `-` plate air capacitor whose electrodes are shaped as discs of radius `R=6.0 cm` is connected to a source of an alternating sinusoidal votage with frequency `omega=1000s^(-1)`. Find the ratio of peak values of magnetic and electric energies within the capacitor. |
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Answer» Inside the condenser the peak electrical enegry `W_(e) = (1)/(2)CV_(m)^(2)` `= (1)/(2)V_(m)^(2)(epsilon_(0)piR^(2))/(d)` `(d=` separation between the plates, `piR^(2) =` area of each plate.) `V = V_(m) sin omegat, V_(m)` is the maximum voltage Changing electric field causes a displacement current `j_(dis) = (delD)/(delt) = epsilon_(0)E_(m) cos omega t` `= (epsilon_(0)omega V_(m))/(d) cos omegat` This gives rise to a magnetic field `B(r)` (at a radial distance `r`from the centre of the plate) `B(r).2pir = mu_(0)pir^(2) j_(dis) = mu_(0)pir^(2) (epsilon_(0) omega V_(m))/(d) cos omega t` `B = (1)/(2)epsilon_(0)mu_(0) omega (r )/(d) V_(m) cos omega t` Energy associated with this field is `= intd^(3) r(B^(2))/(2mu_(0)) = (1)/(8) epsilon_(0)^(2) (omega^(2))/(d^(2)) pi underset(0)overset(R )int r^(2) rdr xx d xx V_(m)^(2) cos^(2) omegat` `=(1)/(16) pi epsilon_(0)^(2)mu_(0)(omega^(2)R^(4))/(d) V_(m)^(2)cos^(2) omegat` Thus the maximum magnetic enegry `W_(m) = (epsilon_(0)^(2)mu_(0))/(16)(omegaR)^(2) (piR^(2))/(d)V_(m)^(2)` Hence `(W_(m))/(W_(e)) = (1)/(8)epsilon_(0)mu_(0)(omegaR)^(2) = (1)/(8) ((omega R)/(c))^(2) = 5xx10^(-15)` The approximation are valid only if `omega R lt lt c`. |
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| 47. |
In a sinusoidal wave the time required for a particular point to move from equilibrium position to maximum displacement is `0.17s`, then the frequency of wave is:A. `1.47Hz`B. `0.36Hz`C. `0.73Hz`D. `2.94Hz` |
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Answer» Correct Answer - A `T=4xx` time from max. displacemetn to zero displacement `=4xx0.170=0.68s` `v=(1)/(T)=(1)/(0.68)=1.47Hz` |
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| 48. |
The frequency of the first overtone of a closed organ pipe is same as that of first overtone of an open organ pipe. What is the ration of their lengths? |
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Answer» Frequency of first overtone of closed pipe, `n_(1)=(3v)/(4l_(1))` Frequency of 1st overtone of open pipe, `n_(2)=(2v)/(2l_(2))` As `n_(1)=n_(2) :. (3v)/(4l_(1))=(2v)/(2l_(2)) :. (l_(1))/(l_(2))=(3)/(4)` |
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| 49. |
How is second resonance length `(l_(2))` in the resonance apparatus related to first resonance length `(l_(1))` ?A. `l_(2)=l_(1)`B. `l_(2)=2l_(1)`C. `l_(2)=3l_(1)`D. `l_(2) ` is slightly greater than `3l_(1)` |
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Answer» Correct Answer - D As is known for theory, `l_(2)gt3l_(1)`. |
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| 50. |
The resonance lengths obtained by a student using a fork in resonance apparatus are `15.6 cm` and `47.6 cm`. If velocity of sound in air at room temperature is `327.5 m//s,` the frequency of fork isA. `512 Hz`B. `256Hz`C. `520 Hz`D. `250Hz` |
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Answer» Correct Answer - A Here, `l_(1)=15.6cm, l_92)=47.6cm` `upsilon=327.5 m//s, v=?` From` upsilon=2v(l_(2)-l_(1))` `v=(upsilon)/(2(l_(2)-l_(1)))=(327.5xx100)/(2(47.6-15.6))=512Hz` |
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