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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.
| 51. |
A balloon is filled with `CO_(2)`. How will it behave as a lens for sound waves? What happens if `CO_(2)` is replaced by hydrogen? |
| Answer» A balloon filled with `CO_(2)` behaves like a convex lens. This is because, velocity of sound in `CO_(2)` is less than that in air. When `CO_(2)` is replaced by hydrogen, the balloon behaves as a concave lens as velocity of sound in hydrogen is greater than that in air. | |
| 52. |
Where to pluck and where to touch a stretched string to excite its first overtone ? |
| Answer» To excite the first overtone, the string must vibrate in two segments of equal length. Therefore, the string must be touched at its centre and plucked at one fourth of the length from one end. | |
| 53. |
In resonance apparatus, if water is replaced by some oil of higher density, how will the frequency of note produced change? |
| Answer» The frequency of note produced will not change as it depends only on length of air column. Water`//`oil surface acts as a reflector. | |
| 54. |
The equation of a plane progressive wave travelling along positive direction of `x-`axis is `y=r sin [(2pit)/(T)-(2pix)/(lambda)]` where `y=` displacement of particle at `(x,t),r=` amplitude of vibratio of particle, `T=` time period of wave motion, `lambda=` wavelength of wave ,` x=` starting distance of wave from the origin. Velocity of wave, `upsilon=vlambda=(lambda)/(T)=` constant. Acceleration of wave, `a=0`. Velocity of particle at time `t=(dy)/(dt)` Acceleration of particle at time `t=(d^(2)y)/(dt^(2))` A harmonic wave travelling along positive direction of x axis is represented by `y=0.25xx10^(-3)sin (500t-0.025x)` where `x` and `y` are in metre and `t` is in second The amplitude of vibration of particle isA. `0.25xx10^(-3)cm`B. `0.25xx10^(-3)m`C. `500m`D. `0.025m` |
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Answer» Correct Answer - B Compare the given equation with the standard form `y=rsin ((2pit)/(T)-(2pix)/(lambda))` Amplitude of vibration of particle, `r=0.25xx10^(-3)m` |
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| 55. |
A body of mass 10 kg is suspended by a massless coil spring of natural length 40 cm and force constant `2.0xx10^(3)Nm^(-1)`. What is stretched length of the spring.? If the body is pulled down further stretching the spring to a length 48cm and then released, what is the frequency of oscillations of the suspended mass ? `g=10ms^(2)` |
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Answer» Here, `m=10kg, k=2.0xx10^(3)Nm^(-1)` Natural length `l=40cm` Extension produced in the spring dur to 10kg mass `x=(mg)/(k)=(10xx10)/(2.0xx10^(3))=5xx10^(-2)m=5cm` Stretched length of spriing `=l+x=40+5` `=45cm` When the loaded spring is stretched further, there will be no change in the frequency of oscillation of the loaded spring. If the loaded spring after further stretching is left free, it will execute linear SHM. Its frequency of oscillation `v=(1)/(2pi)sqrt((k)/(m))=(1)/(2xx(22//7))sqrt((2.0xx10^(3))/(10))=(7xx10sqrt(2))/(2xx22)` `=2.25Hz` |
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| 56. |
Two spring have force constants `k_(1)` and `k_(2)` respectively. They are attached to a mass m and two fixed supports as shwon in figure. If the surface is frictionless, fing the time period of oscillations. What is the spring factore of this combination. |
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Answer» If the mass m is displaced a little through distance x towards right hand side, the spring `k_(1)` gets compressed and spring `k_(2)` gets stretched. Due to it, the restoring froces `F_(1)` and `F_(2)` developed in two springs will be towards left, i.e., in the same direction. Since `k_(1)` and `k_(2)` are the spring constants fo the two springs. hence, `F_(1)=-k_(1)x` and `F_(2)=-k_(2)x` Total restoring force, `F=F_(1)+F_(2)=(-k_(1x)+(-k(2)x)` `=(k_(1)+k_(2))x` ...(i) i.e., `Fprop x`. This force F is directed towards equilibrium position of th ebody. If body is left free, it will execute linear SHM. If k is the force constant of a spring which is equivalent to the combination of the two spring constants as given above, then `F=-ky` ...(ii) Form (i) and (ii), `k=k_(1)+k_(2)` Here, spring factor`=k_(1)+k_(2)` Inertia factore `=` mass of the body`=m` `:.` Time period , `T=2pisqrt(("inertia factor")/("spring factor"))` `=2pisqrt((m)/(k_(1)+k_(2)))` |
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| 57. |
A body of mass `m` was suspended by a non-stretched spring, and then set fre without push. The stiffness of the spring is `x`. Neglecting the mass of the spring , find `:` `(a)` the law of motion `y(t)`, whee `y` is displacement of the body from the equilibrium position, `(b)` the maximum and minimum tensions of the spring in the process of motion. |
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Answer» Let `y(t)=` displacement of the body from the end of the unstretched position of the spring (not the equilibrium position). Then `m ddot(y)=-ky+mg` This equation has the soloution of the form `y=A+B cos (omegat+alpha)` if `- m omega^(2)B cos ( omegat +alpha)=-k[A+B cos(omegat+ alpha)]+ mg` Then ` omega^(2)=(k)/(m)` and `A=(mg)/(k)` we have `y=0` and `y=0` at `t=0`. So `- omega B sin alpha=0` ` A+Bcos alpha=0` Since `B gt 0` and `A gt 0` we have `alpha=pi` `B=A=(mg)/(k)` and `y=(mg)/(k)(1-cos omegat)` `(b)` Tension in the spring is `T=ky=mg(1-cos omegat)` so `T_(max)=2 mg, T_(m i n)=0` |
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| 58. |
An alternating current of frequency `omega=314s^(-1)` is fed to a circuit consisting of a capacitor of capacitance `C=73 muF` and an active resistance `R=100 Omega` connected in parallel. Find the impedance of the circuit. |
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Answer» `(1)/(cancel(Z))=(1)/(r)+(1)/((1)/(iomegaC))=(1)/(R)+iomegaC=(1+iomegaRC)/(R)` `|cancel(Z)|=(R)/( sqrt(1+(omegaRC)^(2)))=40 Omega` |
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| 59. |
A particle of mass (m) is attached to a spring (of spring constant k) and has a narural angular frequency omega_(0). An external force `R(t)` proportional to cos omegat(omega!=omega)(0) is applied to the oscillator. The time displacement of the oscillator will be proprtional to. |
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Answer» Given the natureal angular frequency `=omega_(0)`. If the displacement of the particle is y, then acceleration of the particle is, `a_(0)=-omega_(0)^(2)y` The external force `F(t)propcosomegat.` It has an angular frequency `omega`. For the displacement y, the acceleration produced by this force is `a_(1)=omega^(2)y` Net acceleration of the particle at displacement y is `a=a_(0)+a_(1)=-omega_(0)^(2)y+omega^(2)y=-(omega_(0)^(2)-omega^(2))y` Net force on the particle at displacement y is `F=ma=-m(omega_(0)^(2)-omega^(2))y` or `y=(F)/(m(omega_(0)^(2)-omega^(2)))` So`yprop (1)/(m(omega_(0)^(2)-omega^(2)))` |
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| 60. |
A plane electromagnetic wave with frequency `omega` falls upon an elastically bonded electron whose natural frequency equals `omega_(0)`. Neglecting the damping of oscillations, find the ratio of the mean enegry dissipated by the electron per unit time to the mean value of the enrgy flow density of the incident wave. |
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Answer» For the elastically bound electron `mddotoversetrarr(r) =m omega_(0)^(2) overset(r ) = eoverset(E_(0)) cos omegat` This equation has the particular intergal (i.e. neglecting the part which does not have the frequency of the impressed force ) `oversetrarr(r) = (eoversetrarr(E_(0)))/(m) (cos omegat)/(omega_(0)^(2) - omega^(2))` so and `ddotoversetrarr(p) =- (e^(2)oversetrarr(E_(0))omega^(2))/((omega_(0)^(2)-omega^(2))m)cos omegat` Hence `P =` mean radiated power `= (1)/(4piepsilon_(0)) (2)/(3c^(3)) ((e^(2)omega^(2))/(m(omega_(0)^(2)-omega2)))^(2)(1)/(2)E_(0)^(2)` The mean incident poynting flux is `lt S_("inx") gt = sqrt((epsilon_(0))/(mu_(0)) (1)/(2)E_(0)^(2)` Thus `(P)/( lt S_("inc") gt ) = (mu_(0)^(2))/(6pi) ((e^(2))/(m))^(2) (omega^(4))/((omega_(0)^(2) - omega^(2))^(2))`. |
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| 61. |
The equation for a wave travelling in x-direction on a string is `y =(3.0cm)sin[(3.14 cm^(-1) x - (314s^(-1))t]` (a) Find the maximum velocity of a particle of the string. (b) Find the acceleration of a particle at x =6.0 cm at time t = 0.11 s. |
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Answer» The eqn. of the wave is `y=3sin[3.14x=314t]` `upsilon=(dy)/(dt)=3cos[3.14x-314t]xx314` `v_(max)=3xx314xx1cm//s=9.4m//s` (b) `a=(dupsilon)/(dt)=-3 sin [3.14x-314t]xx(314)^(2)` `=-3(314)^(2)sin[3.14xx6-314x0.11]` `=3(314)^(2)sin(6pi-11)=Zero` |
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| 62. |
The velocity amplitude of a particle is equal to half the maximum value at the frequencies `omega_(1)` and `omega_(2)` of external harmonic force. Find`:``(a)` the frequency corresponding to the velocity resonance, `(b)` the damping coefficient `beta` an dthe damped oscillation frequency `omega` of the particle. |
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Answer» `x=(F_(0))/(m)((omega_(0)^(2)-omega^(2)) cos omegat + 2 beta sinomegat )/(sqrt((omega^(2)- omega_(0)^(2))^(2)+ 4 beta^(2) omega^(2)))` Then `dot(x)=(F_(0)omega)/(m)(2 beta omega cos omegat + ( omega^(2) - omega_(0)^(2)) sin omegat)/( ( omega_(0)^(2)- omega^(2))^(2) + 4 beta^(2) omega^(2))` Thus the velocity amplitude is `V_(0)=( F_(0 ) omega)/( m sqrt(( omega_(0)^(2)- omega^(2))^(2) + 4 beta^(2) omega^(2)))` ` = ( F_(0))/(msqrt(((omega_(0)^(2))/( omega)-omega)^(2)+ 4 beta^(2)))` This is maximum when `omega^(2)=omega_(0)^(2)=omega_(res)^(2)` and then `V_(0 res) =(F_(0))/(2 m beta)` Now at half maximum `((omega_(0)^(2))/( omega)-omega)^(2)=12 beta^(2)` or ` omega^(2)+-2 sqrt(3) beta omega - omega_(0)^(2)=0` `omega=+- betasqrt(3)+ sqrt(omega_(0)^(2)+ 3 beta^(2))` where we have rejected a solution with `- ve ` sign before there dical . Writing `omega_(1)=sqrt(omega_(0)^(2)+3 beta^(2))+ betasqrt(3), omega_(2)=sqrt(omega_(0)^(2)+ 3 beta^(2))- betasqrt(3)` we get `(a) omega_(res)=omega_(0)=sqrt(omega_(1)omega_(2))` ( Velocity resonance frequency ) (b) `beta=(| omega_(1)-omega_(2)|)/( 2 sqrt(3))` and damped oscillation frequency `sqrt(omega_(0)^(2)-beta^(2))=sqrt(omega_(1)omega_(2)-((omega_(1)-omega_(2))^(2))/( 12))` |
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| 63. |
The forced harmonic oscillations have equal displacement amplitude at frequencies `omega_(1)=400 s^(-1)` and `omega_(2)=600 s^(-1)`. Find the resonance frequency at which the displacement amplitude is maximum. |
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Answer» Since `a=(F_(0)//m)/(sqrt((omega^(2)-omega_(0)^(2)+2 beta^(2))^(2)+ 4 beta^(2)( omega_(0)^(2)- beta^(2))))` we must have `omega_(1)^(2)-omega_(0)^(2)+2 beta^(2)=-( omega_(2)^(2)-omega_(0)^(2)+ 2 beta ^(2))` or ` omega_(0)^(2)-2 beta^(2) =( omega_(1)^(2)+ omega_(2)^(2))/( 2) = omega_(res)^(2)` |
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| 64. |
A particle is performing SHM along `x-`axis with amplitude `4.0cm` and time period `1.2s` . What is the minimum time is deci`-` second taken by the particle to move from `x=+2cm `to `x=+4cm` and back again. |
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Answer» Correct Answer - 4 As `x=asin omegat =asin((2pi)/(T))t` so, `t=(T)/(2pi)sin^(-1)((x)/(a)),`where`a=4cm.` At `x=2,` `t=(T)/(2pi)sin^(-1)((2)/(4))=(T)/(2pi)xx(pi)/(6)=(T)/(12)=(1.2)/(12)=(1)/(10)` At `x=4` `t=(T)/(2pi)sin^(-1)((4)/(4))=(T)/(2pi)xx(pi)/(2)=(T)/(4)=(1.2)/(4)=(3)/(10)` so, time taken in going from `x=+2cm` to `x=+4cm` will be `(3)/(10)-(1)/(10)=(2)/(10)s=2` deci second, and same time will be taken for coming back. So, total time taken `=2+2=4` deci second |
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| 65. |
Two physical pendulums perform small oscillations about the same horizontal axis with frequencies `omega_(1)` and `omega_(2)` . Their moments of inertia relative to the given axis are equal to `I_(1)` and `I_(2)` respectively. In a state of stable equilibium the pendulums were fastened rigifly together. What will be the frequency of small oscillations of the compound pendulum ? |
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Answer» When the two pendulums are joined rigidly and set to osciallate, each exert torques on the other, these torques are equal and opposite. We write the law of motion for the two pendulums as `I_(1)ddot(theta)=- omega_(1)^(2)I_(1)theta+G` `I_(2)ddot(theta)=- omega_(2)^(2)I_(2)theta-G` where `+-G` is the torque of mutual interactions. We have written forces on each pendulum in the absence of the other as `-omega_(2)^(2)I_(1)theta` and `-omega_(2)^(2)I_(2)theta` respectively. Then `ddot(theta)=-(I_(1)omega_(1)^(2)+I_(2)omega_(2)^(2))/(I_(1)+I_(2))theta=-omega^(2)theta` Hence `omega=sqrt((I_(1)omega_(1)^(2)+I_(2)omega_(2)^(2))/(I_(1)+I_(2))` |
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| 66. |
Two simple pendulums of length `1m` and `16m` respectively are both given small displacements in the same direction at the same instant. After how many oscillations of the shorter pendulum will, the two pendulums vibrate in the same phase ? |
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Answer» Correct Answer - 4 Time period, `T=2pisqrt((l)/(g))` or `T propsqrt(l)` `:. (T_(2))/(T_(1))=sqrt((l_(2))/(l_(1)))=sqrt((16)/(1))=4` or `T_(2)=4T_(1)` It means, when a pendulum of smaller length will complete 4 oscillations, the pendulum of large length will complet 1 oscillation. It means, the two pendulums will be in the same phase, when shorter has completed 4 oscillations. |
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| 67. |
A light wave is reflected from a mirror. The incident and reflected waves superimpose to form stationary waves. But no nodes and antinodes are seen. Why? |
| Answer» As is known, the distance between two sudccessive nodes or two successive antinodes is `lambda//2`. The wavelength of visible light is of the order of `10^(-7)`m. As such , a small distance cannot be detected by the eye or by an ordinary optical instrument. Therefore, nodes and antinodes are not seen. | |
| 68. |
Assertion `:` In the `nth` normal mode of vibration of a string, there are `(n+1)` nodes and `n` antinodes. Reason `:` This is because both the ends of the string are fixed and from nodes.A. both, Assertion and Reason are true and Reason is the correct explanation of the Assertion.B. both, Assertion and Reason are true but Reason is not a correct explanation of the Assertion.C. Assertion is true but the Reason is false.D. both,Assertion and Reason are false. |
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Answer» Correct Answer - A Reason is correct explanation of the assertion, which it true. Position of nodes is given by `x=0,(L)/(n), (2L)/(n), (3L)/(n)......(nL)/(n)` Antinodes are at `x=(L)/(2n), (3L)/(2n), (5L)/(n),......` |
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| 69. |
Assertion : Sound produced by an open organ pipe is richer than the sound produced by pipe from both ends, in case of oper organ pipe.A. both, Assertion and Reason are true and Reason is the correct explanation of the Assertion.B. both, Assertion and Reason are true but Reason is not a correct explanation of the Assertion.C. Assertion is true but the Reason is false.D. both,Assertion and Reason are false. |
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Answer» Correct Answer - B Sound produced by an open organ pipe is richer because (i) it contains all harmonics. No harmonic is missing. (ii) frequency of fundamental note in an open organ pipe is twice the fundamental frequency in a closed organ pipe of same length. The assertion is true. The statement of reason is not wrong.But it does not explain the assertion. |
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| 70. |
If the intensity increased by a factor of 20, byhow may decibels is the sound level increased?A. 18B. 13C. 9D. 7 |
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Answer» Correct Answer - B Suppose for the intensity `I`, the correspongind decibel level is `beta_(1)`. When the intensity is increased to `20I`, then corresponding increase in decibels level be `beta_(2)`. Then`beta_(1)=10log(I//I_(0))` and `beta_(2)=10 log (20I//I_(0))` `:.beta_(2)-beta_(1)=10log(20I//I_(0))-10log(I//I_(0))` `=10log (20I//I)=10log 20` `=10xx1.3142=13.142dB~~13dB` |
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| 71. |
A transverse harmonic wave on a string is described by `y(x,t)=3.0 sin (36t+0.018x +pi//4)`, where x and y are in cm and t is in s. The positive direction of x is from left to right.A. The wave is travelling from right to left.B. The speed of the wave is `20m//s`C. Frequency of the wave is 5.7 HzD. The least distance betweent two successive crests in the wave is 2.5cm |
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Answer» Correct Answer - A::B::C The given equation is `y(x,t)=3.0sin [36t+0.018x+(pi)/(4)]` (a As positive direction is from left to right and x positive, therefore, wave is travelling from right to left. (b) Compare the given equation with the standard form `y=rsin[(2pit)/(T)+(2pix)/(lambda0+phi)]` `(2pi)/(T)=36, (2pi)/(lambda)=0.018` speed of wave, `upsilon=(lambda)/(T)=(36)/(0.018)=2000cm//s=20m//s` (c) Again, `T=(2pi)/(36)=(pi)/(18)` Frequency, `v=(1)/(V)=(18)/(pi)Hz=5.7Hz` |
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| 72. |
If x, and a denote the displacement, the velocity and the acceler of a particle executing simple harmonic motion of time period T, then, which of the following does not change with time?A. `aT//x`B. `At+2piupsilon`C. `aT//upsilon`D. `a^(2)T^(2)+4pi^(2)upsilon^(2)` |
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Answer» Correct Answer - A `x=rsiniomegat, upsilon=(dx)/(dt)=romegacosomegat,` `a=(dupsilon)/(dt)=-romega^(2)sinomegat =-omega^(2)x.` So, `(aT)/(x)=(-omega^(2)xxxT)/(x)=-omega^(2)T=-(4pi^(2))/(T^(2))xxT` `=-(4pi^(2))/(T)=a ` constant `(aT)/(upsilon)=(-omega^(2)rsinomegat xxT)/(omegar cos omegat )=-omegaTtanomegat` `=(2pi)/(T)xxTtanomegat =(`not constant ) `aT+2piupsilon` is not constant. `a^(2)T^(2)+4pi^(2)upsilon^(2)` is also not constant. |
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| 73. |
A particle executes SHM with a time period of 2s and amplitude 5 cm. Find (i) displacement (ii) velocity (iii) acceleration, after `1//3` second, starting from mean position. |
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Answer» Here, `T=2s,a=5cm,t=(1)/(3)s`. (a) Displacement , `y=asin((2pi)/(T))t=5sin((2pi)/(2))xx(1)/(3)=5xx(sqrt(3)/(2))=4.33cm` `(b)` Velocity , `V=(dy)/(dt)=a(2pi)/(T)cos((2pi)/(T))t=5xx(2pi)/(2)xos((2pi)/(2))xx(1)/(3)=7.86cm//s` `(c)` Acceleration, `A=(dV)/(dt)=-(4ppir^(2))/(T^(2))asin((2pi)/(T))t=(4xxpi^(2))/(4)xx5xxsin((2pi)/(2))xx(1)/(3)` `=(4)/(4)xx((22)/(7))^(2)xx5xx(sqrt(3))/(2)=-42.77cm//s^(2)` |
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| 74. |
A particle executes SHM with an amplitude 4 cm. Locate the position of point where is speed is half its maximum speed. At what displacement is potential energy equal to kinetic energy? |
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Answer» Here, `a=4cm, V=omegasqrt(a^(2)-y^(2))` `(aomega)/(2)= omega sqrt(a^(2)-y^(2)) or (a^(2))/(4)=a^(2)-y^(2)` or `y^(2)=a^(2)-a^(2)//4=3a^(2)//4` or `y=sqrt(3)a//2 =sqrt(3)xx4//2=2sqrt(3)cm` As, `P.E.=K.E. ,` therefore, `(1)/(2)momega^(2)y^(2)=(1)/(2)momega^(2)(a^(2)-y^(2))` or `y^(2)=a^(2)-y^(2) or 2y^(2)=a^(2)` or `y=a//sqrt(2)=4//sqrt(2)=2sqrt(2)cm` |
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| 75. |
A particle executes SHM of period 8 seconds. After what time of its passing through the mean position will the energy be half kinetic and half potential ? |
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Answer» Here, `T=8s,` As, `PE=KE` `:. (1)/(2)ky^(2)=(1)/(2)k(A^(2)-y^(2)) or y^(2)=A^(2)-y^(2)` `y=(A)/(sqrt(2))` Now, `y=Asin omega t =A sin ((2pi)/(T))t` `:. (A)/(sqrt(2))=A sin ((2pi)/(8))t` or `sin ((pit)/(4))=(1)/(sqrt(2))=sin ((pi)/(4))` or ` (pit)/(4)=(pi)/(4) or t=1s` |
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| 76. |
A body is dropped in a hole drilled across diameter of the earth. Show that it executes SHM Asuume earth to be a homogenous sphere, find the time period of its motion. Given density of earth is `5.51 xx10^(3)Kgm^(-3)` and `G=6.67xx10^(-11)Nm^(2)kg^(-2)` |
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Answer» `F=Gmxx(4)/(3)pi((R-d)^(3)rho)/((R-d)^(2))-(4)/(3)piGrhomy` where, `y=(R-d)` `:.F prop y` and is directed towards centre O of earth. Hence, the body will execute linear SHM with O as mean position. Here, inertia factor `=` m Spring factor `=(4)/(3)piGrhom` Time period, `T=2pisqrt((i n ertia fact o r)/(spri ng fact o r))` `=2pisqrt ((m)/((4)/(3)piGrhom))=2pisqrt((3)/(4pirGrho))` |
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| 77. |
A point moves in the plane `xy` according to the law `x=a sin omega t, y=b cos omegat , ` where `a,b` and `omega` are positive constants. Find `:` `(a)` the trajectory equation `y(x)` of the point and the direction of its motion along this trajectory , `(b)` the acceleration `w` of the point as a function of its radius vector `r` relative to the orgin of coordinates. |
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Answer» From the Eqn `: x= a sin omega t` `sin^(2)omegat=x^(2)//a^(2)` or `cos^(2) omegat =1-(x^(2))/(a^(2)) ....(1)` And from the equation `: y=b cos omegat ` `cos^(2)omegat =y^(2)//b^(2) ...(2)` From Eqns `(1)` and `(2)`, we get `:` `1-(x^(2))/(a^(2))=(y^(2))/(b^(2))` or `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` Which is the standard equation of the ellipse shown in the figure. we observe that, at `t=0, x=0` and `y=b` and at `t=(pi)/(2 omega), x=+a` and `y=0` Thus we observe that at `t=0`, the point is at point (figure) and at the following moments, the co- ordinat `y` diminishes and `x` beomes positive. Consequenctly the motion is clock-wise `(b)` As `x=a sin omegat ` and `y=b cos omegat` so we amy write `vec(r)= a sin omega t vec(i+)b cos omegat vec(j)` Thus `ddot(r)=vec(w)=-omega^(2)vec(r)` |
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| 78. |
A point moves along th e`x` axis according to the law `x=a sin^(2)(omegat-pi//4)` Find. `(a)` the amplitude and period oscillations, draw the plot `x(t),` `(b)` the velocity projection `upsilon_(x)` as a function of the coordination `x`, draw the plot `upsilon_(x)(x)`. |
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Answer» From the motion law of particle `x=asin^(2)(omegat-pi//4)=(a)/(2)[a-cos(2omegat-(pi)/(2))]` or, `x-(a)/(2)=-(a)/(2)cos (2omegat-(pi)/(2))=-(a)/(2)sin 2 omegat =(a)/(2)sin (2omegat+pi)` i.e. `x=-(a)/(2)=(a)/(2)sin (2omegat+pi) .......(1)` Now compairing this equation with the general equaltion of harmonic oscillations `:` `X=Asin (omega_(0)t+alpha)` Amplitude, `A=(a)/(2)` and angular frequency , `omega_(0)=2omega`. Thus the period of one full oscillation, `T=(2pi)/(omega_(0))=(pi)/(omega)` `(b)` Differentiating Eqn (1) w.r.t. time `v_(x)=a omega cos (2omegat+pi)` or ` v_(x)^(2)=a^(2)omega^(2)cos^(2)(2omegat+pi)=a^(2)omega^(2)[1-sin^(2)(2omegat+pi)] .......(2)` From Eqn (1) `(x-(a)/(2))^(2)=(a^(2))/(4)sin ^(2)(2omegat+pi)` or,` 4(x^(2))/(a^(2))+1-(4x)/(a)=sin^(2)(2omegat+pi)` or `1-sin^(2)(2 omegat+pi)= (4x)/(a)(1-(x)/(a)) ....(3)` From Eqns `(2)` and (3), `v_(x)=a^(2)omega^(2)(4x)/(a)(1-(x)/(a))=4 omega^(2)x(a-x)` Plot of `v_(x)(x)` is as shown in the answersheet. |
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| 79. |
An arrangedment illlustrated in figure consists of a horizontal uniform disc `D` of mass `m` and radius `R` and a thin rod `AO` whose torsional coefficient is equal to `k`. Find the amplitude and the energy of small torsional oscillationa if at the initial momentu the disc was deviated through an angle `varphi_(0)` from the equilibrium position and then imparted an angular velocity `varphi_(0)`. |
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Answer» The K.E. of the disc is `(1)/(2)Idot(varphi)^(2)=(1)/(2)((mR^(2))/(2))dot(varphi^(2))=(1)/(4)m R^(2) dot(varphi^(2))` The torsional potential energy is `(1)/(2)k varphi^(2)`. Thus the total energy is `:` `(1)/(4)m R^(2)dot(varphi^(2))+(1)/(2)k varphi^(2)=(1)/(4)m R^(2) ddot(varphi_(0)^(2))+(1)/(2)k varphi_(0)^(2)` By definition of the amplitude `varphi_(m), dot(varphi)=0` when `varphi=varphi_(m)` . Thus total energy is `(1)/(2)k varphi_(m)^(2)=(1)/(4)m R^(2)dotvarphi_(0)^(2)+(1)/(2)k varphi_(0)^(2)` or` varphi_(m)=varphi_(0)sqrt(1+(mR^(2))/(2k)(varphi_(0)^(2))/(varphi_(0)^(2)))` |
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| 80. |
A thin uniform disc of mass `m` and radius `R` suspended by an elastic thread in the horizontal plane performs torsional oscillations in a liquid. The moment of leastic forces emerging in the thread is equal to `N=alpha varphi`, where `alpha` is a constant and `varphi` is the angle of rotation from the equilibrium position. The resistance force acting on a unit area of the disc is equal to `F_(1)eta v`, where` eta` is a constant and `v` is the velocity of the given element of the disc relative to the liquid. Find the frequency of small oscillation. |
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Answer» Let us calculate the moment `G_(1)` of all the resistive forces on the disc. When the disc rotates an element `(r d r d theta ) ` with coordinates `( r, theta)` has a velocity `r dot ( varphi)` where `varphi` is the intantaneious angle of rotation from the equilibrium position and `r` is measured from the centre. Then `G_(1)=int_(0)^(2pi)d theta int _(0)^(R) dr. r. (F_(1)xxr)` `=int _(0)^(R) eta r dot(varphi) r^(2) d gammaxx2pi =( eta pi R^(4))/( 2) dot ( varphi)` Also moment of inertia `=( mR^(2))/( 2)` Thus `( mR^(2))/( 2) ddot( varphi)+ ( pi etaR^(4))/( 2) dot (varphi) + alpha varphi=0` or` ddot(varphi) +2(pi eta R^(2))/(2m) dot( varphi)+(2 alpha)/(m R^(2))varphi=0` Hence `omega_(0)^(2)=(2alpha)/( mR^(2))` and ` beta =( pi eta R^(2))/( 2m )` and angular frequency `omega=sqrt(((2 alpha)/( mR^(2)))-((pi etaR^(2))/(2m))^(2))` Note `:-` normally by frequency we mean `( omega)/( 2pi)`. |
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| 81. |
A uniform horizontal disc fixed at its centre to an elastic vertical rod performs forced torsional oscillations dur to the moment of forces `N_(z)=N_(m)cos omegat`. The oscialltions obey the law `varphi=varphi_(m) cos ( omega t- alpha)`. `(a)` the work performed by friction forces acting on the disc during one oscillation period , `(b)` the quality factor of the given oscillator if th emoment of inertia of the disc relative to the axis is equal to `I`. |
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Answer» The equation of the disc is `ddot(varphi)+ 2 beta dot (varphi)+ omega_(0)^(2) varphi=(N_(m) cos omegat)/( I)` Then as before `varphi=varphi_(m) cos ( omegat- alpha)` where `varphi(m)=(N_(m))/(I[( omega_(0)^(2)-omega^(2))^(2)+ 4 beta^(2)omega^(2)] ^(1//2)), tan alpha=(2 beta omega)/( omega_(0)^(2)- omega^(2))` `(a)` Work performed by frictional forces `=-int N_(r) d varphi` where`N_(r)=-2 I beta dot(varphi)=- int_(0)^(T) 2 betaI dot(varphi^(2))dt=-2 pi beta omega I varphi_(m)^(2)` `=- pi I varphi_(m)^(2)[( omega_(0)^(2)- omega^(2))^(2)+4 beta^(2)omega^(2)]^(1//2)sin alpha=- pi N_(m) varphi_(m) sin alpha` `(b)` The quality factor `Q=(pi)/(lambda)=(pi)/(betaT)=(sqrt(omega_(0)^(2)-beta^(2)))/( 2 beta)=( omegasqrt(omega_(0)^(2)-beta^(2)))/( ( omega_(0)^(2)- omega^(2) ) tan alpha) =(1)/(tan alpha){( 4 omega^(2) omega_(0)^(2))/((omega_(0)^(2)-omega^(2))^(2))-(4 beta^(2)omega^(2))/((omega_(0)^(2)-omega^(2))^(2))}` `=(1)/(2 tan alpha) { ( 4 omega^(2) omega_(0)^(2) I^(2) varphi_(m)^(2))/( N_(m)^(2)cos^(2) alpha)- tan ^(2) alpha}`since` omega_(0)^(2)= omega^(2)+(N)/( Ivarphi_(m))cos alpha` `=(1)/(2 sin alpha){(4 omega^(2) omega_(0)^(2) I^(2) varphi_(m)^(2))/( N_(m)^(2)) - sin ^(2) alpha}^(1//2)` `=(1)/(2 sin alpha){ ( 4 omega^(2) I^(2) varphi_(m)^(2))/( N_(m)^(2))(omega^(2)+(N_(m) cos alpha)/( I varphi_(m)))+ 1- cos^(2) alpha}^(1//2)` `=(1)/( 2sin alpha){( 4 I^(2) varphi_(m)^(2))/( N_(m)^(2))omega^(4)+ ( 4Ivarphi_(m))/( N_(m))omega ^(2) cos alpha+ cos ^(2) alpha-1}^(1//2)=(1)/( 2sin alpha ){((2Ivarphi_(m)omega^(2))/( N_(m))+ cos alpha)^(2)-1}^(1//2)` |
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| 82. |
Find the frequency of small oscillatinos of a thin uniform vertical rod of mass ``m and length `l` hinged at the point `O` (figure). The combined stiffness fo the springs is equal to `x`. The mass of the springs is negligible. |
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Answer» Let us locate the rod at the position when it makes an angle `theta` from the vertical . In this problem both, the gravity and spring forces are restoring conservative forces, thus from the conservation of mechanical energy of oscillation of the oscillation system `:` `(1)/(2)(ml^(2))/(3)(theta)^(2)+mg(l)/(2)(1-cos theta)+(1)/(2)k(l theta)^(2)=` constant Differentiating w.r.t. time, we get `: ` `(1)/(2) (ml^(2))/(3)2 dot (theta)ddot(theta)+(mgl)/(2)sin thetadot(theta)+(1)/(2)k l^(2)2 theta dot (theta)=0` Thus for very small `theta` `ddot(theta)=-(3g)/(2l)(1+(kl)/(mg))theta` Hence, `omega_(0)=sqrt((3g)/(2l)(1+(kl)/(mg)))` |
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| 83. |
A set of 25 tuning forks is arranged in order of decreasing frequency. Each fork gives 3 beats with succeding one. The first fork is octave of the last. Calculate the frequency of the first and 16th fork. |
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Answer» Let frequency of last fork be n. `:.` Freq. of 1st fork `=2n,` freq. of 2nd fork `=[2n-3],` freq. of 3rd fork `(2n-6)` `=2n-3(3-1) ` and so on `:.` freq. of 25th fork `=2n-3(25-1)=n` `:.` `n=72` Hence, freq. of 1st fork `=2n=144Hz` and freq. of 16th fork `=144-3(16-1)` `=99Hz` |
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| 84. |
A speed ign motorcyclist sees traffic ham ahead of him. He slows doen to `36 km//h` He finds that traffic has eased and a car moving ahead of him at `18 km//h` is honking at a frequency of 1392 Hz. If the speed of sound is `343m//s`, the frequency of the honk as heard by him will beA. `1332Hz`B. `1372Hz`C. `1412Hz`D. `1454Hz` |
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Answer» Correct Answer - C As listenerg is behind the source of source of sound hence `upsilon_(L)=-36km//h=-10m//s` and`upsilon_(s)=-18km//h=-5m//s` `v=1392Hz, upsilon=343m//s` `=(353)/(348)xx1352~~1412Hz` |
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| 85. |
A uniform rope of legnth `L` and mass `m_1` hangs vertically from a rigid support. A block of mass `m_2` is attached to the free end of the rope. A transverse pulse of wavelength `lamda_1` is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope is `lamda_2`. The ratio `(lamda_2)/(lamda_1)` isA. `sqrt((m_(1)+m_(2))/(m_(2)))`B. `sqrt((m_(2))/(m_(1)))`C. `sqrt((m_(1)-m_(2))/(m_(2)))`D. `sqrt((m_(1))/(m_(2)))` |
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Answer» Correct Answer - A As is known , wavelength of pulse `lambda=(upsilon)/(v)` and `upsilon=sqrt((T)/(m))` `:. lambda=(1)/(v) sqrt((T)/(m))` `i.e., lambda prop sqrt(T)` or `(lambda_(2))/(lambda_(1))=sqrt((T_(2))/(T_(1)))` At the lower end of the rope, `T_(1)=m_(2)g` At the top of the rope, `T_(2)=(m_(1)+m_(2))g` `:. (lambda_(2))/(lambda_(1))=sqrt(((m_(1)+m_(2))g)/((m_(2))g))=sqrt((m_(1)+m_(2))/(m_(2)))` |
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| 86. |
An air column, closed at one end and open at the other resonates with a tuning fork when the smallest legnth of the column is 50 cm. The next larger length of the column resonating with the same tuning fork isA. `100cm`B. `150cm`C. `200cm`D. `66.7cm` |
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Answer» Correct Answer - B Here, `l_(1)=50cm, l_(2)=?` For smalles resonance length `l_(1)`, `(lambda)/(2)=l_(1):. lambda=2l_(1)=2xx50=100cm` The next higher length of resonance column is `l_(2)=(3lambda)/(2)=(3xx100)/(2)=150cm` |
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| 87. |
Assertion `:` In a progressive wave motion, particle velocity is constant at all time. Reason `:` This is because density of medium is uniform throughout.A. both, Assertion and Reason are true and Reason is the correct explanation of the Assertion.B. both, Assertion and Reason are true but Reason is not a correct explanation of the Assertion.C. Assertion is true but the Reason is false.D. both,Assertion and Reason are false. |
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Answer» Correct Answer - d Here, both assertion and reason are false, because, velocity of every particle changes with time as particles executes S.H.M. |
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| 88. |
Damped oscillations are induced in a cirucuit whose quality factor is `Q=50` and natural oscillation frequency is `v_(0)=5.5 kHz`. How soon will the energy stored in the circuit decrease `eta= 2.0 ` times ? |
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Answer» `Q=(pi)/(betaT)=(piv)/( beta)=(omega)/( 2 beta)=(sqrt(omega_(0)^(2)-beta^(2)))/( 2 beta)` or ` (omega_(0))/( beta)=sqrt(1+ 4 Q^(2))` or `beta=( omega_(0))/( sqrt(1+Q^(2)))` Now `W=W_(0)e^(-2 beta t)` Thus energy decreases `eta` times in `(1n eta)/( 2 beta)` sec. `=1n eta(sqrt(1+4 dot (Q^(2))))/(2 omega_(0))~~(Q1n eta)/( 2 pi v_(nn))sec. = 1.033ms ` |
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| 89. |
Find the quality factor of a circuit with capacitance `C=2.0 mu F` and inductance `L=5.0 mH` if the maintenance of undamped oscillations in the circuit with the voltage amplitude across the capacitor being equal to `V_(m)=1.0 V` requires a power `( :P: )=0.10m W`. The damping of oscillations is sufficiently low |
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Answer» Given `V=V_(m) e^(-betat)sin omegat, omega~= omega_(0) betaTlt lt 1` Power loss `=(" Energy loss per cylcle")/(T)` `~=(1)/(2) CV_(m)^(2)xx2 beta` `(` energy decreasesas `W_(0) e^(-2betat) ` so loss per cycle is `W_(0)xx2 betaT)` Thus `lt P gt =(1)/(2)CV_(m)^(2)xx(R)/(L)` or `R=(2lt P gt)/( V_(m)^(2))(L)/(C)` Hence `Q=(1)/(R) sqrt((L)/(C))=sqrt((C)/(L))(V_(m)^(2))/(2lt P gt)=100` on putting the vales. |
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| 90. |
If the elctron `(` charge of each electron `=-e)` are shifted by a small distance `x,a` net `+ ve` charge density `(` per unit area `)` is induced on the surface. This will result in an electric field `E=n ex //epsilon_(0)` in the direction of `x` and a restoring force on an electron of `-(n e^(2)x)/(epsilon_(0))`, Thus `m ddot(x) =-(n e^(2)x)/(epsilon_(0))` or `ddot(x) + ( n e^(2))/( m epsilon_(0))x=0` This gives `omega_(p)= sqrt((n e^(2))/( m epsilon_(0)))=1.645xx10^(16)s^(-1)` as the plasma frequency for the problem,. |
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Answer» If the electron (charge of each electron =`-e`) are shifted by a small distance `x`, a net `+ve` charge density (per unit area) is induced on the surface. This will result in an electric field `E=n e x//epsilon_(0)` in the direction of `x` and `a` restoring force on an electron of `-(n e^(2)x)/(epsilon_(0))`, Thus, `mddot(x)=-(n e^(2)x)/(epsilon_(0))` or `ddot(x)+(n e^(2))/(m epsilon_(0))x=0` This gives `omega_(p)=sqrt(n e^(2))/(mepsilon_(0))=1.645xx10^(16) s^(-1)`. as the plasma frequency for the problem. |
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| 91. |
An oscillating cirucuit consits of capacitance `C=10 mu F`, inductance `L=25 Mh`, and active resistance `R=1.0 Omega. ` How mancy oscilaltion periods does it take for the current amplitude to decrease e`-` fold ? |
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Answer» Current decreases `e` fold in time `t=(1)/(beta)=(2L)/(R) sec=(2L)/(RT) ` oscillations `=(2L)/(R)(omega)/( 2pi)` `=(L)/(piR)sqrt((1)/(LC)-(R^(2))/(4L^(2)))=(1)/(2pi)sqrt((4L)/( R^(2)C)-1)=15.9` oscillations |
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| 92. |
An oscillating circuit consists of a capacitor with capacitance `C`, a coil of inductance `L` with negligible resistance, and aswitch. With the switch disconnected , the capacitor was charged to a voltage `V_(m)` and then at the moment `t=0` the switch was closed . Find `:` `(a)` the current `(I(t))` in the circuit as a function of time, `(b)` the emf of self- inductance in the coil at the moments when the electric energy of the capacitor is equal to that of the current in the coil. |
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Answer» After the switch was closed, the circuit satisfies `-L(dI)/( dt)=(q)/(C)` or `(d^(2)q)/(dt^(2))+ omega_(0)^(2)q=0implies q=CV_(m) cos omega_(0)t` where we have used the fact that when the switch is closed we must have `V=(q)/(C)=V_(m), I=(dq)/( dt)=0` at `t=0` Thus `(a)` `I=(aq)/( dt)=-CV_(m)omega_(0) sin omega_(0)t` `=-V_(m) sqrt((C)/(L)) sin omega_(0)t` `(b)` The electrical energy of the capacitor is `(q^(2))/( 2 C) alpha cos^(2) omega_(0)t` and of the inductor is `(1)/(2) L I^(2) alphasin^(2) omega_(0)t.` The two are equal when ltbr` ` omega_(0)t=(pi)/( 4)` At that instant the emf of the self `-` inductance is `-L(di)/(dt)=V_(m) cos omega_(0)t =V_(m) //sqrt(2)` |
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| 93. |
A series circuit consisting of a capacitor with capacitance with capacitance `C`, a rasistance `R`, and a coil with inductance `L` and negligible active resistance is connected to an oscillator whose frequency can be varied without changing the voltag amplitude. Find the frequency at which the voltage amplitude is maximum `(a)` across the capacitor , `(b)` across the coil. |
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Answer» `(a) v_(c)=(1)/( omegaC) (V_(m))/( sqrt(R^(2)+(omegaL-(1)/( omegaC))^(2)))` `=( V_(m))/( sqrt( (omegaRC)^(2)+ ( omega^(2) LC-1)^(2)))=(V_(m))/( sqrt(((omega^(2))/( omega_(0)^(2))-1)^(2)+ 4 beta^(2) omega^(2) // omega_(0)^(4)))` `=(V_(m))/( sqrt(((omega^(2))/( omega_(0)^(2))-1-( 2 beta)/( omega_90)^(2))^(2)+(4 beta^(2))/( omega_(0)^(2))-( 4 beta^(4))/( omega_(0)^( 4))))` This is maximum when `omega^(2)=omega_(0)^(2)- 2 beta^(2) = (1)/( LC)-(R^(2))/( 2L^(2))` `(b)` `V_(L)=I_(m) omegaL=Vm(omegaL)/( sqrt(R^(2)(omegaL-(1)/( omegaL))^(2)))` `=( V_(m)L)/( sqrt((R^(2))/(omega^(2))+(L-(1)/( omega^(2)C))^(2)))=(V_(m)L)/(sqrt(L^(2)-(1)/( omega^(2))((2L)/(C)-R^(2))+(1)/( omega^(4)C^(2))))` `=(V_(m)L)/( sqrt(((1)/( omega^(2)C)-(L- ( CR^(2))/( 2)) )^(2)+L^(2)- ( L- (1)/( 2) CR^(2))^(2)))` This is maximum when `(1)/( omega^(2)C)=L-(1)/( 2) CR^(2)` or ` omega^(2) =( 1)/( LC-(1)/(2) C^(2) R^(2))=(1)/( (1)/( omega_(0)^(2))-( 2 beta^(2))/( omega_(0)^(4)))` `=( omega_(0)^(4))/( omega_(0)^(2)-2 beta^(2)) `or ` omega=( omega_(0)^(2))/( sqrt( omega_(0)^(2)-2 beta^(2)))` |
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| 94. |
An oscillating circuit consists of a capacitor with capacitance `C=1.2 nF` and a coil with inductance `L=6.0 mu H` and active resistance `R=0.50 Omega`. What mean power should be fed to the circuit to maintain undamped harmonic oscillations with volage amplitude acroos the capacitor being equal to `V_(m) =10 V`? |
| Answer» `lt P gt =(RCV_(m)^(2))/(2L)` . We get `lt P gt =5 mW.` | |
| 95. |
In an oscillating circuit consisting of a parallel- plate capacitor and an inductance coil with negligible active resistance the oscillations with energy `W` are sustained. The capacitor plates were slowly drawn aparto to increase the oscillation frequency `eta-` fold. What work was doen in the process ? |
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Answer» In the oscillating circuit, let `q=q_(m) cos omegat ` be the change on the condenser where `omega^(2)=(1)/(LC)` and `C` is the intantaneous capacity of the condenser `(S=` area of plates `)` `C=(epsilon_(0)S)/( y)` `y=` distance between the plates. Since the oscillation frequency increases `eta` fold, the quancity `omega^(2)=(y)/( epsilon_(0) SL)` change `eta^(2)` fold and so does `y` i.e. changes from `y_(0)` initially to `eta^(2) y_(0)` finally. Now the `P.D.` across the condenser is `V=(q_(m))/(C) cos omegat=(yq_(m))/( epsilon_(0)S) cos omegat` and hence the electric field between the plates is `E=(q_(m))/( epsilon_(0)S) cos omegat` Thus, the charge on the plate being `q_(m)cos omegat` the force on the plate is `E=(q_(m)^(2))/( epsilon_(0)S) cos^(2) omegat` Since this force is always positive and the plate is pulled slowly we can use the average force `bar(F)=(q_(m)^(2))/(2 epsilon_(0)S)` and work done is `A=bar(F)( eta^(2)y_(0)-y_(0))=( eta^(2)-1)( q_(m)^(2)y_(0))/( 2 epsilon_(0)S)` But `(q_(m)^(2)y_(0))/(2 epsilon_(0)S)=(q_(m)^(2))/(2 C_(0))=W` the initial stored energy. Thus. `A=( eta^(2)-1) W`. |
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| 96. |
A source of sinusoidal emf with constant voltage is connected in series with an oscillating circuit with quality factor `Q=100`. At a certain frequency of the external voltage the heat power generated in the circuit reached the maximum value. How much `(` in per cent`)` should this frequency be shifted to decrease the power generated `n=2.0` times ? |
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Answer» `P=(V^(2)R)/(R^(2)+(X_(L)-X_(C))^(2))` At resonance `X_(L)=X_(C)implies omega_(0)=(1)/( sqrt(LC))` Power generated will decrease `n` times when `(X_(L)-X_(C))-( omegaL-(1)/( omegaC))^(2)=(n-1)R^(2)` or ` omega-(omega_(0)^(2))/(omega)=+-sqrt(n-1)(R)/(L)=+-sqrt(n-1) 2 beta` Thus `omegabar(+)2 sqrt(n-1)beta omega - omega_(0)^(2)=0` `( omegabar+sqrt(n-1)beta)^(2)=omega_(0)^(2)+(n-1) beta^(2)` or `(omega)/(omega_(0)=sqrt(1+(n-1)beta^(2)//omega_(0)^(2)))+-sqrt(n-1)beta//omega_(0)` `(` taking only the positive sign in the first term to ensure positive valur for `(omega)/( omega_(0)).)` Now `Q=(omega)/( 2 beta)=(1)/(2) sqrt(((omega_(0))/( beta))^(2) -1)` `(omega_(0))/( beta)=sqrt(1+ 4 Q^(2))` Thus `(omega)/( omega_(0))=sqrt(1+(n-1)/( ( 1+4Q)))+-sqrt(n-1)//sqrt(1+ 4 Q)` For large `Q` `|( omega-omega_(0))/( omega_(0))|=(sqrt( en -1))/( 2Q)-(sqrt(en -1))/(2Q)xx100%=0.5%` |
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| 97. |
A wave is represented by the equation `y=(0.001mm)sin[(50s^-10t+(2.0m^-1)x]`A. the frequency `=25//piHz`B. the amplitude `=0.01mm`C. the wave velocity `=100m//s`D. the wavelength `=2.0m` |
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Answer» Correct Answer - A Compare the given equation with standard equation `y=rsin[(2pi)/(T)t+(2pi)/(lambda)x]` `(2pi)/(T)=50` or `(1)/(T)=v=(50)/(2pi)=(25)/(pi)Hz` `(2pi)/(lambda)=2` or `lambda=pi` `upsilon=vlambda=(25//pi)xxpi=25ms^(-1)` |
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| 98. |
Two particles A and B have a phase diference of `pi` when a sine wave passes through the reginA. The displacements at `A` and `B` have equal magnitudesB. `A` and `B` move in opposite directionsC. `A` and `B` must be separted by half of the wavelengthD. `A` oscillates at half the frequency of `B` |
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Answer» Correct Answer - A::B Phase difference of `pi` means that the displacement of two particles `A` and `B` are equal in magnitude and opposite in direction. `A` and `B` may be separated by half wavelength if they have same frequency of oscillation. |
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| 99. |
What type of motion particles of a medium execute when a wave passes through the medium? |
| Answer» Particles of the medium execute simple harmonic motion about their mean position. | |
| 100. |
Two particles execute simple harmonic motion of same amplitude and frequency along the same straight line. They pass on another, when going in opposite directions, each time their displacement is half of their amplitude. What is the phase difference between them?A. `120^(@)`B. `180^(@)`C. `60^(@)`D. `30^(@)` |
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Answer» Correct Answer - A Let `theta` be the phase difference betweent two simple harmonic motions, r and `omega` be the amplitude and angular frequency of motion of the particles. The two SHMs are represented by the equations. `y_(1)=rsin omegat and y_(2)=rsin (omegat+theta)`. Let the two particles cross one another, moving in opposite directions at time `t=t_(1)` (say) when displacement of each particle is `r//2`, then `y_(1)=y_( 2)=r//2` at time ` t=t_(1)`. `:. (r)/(2)=rsin omegat_(1) or sinomegat_(1)=(1)/(2):. cos omega t_(1)=(sqrt(3))/(2)` Also `(r)/(2)=rsin(omegat_(1)+theta) or (1)/(2)=sin (omegat_(1)+theta)` `(1)/(2)=sin omegat_(1)costheta+cosomegat_(1)sintheta` or `(1)/(2)=(1)/(2)costheta+(sqrt(3))/(2)sintheta` or `1-costheta+sqrt(3) sin theta or 1-cos theta =sqrt(3) sin theta` squaring both sides, we get `1+cos^(2)theta-2cos theta=3sin^(2)theta=3(1-cos^(2)theta)` `4cos^(2)theta-2cos theta-2=0` or `(4cos theta +2)(cos theta-1)=0` when `4cos theta -1=0, cos theta =1, theta=0^(@)`, this is true when both particles start together. `:. `Phase difference, `theta=120^(@)` |
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