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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.
| 701. |
A simple pendulum hangs the celling of a car if the car acceleration with a uniform acceleration , the frequency of the simple pendulum willA. IncreaseB. DecreaseC. Become infiniteD. Remain constant |
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Answer» Correct Answer - A |
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| 702. |
When the displacement is one-half the amplitude, what fraction of the total energy is kinetic and what fraction is potential? At what displacement is the energy half kinetic and half potential? |
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Answer» Correct Answer - `(3)/(4),(1)/(4),(a)/sqrt(2)` |
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| 703. |
An imaginary particle has a charge equal to that of an electron and mass 100 times the mass of the electron. It moves in a circular orbit around a nucleus of charge + `4 e`. Take the mass of the nucleus to be infinite. Assuming that the Bhor model is applicable to this system. (a)Derive an expression for the radius of `n^(th)` Bhor orbit. (b) Find the wavelength of the radiation emitted when the particle jumps from fourth orbit to the second orbit. |
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Answer» Correct Answer - (i) `(n^(2)h^(2)epsi_(0))/(400pime^(2))` (ii) `408 eV` (i) `mvr_(n)=(nh)/(2pi)` and `(mv^(2))/r_(n)=(4e^(2))/(4pi in_(0) r_(n)^(2))` `implies r_(n)=(in_(0) h^(2))/(4pi e^(2))(n^(2)/m)=(n^(2)h^(2)in_(0))/(400 pi me^(2))` (iii) `E_(n^(th))=-(13.6) ((Z^(2)m)/n^(2))` `DeltaE=13.6xx4^(2)xx100 [1/2^(2)-1/4^(2)]=408 eV` |
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| 704. |
A classical model for the hydrogen atom consists of a singal electron of mass `m_(e)` in circular motion of radius r around the nucleous (proton). Since the electron is accelerated, the atom continuously radiates electromagnetic waves. The total power P radiated by the atom is given by `P = P_(0)//r^(4)` where `P_(0) = (epsi^(6))/(96pi^(3)epsi_(0)^(3)c^(3)m_(epsi)^(2))` (c = velocity of light) (i) Find the total energy of the atom. (ii) Calculate an expression for the radius r(t) as a function of time. Assume that at `t = 0`, the radiys is `r_(0) = 10^(-10)m`. (iii) Hence or otherwise find the time `t_(0)` when the atom collapses in a classical model of the hydrogen atom. Take: `[2/(sqrt(3))(e^(2))/(4piepsi_(0))1/(m_(epsi)c^(2)) = r_(e) ~~ 3 xx 10^(-15) m]` |
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Answer» Correct Answer - (i) `- (1)/(8piepsi_(0)) (e^(2))/(r)` (ii) `r_() (1-(3cr_(e)^(2)t)/(r_(0)^(3)))^(1//3)` (iii) `10^(-10) xx (100)/(81) sec` (i) `(mv^(2))/r=1/(4pi in_(0)) e^(2)/r^(2) implies mvr=(h)/(2pi) implies E_("total")=e^(2)/(8pi in_(0) r)` (ii) `(dE)/(dt)=-P` (loss of energy per sec) `implies d/(dt) (- e^(2)/(8pi in_(0) r))= - P_(0)/r^(4) implies (e^(2)/(8pi in_(0) r^(2))) (dr)/(dt)= - P_(0)/r^(4)` `implies e^(2) underset(r_(0))overset(r)(int) r^(2) dr= -8 pi in_(0) P_(0) underset(0)overset(t) (int) dt` `implies r^(3)=r_(0)^(3) - (6P_(0) (4pi in_(0))t)/e^(2) implies r=r_(0) [1-(3cr_(e)^(2) t)/r_(0)^(3)]^(1//3)` (iii) For `r=0`, (to collapse and fall into nucleus) `implies 1- (3cr_(e)^(2) t)/r_(0)^(3)=0` `implies t=r_(0)^(3)/(3cr_(e)^(2))=10^(-30)/(3xx3xx10^(8)xx9xx10^(-30))=(10^(-10)xx100)/81 sec` |
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| 705. |
Passage V) In SHM displacement, velocity and acceleration all oscillate simple harmonically with same angular frequency `omega`. Phase difference between any two is `pi/2` except that between displacement and acceleration which is `pi`. v-t graph of a particle is SHM is as shown in figure Choose the wrong option. A. At A particle is at extreme positionB. At B acceleration of particle is zeroC. At C acceleration of particle is maximum and in negative direction.D. None of the above. |
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Answer» Correct Answer - C |
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| 706. |
Amplitude of a harmonic oscillator is `A` , when velocity of particle is half of maximum velocity, then determine posiition of particle. |
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Answer» `v = omegasqrt(A^(2)-x^(2))` but `v = (v_("max"))/(2)=(Aomega)/(2)=omegasqrt(A^(2)-x^(2))` `rArr A^(2) = 4[A^(2)-x^(2)] rArr x^(2) = (4A^(2)-A^(2))/(4) rArr x = +-(sqrt(3A))/(2)` |
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| 707. |
Statement I: If the amplitude of a simple harmonic oscillator is doubled, its total energy becomes four times. Statement II: The total energy is directly proportional to the square of the amplitude of vibration of the harmonic oscillator.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false |
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Answer» Correct Answer - A |
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| 708. |
Assertion : For an oscillating simple pendulum, the tension in the string is maximum at the mean position and minimum at the extreme position. lt brgt Reason : The velocity of oscillating bob in simple harmonic motion is maximum at the mean position.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false |
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Answer» Correct Answer - B |
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| 709. |
The amplitude of an oscillating simple pendulum is 10 cm and its period is 4 sec . Its speed after 1 sec after it passes its equilibrium position, isA. zeroB. 0.57 m/sC. 0.212 m/sD. 0.32 m/s |
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Answer» Correct Answer - A |
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| 710. |
The maximum tension in the string of an oscillating pendulum is double of the minimum tension. Find the angular amplitude. |
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Answer» Correct Answer - A::C::D The tension in the pendulum is maximum at the mea position and minimum on th extreme postion. `Here `1/2mv^2-0=mgl(1-costheta)` `v^2=2gl(1-costheta)` ……….1 `-Now, T_(ma)=mg+2mg(1-costheta)` `[T=mg+(mv^2/l)]` `Again T_(min)=mgcostheta` According to question `T_(max)=2Tmin` `rarr mg+2mg-2mgcostheta=2mgcostheta` `rarr 3mg=4mgcostheta` `costheta=3/4` `rarr thetas=cos^-1 3/4` |
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| 711. |
A simple pendulum is constructed by hanging a heavy ball by a 5.0 long string. It undergoes small oscillation. a. How many oscillations does it make per second? b. What will be the frequency if the system is taken on the moon where acceleration due to gravitation of the moon is `1.67 ms^-2?` |
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Answer» Correct Answer - A::B::C `L=5m` a. `T=2pisqrt(l/g)` `=2pisqrt(5/10)` `=2pisqrt0.5=2pi(0.7)` In `2pi(0.7)sec`, the body will complete `1/(2pi) (.7)` oscillation `:.f=1/(2pi)(0.7)` `=10/(14pi)=0.71/pi` times b. When it is taken to the moon `T=2pisqrt((l/g))` Where `g` is Acceleration on the moon `=2pisqrt(5/1.67)` `f=1/T` `1/(2pi)sqrt(1.67/5)=1/(2pi)(0.557)` `=1/(2pisqrt3)` |
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| 712. |
In `SHM`, acceleration versus displacement (from mean position) graph:A. is always a straight line passing through origin and slope-`1`B. is always a stright line passing through origin and slope`+1`C. is always a stright line not necessarily passing through oringinD. none of the above |
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Answer» Correct Answer - D |
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| 713. |
Acceleration A and time period T of a body in S.H.M. is given by a curve shown below. Then corresponding graph, between kinetic energy (K.E.) and time t is correctly represented by A. B. C. D. |
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Answer» Correct Answer - A |
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| 714. |
Acceleration versus time graph of a body in SHM is given by a curve shown below. T is the time period. Then corresponding graph between kinetic energy KE and time t is correctly represented by A. B. C. D. |
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Answer» Correct Answer - A From the graph, equation of acceleration can be writtern as `a = - a_(max) cos omegat` `:.` velocity can be written is `v = -v_(max) sin omegat`. `KE = 1/2 mv^(2) = 1/2mv_(max)^(2) sin^(2)omegat` Hence the graph is as shown in A |
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| 715. |
The dispalcement of an object attached to a spring and excuting simple harmonic motion is given by `x = 2 xx 10^(-2) cospit` metres. The time at which at maximum speed first occurs is :A. ` 0.5 s`B. `0.75 s`C. `0.125 s`D. `0.25 s` |
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Answer» Correct Answer - A `|v| = (2 xx 10^(2-2))(pi)sinomegat` For `|v|` to be maximum `sinpit = 1` `pit = (pi)/(2), (3pi)/(2)…….. t = (1)/(2)s `. |
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| 716. |
A particle of mass `m` excuting `SHM` make `f` oscillation per second. The difference of its kinetic energy when at the centre, and when at distance x from the centre isA. `pi^(2)f^(2)x^(2)m`B. `2pi^(2)f^(2)x^(2)m`C. `(1)/(2)pi^(2)f^(2)x^(2)m`D. `f^(2)x^(2)m` |
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Answer» Correct Answer - B `KE` at centre `= 1/2 momega^(2)(A)^(2) = 1/2m4pi^(2)f^(2)A^(2)` KE at distance `x` `= 1/2 m4pi^(2)f^(2)(A^(2) - x^(2))` Difference `= 1/2 m xx 4pi^(2)f^(2)x^(2) = 2pi^(2)f^(2)x^(2)m` |
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| 717. |
A particle of mass 0.2 kg is excuting SHM of amplitude 0.2 m. When it passes through the mean position its kinetic energy is `64 xx 10^(-3) J` . Obtain the equation of motion of this particle if the initial phase of oscillation is `pi//4`. |
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Answer» Here, A=0.2 m `phi =(pi)/(4)` As we know , KE`=(1)/(2)momega^(2)A^(2)` where , m=mass of particle =0.2 kg `64 xx10^(-3) =(1)/(2)xx0.2xx omega^(2)xx0.2^(2)` `omega^(2)=(128 xx10^(-3))/(0.2xx0.2xx0.2) "or" omega=4s^(-1)` `therefore` Equation of motion can be written as `x=A"sin"(omegat+phi)` `=0.2 "sin"(4t+(pi)/(4))` |
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| 718. |
A particle excuting `S.H.M.` of amplitude `4 cm `and `T = 4` sec .The time take by it to move position extreme position to half the amplitude isA. 1 secB. 1/3 secC. 2/3 secD. `sqrt3//2 sec` |
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Answer» Correct Answer - C |
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| 719. |
The maximum velocity of a particle, excuting simple harmonic motion with an amplitude `7 mm`, is `4.4 m//s` The period of oscillation is.A. `100 s`B. `0.01 s`C. `10 s`D. `0.1 s` |
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Answer» Correct Answer - B `Aomega = v_(max)` `T = (2pi)/(omega) = (2piA)/(v_(max)) = 0.01` sec. |
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| 720. |
A particle of mass m executes simple harmonic motion with amplitude a and frequency v. The average kinetic energy during its motion from the position of equilinrium to the end is.A. `2pi^(2)ma^(2)v^(2)`B. `pi^(2)m a^(2) v^(2)`C. `(1)/(4)pi^(2)m a^(2)v^(2)`D. `4pi^(2) ma^(2)v^(2)` |
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Answer» Correct Answer - B |
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| 721. |
A particle of mass m executes simple harmonic motion with amplitude a and frequency v. The average kinetic energy during its motion from the position of equilinrium to the end is.A. (a) `2pi^(2)ma^(2)v^(2)`B. (b) `pi^(2)ma^(2)v^(2)`C. (c ) `1/2ma^(2)v^(2)`D. (d) `4pi^(2)ma^(2)v^(2)` |
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Answer» Correct Answer - B (b) KEY CONCEPT: The instantaneous kinetic energy of a particle executing S.H.M. is given by. `K=1/2ma^(2) omega^(2) sin^(2)omegat` :. Average K.E. `=ltKgelt1/2momega^(2)a^(2) sin^(2)omegatgt` `=1/2momega^(2)a^(2)ltsin^(2)omegatgt` `=1/2momega^(2)a^(2)(1/2) (:. ltsin^(2)thetage1/2)` `=1/2momega^(2)a^(2)=1/2ma^(2)(2piv)^(2) (:. omega=2piv)` `=1/2momega^(2)a^(2) =1/2ma^(2) (2piv)^2) (:. omegapiv)` or, `ltKgtpi^(2)ma^(2)v^(2)`. |
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| 722. |
A particle executes simple harmonic motion with an amplitude of 10 cm. At what distance from the mean position are the kinetic and potential energies equal? |
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Answer» Correct Answer - B::C `r=10cm ` Because K.E. =P.E. `=so, (1/2)momega^2(r^2-y^2)=(1/2)momega^2y^2` `r^2-y^2=y^2` `2y^2=r^2` `rarr y=r/sqrt2=10/sqrt2 =5sqrt2` from the mean position |
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| 723. |
A simple harmonic motion has an amplitude `A` and time period `T`. Find the time required bu it to trvel directly form (a) `x = 0` to `x = A//2` (b) `x = 0` to `x = (A)/(sqrt(2))` (c) `x = A` to `x = A//2` (d) `x = -(A)/(sqrt(2))` "to" `x = (A)/(sqrt(2))` (e) `x = (A)/(sqrt(2))` to `x = A`. |
| Answer» Correct Answer - (a) `T//12`, (b) `T//8`, (c) `T//6`, (d) `T//4`, (e) `T//8` | |
| 724. |
A particle of mass m executes simple harmonic motion with amplitude a and frequency v. The average kinetic energy during its motion from the position of equilinrium to the end is.A. `pi^(2)ma^(2)epsi^(2)`B. `(1)/(4)ma^(2)epsi^(2)`C. `4pi^(2)ma^(2)epsi^(2)`D. `2pi^(2)ma^(2)epsi^(2)` |
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Answer» Correct Answer - A `K_(av)=(int_(0)^(T//4)(1)/(4)m[aomega cos(omegat)]^(2)dt)/(int_(0)^(T//4)dt)=(ma^(2)omega^(2))/(2(T)/(4))int_(0)^(T//4) cos^(2)(omegat)dt=(2ma^(2)omega^(2))/(T).(T)/(8)=(1)/(4)ma^(2)omega^(2)` `=(1)/(4)ma^(2)(2piepsi)^(2) , =pi^(ma^(2)epsi^(2)`. |
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| 725. |
At particle is executing `SHM` with amplitude `A` and has maximum velocity `v_(0)`. Find its speed when it is located at distance of `(A)/(2)` from mean position. |
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Answer» Correct Answer - `(sqrt(3)v_(0))/(2)` `v_(0) = Aomega rArr omega = (v_(0))/(A)` `v^(2) = omega^(2) (A^(2) - x^(2))` `v^(2) = ((v_(0))/(A))^(2) (A^(2) - (A^(2))/(2^(2)))` `v^(2) = v_(0)^(2) (3)/(4), v = (sqrt(3t))/(2)v_(0)` |
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| 726. |
A particle executes simple harmonic motion with an amplitude of `10 cm` and time period `6 s`. At `t = 0` it is position `x = 5 cm` from mean postion and going towards positive `x-`direaction. Write the equation for the displacement `x` at time `L`. Find the magnitude of the acceleration of the particle at `t = 4 s`. |
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Answer» Correct Answer - `x = (10 cm) sin [((pi)/(3)s^(-1))t + (pi)/(6)]` `(10)/(9)pi^(2) prop 11 cm//s^(2)` |
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| 727. |
A particle is executing `SHM`. Find the positions of the particle where its speed is `8 cm//s`, if maximum magnitudes of its velocity and accleration are `10 cm//s` and `50 cm//s` respectively. |
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Answer» Correct Answer - `+- (6)/(5) cm = +- 1.2 cm` from the mean position `|v| = |A|omega rArr 10 = |A|omega` `|a| = |A|omega^(2) rArr 50 = |A|omega^(2)` `rArr omega = 5 rArr |A|= 2` `v^(2) = omega^(2)(A^(2) - x^(2))` `8^(2) = 5^(2)(2^(2) - x^(2))` `4 - x^(2) = (64)/(25) rArr x^(2) = (36)/(25) rArr x = +- (6)/(5) cm` |
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| 728. |
Fill in the blanks using appropriate word from the list at the end of each statement : `(i)` The magnitude of accleration of a particle in `S.H.M.` is the `"….."` at the end points. (least, greatest) `(ii)` The total energy of a particle in `S.H.M.` equals the `"......"` at the end point and the `"....."` at the mean position, (kinetic energy, potential energy) `(iii)` the restoring force in `S.H.M.` is `"........"` in magnitude, when the particle is intantaneously at rest. (zero, maximum) `(iv)` The time period of a particle in `S.H.M.` depends, in general , on `"........"`, but in independent of `"......."` (amplitude, force constant, initial phase, mass, total energy). `(v)` For a particle in `S.H.M.` with a given mass `m` and force constant `k`, quantities which depends on initial conditions are `"......"` (time period, amplitude, phase, total energy, frequency). `(vi)` A particle in `S.H.M`. has `"......"` speed, and `"....."` magnitude of accleration at its mean position. (minimum, maximum) |
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Answer» Correct Answer - `(i)` greatest `(ii)` potential energy, kinetic energy `(iii)` zero `(iv)` force costant and mass; amplitude, initial phase, and total energy `(vi)` maximum, minimum |
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| 729. |
The magnitude of average accleration in half time period in a simple harmonic motion isA. `(2Aomega^(2))/(pi)`B. `(Aomega^(2))/(2pi)`C. `(Aomega^(2))/(sqrt(2)pi)`D. zero |
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Answer» Correct Answer - A `lt` acceleration `gt = (int_(0)^(T//2)omega^(2)asinomegatdt)/(int_(0)^(T//2)dt) = (2piomega^(2))/(pi)` |
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| 730. |
Some springs are combined in series and parallel arrangement as shown in the figure and a mass m is suspended from them. The ratio of their frequencies will be :- A. `1:1`B. `2:1`C. `sqrt(3):2`D. `4:1` |
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Answer» Correct Answer - C For `1^(st)` condition `k_(eff) = k/2` For `2^(nd)` condition `k_(eff) = ((2k)(k))/(2k+k) = 2/3k` `:. T = 2pisqrt((m)/(k)) :. f = 1/T = 1/2pi sqrt((k)/(m))` `(f_(1))/(f_(2)) = sqrt((k//2)/(2k//3)) = (sqrt(3))/(2)` |
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| 731. |
A plate oscillating on a horizontal plane oscillation with time period `T` suddenly another plate put on the first plate, then time periodA. Will decreaseB. Will increaseC. will be sameD. none of these |
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Answer» Correct Answer - C |
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| 732. |
The acceleration due to gravity at height `R` above the surface of the earth is `g/4`. The periodic time of simple pendulum in an artifical satellie at this height will be :-A. `T = 2pisqrt((2l)/(g))`B. `T = 2pisqrt((l)/(2g))`C. zeroD. infinity |
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Answer» Correct Answer - D In an artificial satellite `g_(eff) = 0 rArr T = oo` |
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| 733. |
Calculate the tiem period of a simple pendulum of length one meter. The acceleration due ot gravity at the place is `pi^2ms^-2`.A. `2/pi sec`B. `2pi sec`C. 2 secD. `pi sec` |
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Answer» Correct Answer - C |
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| 734. |
A radioactive sample consists of two distinct species having equal number of atoms initially. The mean life of one species is `tau` and that of the other is `5 tau`. The decay products in both cases are stable. A plot is made of the total number of radioactive nuclei as a function of time. Which of the following figure best represents the form of this plot? (a), (b), (c), (d) A. B. C. D. |
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Answer» Correct Answer - D The total number of atoms can neither remain constant (as in option a) nor can ever increase (as in option b and c). They will continuously decrease with time. Therefore, (d) is the appropriate option. |
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| 735. |
Two protons are having same kinetic energy. One proton enters a uniform magnetic field at right angles ot it. Second proton enters a uniform electric field in the direction of field. After some time their de Broglie wavelengths are `lambda_1 and lambda_2` then (a) `lambda_1 = lambda_2` (b) `lambda_1 lt lambda_2` (c) `lambda_1 gt lambda_2` (d) some more information is requiredA. `lambda_(1) = lambda_(2)`B. `lambda gt lambda_(2)`C. `lambda_(2) gt lambda_(1)`D. None of these |
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Answer» Correct Answer - D We have work is done by only electric field. Thus if `bar(E)||vec(v),|vec(v)|` decreases, & thus momentum of electron decreases & vice-versa, while in magnetic field `|vec(v)|` remains constant. |
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| 736. |
Let `(n_r) and (n_b) be respectively the number of photons emitted by a red bulb and a blue blub of equal power in a given time.A. `n_(r) = n_(b)`B. `n_(r) lt n_(b)`C. `n_(r) gt n_(b)`D. data in insufficient |
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Answer» Correct Answer - C `P_(r)=(n_(r)hc)/lambda_(r), P_(b)=(n_(b)hc)/lambda_(b)` if `P_(r)=P_(b)` Since `lambda_(r) gt lambda_(b) implies n_(r) gt n_(b)` |
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| 737. |
`10^-3` W of `5000A` light is directed on a photoelectric cell. If the current in the cell is `0.16muA`, the percentage of incident photons which produce photoelectrons, isA. `0.4%`B. `.04%`C. `20%`D. `10%` |
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Answer» Correct Answer - B (a) `P= (overset(.)(N)hc)/(lambda)` (b) `i=overset(.)(n) e` (c) `%=overset(.)(n)/(overset(.)(N))xx100%` |
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| 738. |
The electron in a hydrogen atom makes a transition `n_(1) rarr n_(2)` where `n_(1)` and `n_(2)` are the principal qunatum numbers of the two states. Assume the Bohr model to be valid. The frequency of orbital motion of the electron in the initial state is `1//27` of that in the final state. The possible values of `n_(1)` and `n_(2)` areA. `n_(1) = 4, n_(2) = 2`B. `n_(1) = 3, n_(2) = 1`C. `n_(1) = 8, n_(2) = 1`D. `n_(1) = 6, n_(2) = 3` |
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Answer» Correct Answer - B `T prop n^(3)` `T_(i)/T_(f)=1/27=(n/m)^(3) implies n/m=1/3` |
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| 739. |
The angle made by the string of a simple pendulum with the vertical depends on time as `theta=pi/90sin[(pis^-1)t]`. Find the length of the pendulum if `g=pi^2ms^-2`A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - A Here `2pisqrt((m)/(k)) = (2pi)/(omega) = (2pi)/(pi) = 2pisqrt((l)/(g)) rArr (l)/(g) = (1)/(pi^(2))` But `g = pi^(2)` therefore `l = 1 m` |
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| 740. |
Highly excited states for hydrogen like atom (alos called Ryburg states) with nucleus Charge Ze are defined by their principal qunatum number n, where `n lt lt 1`. Which of the following statement(s) is (are) true?A. Releative change in the radii of two consecutive orbitals does not depend on `Z`B. Relative change in the radii of two consecutive orbitals varies as `1//n`C. Relative change in the energy of two consecutive orbitals varies as `1//n^(2)`D. Relative change in the angular momenta of two consecutive orbitals varies as `1//n` |
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Answer» Correct Answer - A::B::D As radius `r prop n^(2)/z` `implies (Deltar)/r=(((n+1)/z)^(2)-(n/z)^(2))/((n/z)^(2))=(2n+1)/n^(2) ~~2/n prop 1/n` as energy `E prop z^(2)/n^(2)` `implies (DeltaE)/E=(z^(2)/n^(2)-z^(2)/((n+1)^(2)))/(z^(2)/((n+1)^(2)))=((n+1)^(2)-n^(2))/(n^(2).(n+1)^(2)).(n+1)^(2)` `implies (DeltaE)/E=(2n+1)/n^(2) cong (2n)/n^(2) prop 1/n` as angular momentum `L=(nh)/(2pi)` `implies (DeltaL)/(L)=(((n+1)h)/(2pi)-(nh)/(2pi))/((nh)/(2pi))=1/n prop 1/n` |
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| 741. |
The length of a simple pendulum is `39.2//pi^(2)` m. If `g=9.8 m//s^(2)` , the value of time period isA. 4 sB. 8 sC. 2 sD. 3 s |
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Answer» Correct Answer - A (a) `T=2pisqrt((l)/(g)) impliesT=2pisqrt((39.2)/(pi^(2)xx9.8))=4s` |
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| 742. |
Two simple pendulums, having periods of `2s` and `3s` respectively, pass through the mean position simulaneously at a particular instant. They may be in phase after an interval of:A. `5s`B. `6s`C. `1s`D. none of these above |
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Answer» Correct Answer - B `omega_(1) = (2pi)/(2) = pi` and `omega_(2) = ((2pi)/(3))` They will be in phase if `(omega_(t) - omega_(2)) t = 0, 2pi, 4pi`….. `t = (2pi)/((omega_(1) - omega_(2))) = (2pi)/(pi - (2pi)/(3)) = 6` sec |
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| 743. |
The ratio of frequencies of two pendulums are 2 : 3, then their length are in ratioA. `sqrt((2)/(3))`B. `sqrt((3)/(2))`C. `(4)/(9)`D. `(9)/(4)` |
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Answer» Correct Answer - D (d) Frequency , `nprop (1)/(sqrt(l))` `implies (n_(1))/(n_(2))=sqrt((l_(2))/(l_(1)))implies (l_(2))/(l_(1))=(n_(1)^(2))/(n_(2)^(2))implies (l_(2))/(l_(1))=((2)^(2))/((3)^(2))` `(l_(2))/(l_(1))=(4)/(9)implies(l_(1))/(l_(2))=(9)/(4)` |
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| 744. |
A simple pendulum of length L and mass (bob) M is oscillating in a plane about a vertical line between angular limits `-phi` and `+phi` For an angular displacement `theta(|theta|ltphi).`the tension in the string and the velocity of the bob are T and v respectively. The following relations hold good under the above conditions `T-Mgcostheta=(Mv^(2))/(L)` `T costheta=Mg` The magnitude of the tangential acceleration of the bob `|a_(T)|=gsintheta` `T=Mgcostheta`A. `T cod theta = Mg`B. `T= Mg cos theta = (Mv^(2))/(L)`C. the magtiude of the tangenital acceleration of the bob `|a_(T)|g sin theta`D. `T = Mg cos theta` |
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Answer» Correct Answer - B::C |
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| 745. |
The half life of radioactive Radon is `3.8 days` . The time at the end of which `(1)/(20) th` of the radon sample will remain undecayed is `(given log e = 0.4343 ) `A. `3.8` daysB. `1.65` dayC. `33` dayD. `76` day |
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Answer» Correct Answer - B `M = M_(0)e^(-lambdat) rArr (M_(0))/(20) = M_(0)e^(-lambdat) rArr ln (1/20) = -lambdat` `rArr ((ln(20))/(ln2)) T_(1//2) = t rArr ((ln(2) + ln(10))/(ln2))T_(1//2) = t` `rArr t = 16.42` days |
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| 746. |
The stopping potential necessary to reduce the phtoelectric current of zero-A. is directly proprotional to wavelength of incident light.B. uniformly increases with the wavelength of incident light.C. directly proportional to frequency of incident light.D. uniformly increases with the frequnecy of incident light. |
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Answer» Correct Answer - D `eV_(s) = hv - phi` |
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| 747. |
The work functions of tungsten and sodium are `5.06 eV` and `2.53 eV` respectively. If the threshold wavelenth for sodium is `5896 Å`, then the threshold wavelength for the tungsten will beA. `11792 Å`B. `5896 Å`C. `4312 Å`D. `2948 Å` |
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Answer» Correct Answer - D `(phi_(1))/(phi_(2)) = (lambda_(2))/(lambda_(1)) rArr (2.53)/(5.06) = (lambda)/(5896) rArr lambda = 2948 Å` |
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| 748. |
The variation of the acceleration (f) of the particle executing S.H.M. with its displacement (X) is represented by the curveA. B. C. D. |
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Answer» Correct Answer - C |
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| 749. |
In S.H.M., potential energy (`U`) `vs` time (`t`) graph isA. B. C. D. |
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Answer» Correct Answer - B |
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| 750. |
The potential energt of a particle of mass 0.1 kg, moving along the x-axis, is given by `U=5x(x-4)J`, where x is in meter. It can be concluded thatA. the particle is acted upon by a constant forceB. the speed of the particle is maximum at `x = 2 m`C. the particle executes `SHM`D. the period of oscillation of the particle is `(pi//5)` sec |
| Answer» Correct Answer - B::C::D | |