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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.
| 801. |
In case of uniform circular motion which of the following physical quantity do not remain constantA. kinetic energyB. potential energyC. restoring forceD. Frequency |
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Answer» Correct Answer - D (e) A harmonic oscillation of constant amplitude and a single frequency under a restoring force whose magnitude is proportional ot the displacement and always acts towards mean position is known as simple harmonic motion. |
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| 802. |
A block of mass `m` is attached to three springs `A,B` and `C` having force constants `k, k` and `2k` respectively as shown in figure. If the block is slightly oushed against spring `C`. If the angular frequency of oscillations is `sqrt((Nk)/(m))`, then find `N`. The system is placed on horizontal smooth surface. |
| Answer» Correct Answer - 3 | |
| 803. |
Period of small oscillations in the two cases shown in figure is `T_(1)` and `T_(2)` respectively . Assume fluid does not have any viscosity , then A. `T_(1) =T_(2)`B. `T_(1) lt T_(2)`C. `T_(1) gt T_(2)`D. Cannot say anything |
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Answer» Correct Answer - B (b) `T_(1)=2pisqrt((m)/(k+rhoAg))` but `T_(2)=2pisqrt((m)/(k))` Hence , `T_(1) lt T_(2)` |
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| 804. |
Assertion : In spring block system if length of spring and mass of block both are halved, then angular frequency of oscillations will remain unchanged. Reason : Angular frequency is given by `omega = sqrt((k)/(m))` .A. If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.B. If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.C. If Assertion is true but Reason is false.D. If Assertion is false but Reason is true. |
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Answer» Correct Answer - D (d) `k prop (1)/("length of spring")` If length is halved k will become two times. from, `T=2pisqrt((m)/(k))` Time period T will remain half for the given conditions. |
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| 805. |
For given spring mass system, if the time period of small oscillations of block about its mean position is `pisqrt((nm)/(K))`, then find `n`. Assume idela conditions. The system is in vertical plane and take `K_(1) = 2K, K_(2) = K`. |
| Answer» Correct Answer - 12 | |
| 806. |
For given spring mass system, if the time period of small oscillations of block about its mean position is `pisqrt((nm)/(K))`, then find `n`. Assume idela conditions. The system is in vertical plane and take `K_(1) = 2K, K_(2) = K`. |
| Answer» Correct Answer - 12 | |
| 807. |
The time period of small oscillations of mass `m` :- A. `2pisqrt((m)/(6k))`B. `2pisqrt((11m)/(6K))`C. `2pisqrt((6m)/(11K))`D. `2pisqrt((m)/(K))` |
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Answer» Correct Answer - B Equivalent spring constant `(1)/(K_(eq)) = (1)/(3K) + (1)/(2K) + (1)/(K) rArr K_(eq) = (6K)/(11) :. T 2pisqrt((m)/(K_(eq))) = 2pisqrt((11m)/(6K))` |
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| 808. |
Values of the acceleration `A` of a particle moving in simple harmonic motion as a function of its displacement `x` are given in the table below. `|{:(A(mm s^(-2)),16,8,0,-8,-16),(x(mm),-4,-2,0,2,4):}|` The pariod of the motion isA. `1/(pi)s`B. `(2)/(pi)s`C. `(pi)/(2)s`D. `pis` |
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Answer» Correct Answer - D In SHM `a = -omega^(2)x`. So `16 = -omega^(2) (-4) rArr omega = 2 rArr "Time period" T = (2pi)/(omega) = (2pi)/(2) = pis` |
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| 809. |
Values of the acceleration `A` of a particle moving in simple harmonic motion as a function of its displacement `x` are given in the table below. `|{:(A(mm s^(-2)),16,8,0,-8,-16),(x(mm),-4,-2,0,2,4):}|` The pariod of the motion isA. `1/pi s`B. `2/pi s`C. `pi/ 2 s`D. `pi s` |
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Answer» Correct Answer - D |
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| 810. |
A spring with 10 coils has spring constant k. It is exactly cut into two halves, then each of these new springs will have a spring constantA. K/2B. 3k/2C. 2kD. 3 k |
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Answer» Correct Answer - B |
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| 811. |
Four mass less spring whose force constant are `2k , 2k, k` and `2k` respectively are attached to a mass `M` kept on a friction less plate (as shown in figure) if the mass `M` is displaced in the horizontal direction then the frequency of oscillation of the system is A. `1/(2pi)sqrt(k )/(4M)`B. `1/(2pi)sqrt(4k)/(M)`C. `1/(2pi)sqrt(k)/(7M)`D. `1/(2pi)sqrt(7k)/(M)` |
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Answer» Correct Answer - B |
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| 812. |
Time period of spring-block system on surface of earth is `T_(1)` and that of a simple pendulum is `T_(2)`. At height h = R (the radius of earth) the corresponding values are `T_(1)` and `T_(2)` thenA. `T_(1)gtT_(1)`B. `T_(1)=T_(2)`C. `T_(2)gtT_(2)`D. `T_(2)ltT_(2)` |
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Answer» Correct Answer - B::C |
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| 813. |
In simple harmonic motion of a particle, maximum kinetic energy is 40 J and maximum potential energy is 60 J. thenA. minimum potential energy will be 20 JB. potential energy at half the displacement will be 30JC. kinetic energy at half the displacement is 40 JD. potential energy or kinetic energy at some intermediate position cannot be found the given data |
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Answer» Correct Answer - A::B |
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| 814. |
Two particles are in SHM with same amplitude A and same regualr frequency `omega`. At time t=0, one is at x =`+A/2` and the other is at `x=-A/2`. Both are moving in the same direction.A. phase difference between the two particles is `pi/3`B. phase difference between the two particles is `(2pi)/3`C. they will collide after time t= `pi/(2omega)`D. they will collide after time t=`(3pi)/(4omega)` |
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Answer» Correct Answer - A::C |
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| 815. |
A particle executing simple harmonic motion has an amplitude of 6 cm . Its acceleration at a distance of 2 cm from the mean position is `8 cm/s^(2)` The maximum speed of the particle isA. `8 cms^(-1)`B. `12 cms^(-1)`C. `16 cms^(-1)`D. `24 cms^(-1)` |
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Answer» Correct Answer - B `a=omega^(2)yimpliesomega=sqrt((a)/(y))=sqrt((8)/(2))=2"rads"^(-1)` Now `v_("max")=Aomega=6xx 2=12 cms^(-1)` |
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| 816. |
For a particle executing simple harmonic motion, the acceleration is proportional toA. is uniformB. varies linearly with timeC. is non-uniformD. Both (b) and (c) |
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Answer» Correct Answer - C The acceleration of the particle is non-uniform in case of simple harmonic motion. |
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| 817. |
The radioactivity of a sample is `R_(1)` at a time `T_(1)` and `R_(2)` at time `T_(2)`. If the half-life of the specimen is T, the number of atoms that have disintegrated in the time `(T_(2) -T_(1))` is proporational toA. `(R_(1)T_(1) - R_(2)T_(2))`B. `(R_(1)-R_(2))T`C. `(R_(1) - R_(2))//T`D. `(R_(1) - R_(2))(T_(1) - T_(2))` |
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Answer» Correct Answer - B `N_(1)=N_(0)e^(-lambdaT_(1)), N_(2)=N_(0)e^(-lambdaT_(2))` `R_(1)=lambdaN_(1), R_(2)= lambdaN_(2)` `(N_(1)-N_(2))=lambda/lambda (N_(1)-N_(2))=((R_(1)-R_(2)))/(lambda)` `T=(log_(e)2)/(lambda), lambda=(log_(e)2)/(T)` `(N_(1)-N_(2))=((R_(1)-R_(2))T)/((log_(e)2))` `(N_(1)-N_(2)) prop (R_(1)-R_(2)) T` |
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| 818. |
The neutron is a particle with zero charge still it has a non-zero magnetic moment with z-component `9.66 xx 10^(-27) A-m^(2)`. This can be explained by the internal structure of the neutron. The evidence indicates that a neutron is composed of three fundamental particles calleed quarks : an `"up"` (u) quark, of charge `+ (2e)/(3)`, and two `"down"` (d) quarks, each of change `- (e)/(3)`. The combinations of the three quarks prodcues a net charge of `(2e)/(3) - e/3 - e/3 = 0`. If the quarks are in motion they can produce a non-zero magnetic moment. As a very simple model, suppose the u quark moves in a counter clockwise circular path and the d quarks move in a clockwise circular path, all of the radius r and all with the same speed v see figure. Determine the magnitude of the magnetic moment due of the circular u quark :-A. `(evr)/(3pir)`B. `(2evr)/(3)`C. `(4evr)/(3)`D. `evr` |
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Answer» Correct Answer - A `M_(u)=(evr)/3` |
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| 819. |
Neutron in thermal ewuilibrium with matter at `27^(@)C` can be throught to behave like ideal gas. Assuming them to gave speed of `v_("rns")`, what is therir De broglie wavelength `lambda` (in mm). Fill `((156)/(11)) lambda` in the OMR shee. (Take `m_(n) = 1.69 xx 10^(-27) kg. k = 1.44 xx 10^(-23) J//K, h = 6.60 xx 10^(-34) Jsec`] |
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Answer» Correct Answer - 2 `v_(max) = sqrt((3kT)/(m)), lambda = h/p rarr lambda = (h)/(m_(n) xx v_(rms)) = (h)/(sqrt(3kTm_(n))) rArr lambda = (6.6 x 10^(-34))/(sqrt(3xx1.44xx10^(-23)xx1.69xx10^(-27)xx300)) = (2.2 xx 10^(-10))/(1.2 xx 1.3)` `rArr (156)/(11)lambda = (156 xx 2.2 xx 10^(-10))/(11 xx 1.2 xx 1.3) = (220 xx 10^(-10))/(11) = (22)/(11) nm = 2` |
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| 820. |
The neutron is a particle with zero charge still it has a non-zero magnetic moment with z-component `9.66 xx 10^(-27) A-m^(2)`. This can be explained by the internal structure of the neutron. The evidence indicates that a neutron is composed of three fundamental particles calleed quarks : an `"up"` (u) quark, of charge `+ (2e)/(3)`, and two `"down"` (d) quarks, each of change `- (e)/(3)`. The combinations of the three quarks prodcues a net charge of `(2e)/(3) - e/3 - e/3 = 0`. If the quarks are in motion they can produce a non-zero magnetic moment. As a very simple model, suppose the u quark moves in a counter clockwise circular path and the d quarks move in a clockwise circular path, all of the radius r and all with the same speed v see figure. If all Quarks start moving in the same direction then what will be the magnetic moment of the neutron :-A. `(evr)/(3pir)`B. `(2evr)/(3)`C. `evr`D. None of these |
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Answer» Correct Answer - D Net magnetic moment in that case will be zero. |
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| 821. |
Assume that a neutron breaks into a proton and an electron . The energy reased during this process is (mass of neutron `= 1.6725 xx 10^(-27) kg` mass of proton `= 1.6725 xx 10^(-27) kg` mass of electron `= 9 xx 10^(-31) kg )`A. `5.4 MeV`B. `0.73 MeV`C. `7.10 MeV`D. `6.30 MeV` |
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Answer» Correct Answer - B Released energy `=(1.6747xx10^(-27)-1.6725xx10^(-27)-9xx10^(-31))xx (3xx10^(8))^(2)J=0.73 MeV` |
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| 822. |
The neutron is a particle with zero charge still it has a non-zero magnetic moment with z-component `9.66 xx 10^(-27) A-m^(2)`. This can be explained by the internal structure of the neutron. The evidence indicates that a neutron is composed of three fundamental particles calleed quarks : an `"up"` (u) quark, of charge `+ (2e)/(3)`, and two `"down"` (d) quarks, each of change `- (e)/(3)`. The combinations of the three quarks prodcues a net charge of `(2e)/(3) - e/3 - e/3 = 0`. If the quarks are in motion they can produce a non-zero magnetic moment. As a very simple model, suppose the u quark moves in a counter clockwise circular path and the d quarks move in a clockwise circular path, all of the radius r and all with the same speed v see figure. Determine the magnitude of the three-quark system :-A. `(evr)/(3pir)`B. `(2evr)/(3)`C. `evr`D. `2 evr` |
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Answer» Correct Answer - B `M_("net")=M_(u)+M_(d)+M_(d)=(evr)/3+(evr)/6+(evr)/6=(2evr)/3` |
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| 823. |
The count rate meter is used to measure that activity of a given amount of a radio active element. At one instant, the meter shows 475 counts/minute. Exactly 5 minutes later, is shown 270 counts/minute then Half life of the sample is (in minute) |
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Answer» Initial activity `A_(i) = (-dN)/(dt)|_(t=0) = lambdaN_(0) = 4750"….."(i)` , Final activity `A_(f) = (-dN)/(dt) |_(t=5) = lambdaN = 2700"……"(ii)` Dividing (i) by (ii), we get `(4750)/(2700) = (N_(0))/(N_(t))` The decay constant is given by `lambda = (2.303)/(t) "log"(N_(0))/(N_(t)) = (2.303)/(5) "log" (4750)/(2700) = 0.113 "min"^(-1)` Half-life of the sample is `T = (0.693)/(lambda) = (0.693)/(0.113) = 6.14 min` |
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| 824. |
The neutron is a particle with zero charge still it has a non-zero magnetic moment with z-component `9.66 xx 10^(-27) A-m^(2)`. This can be explained by the internal structure of the neutron. The evidence indicates that a neutron is composed of three fundamental particles calleed quarks : an `"up"` (u) quark, of charge `+ (2e)/(3)`, and two `"down"` (d) quarks, each of change `- (e)/(3)`. The combinations of the three quarks prodcues a net charge of `(2e)/(3) - e/3 - e/3 = 0`. If the quarks are in motion they can produce a non-zero magnetic moment. As a very simple model, suppose the u quark moves in a counter clockwise circular path and the d quarks move in a clockwise circular path, all of the radius r and all with the same speed v see figure. The current due to the circular motion of the u quark :-A. `(ev)/(6pir)`B. `(ev)/(3pir)`C. `(ev)/(pir)`D. `(2ev)/(pir)` |
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Answer» Correct Answer - B `i=qf=((2e)/3)(v/(2pir))=(ev)/(3pir)` |
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| 825. |
Let `R_(t)` represents activity of a sample at an insant and `N_(t)` represent number of active nuclei in the sample at the instant. `T_(1//2)` represents the half life. `{:(,"Column I",,"Column II"),((A),t=T_(1//2),(p),R_(t)=(R_(0))/(2)),((B),t=(T_(1//2))/(ln2),(q),N_(0)-N_(t)=(N_(0))/(2)),((C),t=(3)/(2)T_(1//2),(r),(R_(t)-R_(0))/(R_(0)) = (1-e)/(e)),(,,(s),N_(t)=(N_(0))/(2sqrt(2))):}` |
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Answer» Correct Answer - (A) p,r (B) r (C) s (A) In half life active sample reduce `=R_(0)/2` `:.` Decay number of nuclei is `=R_(0)/2` (B) `N=N_(0)e^(-lambdat)` where `lambda=` decay constant, `lambda=(ln(2))/t_(1//2)` `N=N_(0)e^(-(ln2t_(1//2))/(t_(1//2)ln2))implies N=N_(0)/e` `N_(0)-N=N_(0) [(e-1)/e]implies (N-N_(0))/N_(0) =(1-e)/e` (C) `N=N_(0)/((2)^(t//T_(1//2)))implies N=N_(0)/2^(3//2)implies N_(0)/(2sqrt(2))` |
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| 826. |
The figure shows an energy level diagram for the hydrogen atom. Several transitions are marked as `I, II, III` `",______,"` the diagram is only indicative and not to scale. . Which transition involves the longest wavelength line in the visible portion of the hydrogen spectrumA. IIB. IIIC. VID. IV |
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Answer» Correct Answer - D For longest wavelength, energy difference should be minimum. So in visible portion of hydrogen atom, minimum energy is in transition VI & IV. |
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| 827. |
The figure shows an energy level diagram for the hydrogen atom. Several transitions are marked as I, II, III,_____. The diagram is only indicative and not be scale. In which transitions is a Balmer photn absorbed?A. IIB. IIIC. IVD. VI |
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Answer» Correct Answer - D For Balmer series, `n_(1)=2` (lower), `n_(2)=3, 4` (higher) `:.` In transition `(VI)`, Photon of Balmer series is absorbed. |
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| 828. |
The figure shows an energy level diagram for the hydrogen atom. Several transitions are marked as I, II, III,_____. The diagram is only indicative and not be scale. In which transitions is a Balmer photn absorbed?A. `291 nm`B. `364 nm`C. `487 nm`D. `652 nm` |
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Answer» Correct Answer - C In transition II : `E_(2)=-3.4 eV, E_(4) =-0.85 eV` `Delta=2.55 eV, DeltaE=(hc)/lambda implies lambda=(hc)/(Delta E)=487 nm`. |
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| 829. |
The figure shows an energy level diagram for the hydrogen atom. Several transitions are marked as I, II, III,_____. The diagram is only indicative and not be scale. Which transition will occur when a hydrogen atom is irradiated with radiation of wavelength `103 nm`.A. IIB. IIC. IVD. V |
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Answer» Correct Answer - D Wavelength of radiation `=103 nm=1030 Å` `:. DeltaE=(12400)/(1030 Å) cong 12.0 eV` So difference of energy should be `12.0 eV` (approx) Hence `n_(1)=1` and `n_(2)=3` `(-13.6) eV" "(-1.51)eV` `:.` Transition is V. |
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| 830. |
A point mass oscillates along the `x-`acis according to the law `x = x_(0)cos(omegat - pi//4)` if the acceleration of the particle is written as, a `= A cos(omega + delta)`, then :A. `A = x_(0), delta = -pi//4`B. `A = x_(0)omega^(2), delta = -pi//4`C. `A = x_(0)omega^(2), delta = -pi//4`D. `A = x_(0)omega^(2), delta = 3pi//4` |
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Answer» Correct Answer - D `x = x_(0)cos(omegat - (pi)/(4)), v = -x_(0)omegasin(omegat - (pi)/(4))` `a = -x_(0)omega^(2)cos(omegat - (pi)/(4)), a = x_(0)omega^(2)cos(omegat - (pi)/(4) + pi)` `a = x_(0)omega^(2)cos(omegat + (3m)/(4))` |
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| 831. |
A man measures time period of a pendulum (T) in stationary lift. If the lift moves upward with acceleration `(g)/(4)`, then new time period will beA. `(T)/(4)`B. `4T`C. `(2)/(sqrt(5))T`D. `(sqrt(5))/(2)T` |
| Answer» Correct Answer - C | |
| 832. |
Two pendulums at rest swinging together. Their lengths are respectively `1.44 m` and `1 m`. They will again start swinging in same phase together after (of longer pendulum) : A. `1` vibrationB. `3` vibrationsC. `4` vibrationsD. `5` vibrations |
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Answer» Correct Answer - D Let `x_(1) = A_(1)sinomega_(1)t` and `x_(2) = A_(2)sinomega_(2)t` Two pendulums will vibrate in same phase again when there phase difference `(omega_(2) - omega_(1))t = 2pi` `rArr ((2pi)/(T_(2)) - (2pi)/(T_(1)))t = 2pi` `rArr (sqrt((g)/(1)) - sqrt((g)/(1.44)))n xx T_(1) = 2pi` (where `n` is number of vibrations completed by ionger pendulum) `rArr (sqrt((g)/(1)) - sqrt((g)/(1.44)))n xx 2pisqrt((1.44)/(g)) = 2pi rArr n = 5` Thus after `5` vibrations of longer pendulum they will again start swinging in same phase. |
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| 833. |
Two pendulums at rest swinging together. Their lengths are respectively `1.44 m` and `1 m`. They will again start swinging in same phase together after (of longer pendulum) : A. `1` vibrationB. `3` vibrationsC. `4` vibrationsD. `5` vibrations |
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Answer» Correct Answer - D Let `x_(1) = A_(1)sinomega_(1)t` and `x_(2) = A_(2)sinomega_(2)t` Two pendulums will vibrate in same phase again when there phase difference `(omega_(2) - omega_(1))t = 2pi` `rArr ((2pi)/(T_(2)) - (2pi)/(T_(1)))t = 2pi` `rArr (sqrt((g)/(1)) - sqrt((g)/(1.44)))n xx T_(1) = 2pi` (where `n` is number of vibrations completed by ionger pendulum) `rArr (sqrt((g)/(1)) - sqrt((g)/(1.44)))n xx 2pisqrt((1.44)/(g)) = 2pi rArr n = 5` Thus after `5` vibrations of longer pendulum they will again start swinging in same phase. |
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| 834. |
The acceleration of a particle is `a = - 100x + 50`. It is released from `x = 2`. Here, `a` and `x` are in SI unitsA. the particle will perform SHM of amplitude `2m`B. the particle will perform SHM of amplitude `1.5 m`C. the particle will perform SHM of time period `0.63 s`D. the particle will have a maximum velocity of `15 m//s` |
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Answer» Correct Answer - B::C::D `a = 0` at `x = 0.5 m` and particle is released from `x = 2m` Hence, `A = 2 - 0.5` `= 1.5 m` `omega^(2) = 100` `:. omega = 10 rad //s` `T = (2pi)/(omega) = (2pi)/(10)` `= 0.63 s` `v_(max) = omega A = (10)(1.5)` ` = 15m//s` |
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| 835. |
Four massless springs whose force constants are 2k, 2k, k and 2k respectively are attached to a mass M kept on a frictionless plane (as shown in figure). If the mass M is displaced in the horizontal direction. A. `(1)/(2pi)sqrt((k)/(M))`B. `(1)/(2pi)sqrt((4k)/(M))`C. `(1)/(2pi)sqrt((k)/(7M))`D. `(1)/(2pi)sqrt((7k)/(M))` |
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Answer» Correct Answer - B `k_(eq) = 2k + k + (2k xx 2k)/(2k + 2k) = 4k` so, frequency, `f = (1)/(2pi) sqrt(K_(eq))/(M) = (1)/(2pi)sqrt(4K)/(M)` |
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| 836. |
Let `T_(1)` and `T_(2)` be the time periods of two springs A and B when a mass m is suspended from them separately. Now both the springs are connected in parallel and same mass m is suspended with them. Now let T be the time period in this position. ThenA. `T = T_(1)+T_(2)`B. `T=(T_(1)T_(2))/(T_(1)+T_(2))`C. `T^(2)= T_(1)^(2)+T_(2)^(2)`D. `1/T^(2)=1/T_(1)^(2)+1/T(2)^(2)` |
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Answer» Correct Answer - D |
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| 837. |
Find the ratio of the periods of two identical springs if they are first joined in series & then in parallel & a mass `m` is suspended from them:A. `4`B. `2`C. `1`D. `3` |
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Answer» Correct Answer - B |
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| 838. |
A body performs simple harmonic oscillations along the straight line `ABCDE` with `C` as the midpoint of `AE`. Its kinetic energies at `B` and `D` are each one fourth of its maximum value. If `AE = 2R`, the distance between `B` and `D` is A. `sqrt(3)/(2)R`B. `( R)/sqrt(2)`C. `sqrt(3) R`D. `sqrt(2) R` |
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Answer» Correct Answer - C `(1)/(2)k(A^(2) - x^(2)) = (1)/(4)((1)/(2)kA^(2))` `:. x = sqrt(3)/(2) A` `:. CD = CB = sqrt(3)/(2)R` or `BD = 2 (CD)` or `2 (CB) = sqrt(3)R` |
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| 839. |
The pulley shown in figure has a moment of inertias I about its xis and mss m. find the tikme period of vertical oscillastion of its centre of mass. The spring has spring constant k and the string does not slip over the pulley. |
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Answer» Let us first find the equilibrium position. For rotatioN/Al equilibrium of the pulley, the tensions in the two strings should be equal. Only then the torque on the pulley wil be zero. Let this tension be T. The extension of the spring wil be `y=T/k`, as teh tension in the spring will be the same as the tension in the string. For translatioN/Al equilibrium of the pulley. `2T=mgor 2ky=mg, or ly=(mg)/(2k)` The spring is extended by a distance `(mg)/(2k)` when the pulley is in equilibrium. Now suppose the centre of the pulley goes down further by a distance x. The total inrease in tehlengthof the stringplus the springnis `2x(x` on the left of the pulley and x on the right). As the string has a constant length, the extension of the spring is 2x. The energy of the system is `U=1/2iomega^2+1/2mv^2-mgx+1/2k((mg)/(2k)+2x)^2 ` `=1/2(1/r^2+m)v^2+(m^2g^2)/(8k)+2kx^2` As the system is conservative `(dU)/(dt)=0, ` ltbr. giving `0=(I/r^2+m)v(dv)/(dt)+4kxv` `or (dv)/(dt)=-(4kx)/((I/r^2+m))` or `a=-omega^2x, where omega^2=(4k)/((I/r^2+m))` Thus the centre of mass of the pulley executes a simple harmonic motion with time period `T=2pisqrt((I/r^2+m)/(4k))`. |
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| 840. |
Five identical springs are used in the following three configurations. The time periods of vertical oscillations in configurations (i), (ii) and (iii) are in the ratio A. `1:sqrt2 : 1/sqrt2`B. `2:sqrt2 : 1/sqrt2`C. `1/sqrt2:2 : 1`D. `2 :1/sqrt2 : 1` |
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Answer» Correct Answer - A |
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| 841. |
A mass m performs oscillations of period T when hanged by spring of force constant K . If spring is cut in two parts and arranged in parallel and same mass is oscillated by them, then the new time period will be A. 2TB. TC. `T/sqrt2`D. `T/ 2` |
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Answer» Correct Answer - D |
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| 842. |
A body of mass 1.0 kg is suspended from a weightless spring having force constant `600Nm^(-1)`. Another body of mass 0.5 kg moving vertically upwards hits the suspended body with a velocity of `3.0ms^(-1)` and gets embedded in it. Find the frequency of oscillations and amplitude of motion. |
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Answer» Correct Answer - A::C Frequency of oscillation `f = (1)/(2pi) sqrt((k)/(m_(1) + m_(2))) = 1/(2pi) sqrt((600)/(1.5)) = 10/pi Hz` Let maximum amplitude ba A than `v = omegasqrt(A^(2) - x^(2))` where `x =` difference in equilibrium position `= ((m_(1) + m_(2))g)/(k) - (m_(1))g/(k) = 1/120 m` and `v = (0.5 xx 3)/(1.5) = 1 m//s` Therefore `1 = 20 sqrt(A^(2) - (1/120)^(2)) rArr A = (5sqrt(37))/(600) = (5sqrt(37))/(6) cm` |
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| 843. |
A body of mass `1kg` is suspended from a weightless spring having force constant `600N//m`. Another body of mass `0.5 kg` moving vertically upward hits the suspended body with a velocity of `3.0m//s` and get embedded in it. Find the frequency of oscillations and amplitude of motion. |
| Answer» Correct Answer - `10//pi Hz, (5sqrt(37))/(6)cm` | |
| 844. |
A force F=-10x+2 acts on a particle of mass 0.1 kg where m is m and F is in newtons. If F is released from rest at x=0, find: a. amplitude: b. time period: c. equation of motion. |
| Answer» Correct Answer - `(11)/(5)m,(b)(pi)/(5)sec.,(c )X=0.2-(11)/(5)cosomegat` | |
| 845. |
If the metal bob of a simple pendulum is replaced by a wooden bob, then its time period willA. IncreaseB. DecreaseC. Remain the sameD. First increase then decrease |
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Answer» Correct Answer - C |
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| 846. |
Find the length of seconds pendulum at a place where `g = pi^(2) m//s^(2)`. |
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Answer» Correct Answer - `1m` `T = 2pisqrt((l)/(g)), 2 = 2pisqrt((l)/(g)) rArr l = 1m` |
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| 847. |
In a seconds pendulum, mass of bob is 30 gm . If it is replaced by 90 gm mass. Then its time period willA. 1 secB. 2 secC. 4 secD. 3 sec |
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Answer» Correct Answer - B |
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| 848. |
If the length of a seconds pendulum is increased by `2%` then in a day the pendulumA. 3927 secB. 3727 secC. 3427 secD. 864 sec |
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Answer» Correct Answer - D |
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| 849. |
A small block is connected to one end of a massless spring of un - stretched length `4.9 m`. The other end of the spring (see the figure) is fixed. The system lies on a horizontal frictionless surface. The block is stretched by `0.2 m` and released from rest at `t = 0`. It then executes simple harmonic motion with angular frequency `(omega) = (pi//3) rad//s`. Simultaneously at `t = 0`, a small pebble is projected with speed (v) from point (P) at an angle of `45^@` as shown in the figure. Point (P) is at a horizontal distance of `10 m from O`. If the pebble hits the block at `t = 1 s`, the value of (v) is `(take g = 10 m//s^2)`. .A. `sqrt50m//s`B. `sqrt(51)m//s`C. `sqrt(52)m//s`D. `sqrt(53)m//s` |
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Answer» Correct Answer - A For block `x=Asin(omegat+phi)` `t=0`, `x=Aimpliessinphi=1impliesphi=pi//2` `x=Asin(omegat+phi)=Asin(omegat+pi//2)=Acosomegat` At `t=1s`, `x=Acosomegaxx1=0.2cos((2pi)/(3))=0.1m` Distance of block from `O=4.9+0.1=5m` Range for pebble `=10-5=5m` `R=(v^2)/(g)sin(2xx45^@)` `5=(v^2)/(10)impliesv=sqrt(50)m//s` |
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| 850. |
Which of the followign quantities are always zero in a simple harmonic motion?A. (i), (ii)B. (i), (iii)C. (ii), (iii)D. All |
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Answer» Correct Answer - D Cross product of parallel or antiparallel vectors is always zero |
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