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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.
| 351. |
Find the period of oscillation of the system shown in figure. |
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Answer» Correct Answer - B Resultant of `k` and `k` is `2k`. Then, resultant of `2k` and `2k` is `k`. Now `T = 2pi sqrt((m)/(k)) ` |
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| 352. |
Choose the correct statement (s) for hydrogen and deuterium atoms (considering the motion of nucleus)A. The radius of first Bohr orbit of deuterlum is less than that of hydrogenB. The speed of electron in first Balmer line of deuterlum is more than than that of hydrogenC. The wavelength of first Balmer line of deuterium is more than that of hydrogenD. The angular momentum of electron in the first Bohr orbit of deuterium is more than that of hydrogen |
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Answer» Correct Answer - A We have `r prop (1)/(mu) , mu = "reduced mass"` `(Kq^(2))/(r)= (mv_(e)^(2))/(r_(e)) = (mv_(N)^(2))/(r_(N)) , mv_(e)r_(e) = (nh)/(2pi)` |
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| 353. |
A particular hydrogen like atom has its ground state Binding energy `= 122.4 eV`. It is in ground state. ThenA. Its atomic number is 3B. An electrom of `90` eV cam excite itC. An electron of kinetic energy nearly `91.8 eV` can be brought to almost rest by this atomD. An electron of kinetic energy `2.6 eV` may emerge from the atom when electron of kinetic energy 125 eV collides with this atom |
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Answer» Correct Answer - A::C::D `122.4 = (13.6Z^(2))/(1^(2)) rArr Z = 3, 91.8 = 122.4[1-(1)/(4)]` So an electron of KE `91.8 eV` can transfer its energy to this atom. |
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| 354. |
In the hydrogen atom, if the reference level of potential energy is assumed to be zero at the ground state level. Choose the incorrect statement.A. The total energy of the shell increase with increase in the value of nB. The total energy of the shell decrease with increase in the value of nC. The difference in total energy of any two shells remains the sameD. The total energy at the ground state become `13.6eV` |
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Answer» Correct Answer - B `U = - (27.2eV)/(n^(2)) + C , k = (13.6eV)/(n^(2))` `E = - (13.6 eV)/(n^(2)) + C , DeltaE = 13.6eV((1)/(n^(2)) - (1)/(m^(2)))` here in these question `C = +27.2 eV` |
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| 355. |
Which of the following is/are correct.A. If quantum number is very large frequency of photon emitted when jumps from a orbit to next lower orbit is equal to the interger multiple of the frequency of revolution of electron in previous orbitB. Electron debrogile wavelength depends on kinetic energy of elctronC. Energy of elctrons is quantized in `Li^(++)` atomD. Electrons cannot exists inside nucleus because they are altracted by protons inside nucleus |
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Answer» Correct Answer - A::B::C Frequency of photon `= (me^(4)z^(2))/(8in_(0)^(2)h^(3))[(1)/((n-1)^(2)) - (1)/(n^(2))]` Frequency of revolution of electron `= (me^(4)z^(2))/(4in_(0)^(2)h^(3)n^(3))` |
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| 356. |
If radiation of all wavelengths from ultraviolet to infrared ispassed through hydrogen gas at room temperature absorption lines will be observed in theA. Lymann seriesB. Balmer seriesC. Both (A) and (B)D. Neighter (A) nor (B) |
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Answer» Correct Answer - A Room temperature `rArr n = 1` so lyman series |
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| 357. |
Which of the following is/are not true for `x-`rays?A. Wavelength of continuous x-rays does not depends on potential differenceB. Wavelength of discreate x-rays does not depends on potential differenceC. Discrete x-rays have energy of the order of MeVD. Continuous x-rays have energy of the order of KeV |
| Answer» Correct Answer - A::C::D | |
| 358. |
A beam of ultraviolet light of all wavelength passes through hydrogen gas at room temperature, in the x-direction. Assume that all photons emitted due to electron transitions inside the gas emerge in the y-direction. Let A and B denote the lights emerging from the gas in the x-and y-directions respectively. (i) Some of the incident wavelengths will be absent in A (ii) Only those wavelengths will be present in B which are absent in A (iii) B will contain some visible light (iv) B will contain some infrared lightA. Some of the incident wavelengths will be absent in AB. Only those wavelngth will be present in B which are absent in AC. B will contain some visible lightD. B will contain some infrared light |
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Answer» Correct Answer - A::C::D Room temperature `rArr n = 1` Upon absorption excitations takes place to many higher states which upon de-excitation emit all U.V., infrared and visible light. |
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| 359. |
The acceleration a of a particle undergoing S.H.M. is shown in the figure. Which of the labelled points corresponds to the particle being at -x A. 4B. 3C. 2D. 1 |
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Answer» Correct Answer - D |
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| 360. |
The displacement time graph of a particle executing S.H.M. is as shown in the figure The corresponding force-time graph of the particle is A. B. C. D. |
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Answer» Correct Answer - D |
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| 361. |
The displacement time graph of a particle executing S.H.M. (in straight line) is shown. Which of the following statements is true? A. The force is zero at time 3T/4B. The velocity is maximum at time t/2C. the acceleration is maximum at time TD. The P.E is equal to tatol energy at time T/2 |
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Answer» Correct Answer - D |
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| 362. |
The P.E. of a particle executing SHM at a distance x from its equilibrium position isA. `1/2 m omega^(2)x^(2)`B. `1/2 m omega^(2)a^(2)`C. `1/2 m omega^(2)(a^(2)-x^(2))`D. zero |
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Answer» Correct Answer - A |
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| 363. |
A particle of mass 10 gm is describing S.H.M. along a straight line with period of 2 sec and amplitude of 10 cm . Its kinetic energy when it is at 5 cm from its equilibrium position isA. `37.5 pi ^(2)` ergsB. `3.75 pi ^(2)` ergsC. `3.75pi^(2)` ergsD. `0.375 pi ^(2)` ergs |
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Answer» Correct Answer - C |
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| 364. |
When the displacement is half the amplitude, the ratio of potential energy to the total energy isA. `1/2`B. `1/4`C. 1D. `1/8` |
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Answer» Correct Answer - B |
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| 365. |
Statement-1: Davisson-Germer experiment established the wave nature of electron Statement-2: If electrons have wave nature, they can interfere show differaction.A. Statement-1 is true, Statement-2 is true, Statement-2 is not the correct explanation of Statement-1B. Statement-1 is false, Statement-2 is tureC. Statement-1 is true, Statement-2 is falseD. Statement-1 is true, Statement-2 is ture, Statement-2 is the correct explanation of Statement-1 |
| Answer» Correct Answer - D | |
| 366. |
This question has statement - 1 and statement - 2 of the four choice given after the statements choose the one that best describes the two statements statement - 1 : A metallic surface is irradiated by a monochromatic light of frequency ` v gt v_(0)` (the threshold frequency). The maximum kinetic energy and the stopping potential are `K_(max) and V_(0)` respectively if the frequency incident on the surface is doubled , both the `K_(max) and V_(0)` are also doubled statement - 2 : The maximum kinetic energy and the stopping potential of photoelectron emitted from a surface are linearly dependent on the frequency of incident lightA. Statement-1 is true, Statement-2 is true, Statement-2 is not the correct explanation of Statement-1B. Statement-1 is false, Statement-2 is tureC. Statement-1 is true, Statement-2 is falseD. Statement-1 is true, Statement-2 is ture, Statement-2 is the correct explanation of Statement-1 |
| Answer» Correct Answer - B | |
| 367. |
This question contains Statement - 1 and Statement -2 Of the four choice given after the Statements , choose the one that best decribes the two Statements Statement- 1: Energy is released when heavy undergo fission or light nuclei undergo fusion and Statement- 2: for nuclei , Binding energy nucleon increases with increasing `Z` while for light nuclei it decreases with increasing `Z`A. Statement-1 is false, Statement-2 is true.B. Statement-1 is true, Statement-2 is true, Statement-2 is a correct explanation for Statement-1.C. Statement-1 is ture, Statement-2 is a correct explanation for Statement-1.D. Statement-1 is true, Statement-2 is false. |
| Answer» Correct Answer - D | |
| 368. |
Graph between velocity and displacement of a particle, executing S.H.M. isA. A straight lineB. A parabolaC. A hyperbolaD. An ellipse |
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Answer» Correct Answer - D |
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| 369. |
A particle of mass m is moving in a potential well, for which the potential energy is given by `U(x) = U_(0)(1-cosax)` where `U_(0)` and a are positive constants. Then (for the small value of x)A. the time period of small osciallation is `T=2pisqrt(m/(aU_(0))`B. the speed of the particle is maximum at x=0C. the amplitude of oscillations is `pi/8`D. the time period of small osciallations is `T=2pisqrt(m/(a^(2)U_(0))` |
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Answer» Correct Answer - B::C::D |
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| 370. |
Two small particles P and Q each of mass m are fixed along x-axis at points (a,0) and (-a,0). A third particle R is kept at origin. ThenA. if particle R is displaced along x-axis it will start oscillating.B. oscillations of R along x-axis are simple harmonic in natureC. if R is displaced of R along y-axis, it starts oscillatingD. oscillations along y-axis may be simple harmonic in nature. |
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Answer» Correct Answer - C::D |
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| 371. |
In the options given below, let `E` denote the rest mass energy of a nucleus and `n` a neutron. The correct option is:A. `E(._(92)^(236)U) gt E(._(253)^(137)Y)+2E(n)`B. `(E_(92)^(236)U) lt E(._(53)^(137)I)+E(._(39)^(97)Y)+2E(n)`C. `E(._(92)^(236)U) lt E(._(56)^(140)Ba)+E(._(36)^(94)Kr)+2E(n)`D. `E(._(92)^(236)U) lt E(._(56)^(140)Ba)+E(._(36)^(94)Kr)+2E(n)` |
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Answer» Correct Answer - A Rest mass of parent nucleus should be greater than the rest mass of daughter nuclei. |
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| 372. |
A radioactive sample `S_1` having an activity of `5muCi` has twice the number of nuclei as another sample `S_2` which has an activity of `10muCi`. The half-lives of `S_1` and `S_2` can beA. `20` years and `5` years, respectivelyB. `20` years and `10` years, respectivelyC. `10` years eachD. `5` year each |
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Answer» Correct Answer - A Activity `=lambdaN` & `T_(1//2)=(ln2)/lambda` So `5=(ln2)/T_(1) (2N_(0))` & `10 =(ln2)/T_(2) (N_(0)) implies T_(1)=4T_(2)` |
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| 373. |
Which one of the following statement is `WRONG` in the context of X- rays generated from X- rays tube ?A. Wavelength of characteristic X-rays decreases when the atomic number of target increasesB. Cut-off wavelength of the continuous X-rays depends on the atomic number of the targetC. Intensity of the characteristic X-rays depends on the energy of the electron in the X-rays tubeD. Cut-off wavelengths of the continuous X-rays depends on the energy of the electrons in the X-rays tube |
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Answer» Correct Answer - B As for continuous X-rays `lambda_("min")=(hc)/(eV)` so cut off wavelength depends on the accelerating potential and is independent of nature of target. |
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| 374. |
Passage VI Two particles collide when they are at same position at same time. Displacement time equation of two particles moving along x-axis are `x_(1)=(8+3sinomegat)m` and `x_(2)=(4cosomegat)m` Here, `omega=pirad/s` The two particles will collide after time t=...................s.A. 1B. 2C. 4D. None of the above. |
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Answer» Correct Answer - D |
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| 375. |
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`. Displacement -time equation of a particle is given as `x=Asin(omegat+pi/6)-A cos(omega+pi/6)`, then:A. the motion of the particle is not simple harmonicB. at t=0, acceleration of particles is negativeC. at t-0, velocity of particle is negativeD. None of the above. |
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Answer» Correct Answer - D |
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| 376. |
Passage IV) Angular frequency in SHM is given by `omega=sqrt(k/m)`. Maximum acceleration in SHM is `omega^(2)` A and maximum value of friction between two bodies in contact is `muN`, where N is the normal reaction between the bodies. In the figure shown, what can be the maximum amplitude of the system so that there is no slipping between any of hte blocks?A. `2/7`mB. `3/4`mC. `4/9`mD. `10/3`m |
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Answer» Correct Answer - C |
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| 377. |
Passage IV) Angular frequency in SHM is given by `omega=sqrt(k/m)`. Maximum acceleration in SHM is `omega^(2)` A and maximum value of friction between two bodies in contact is `muN`, where N is the normal reaction between the bodies. Now the value of k, the force constant is increased, then the maximum amplitude calcualted in above question willA. remain sameB. increaseC. decreaseD. data is insufficient |
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Answer» Correct Answer - C |
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| 378. |
A small ball of density `rho=(rho_(0))/2(alpha+betah)`. Mass of the ball is m. Select the most appropriate one option.A. The particle will execute SHMB. The maximum speed of the ball is `(2-alpha)/(sqrt(2)beta)`C. Both a and b are correctD. Both a and b are wrong |
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Answer» Correct Answer - A |
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| 379. |
A spring of force constant k is cut into two pieces such that one piece is double the length of the other. Then the long piece will have a force constant ofA. `(2)/(3)k`B. `(3)/(2)k`C. 3kD. 6k |
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Answer» Correct Answer - B `(1)/(k)=(1)/(k_(1))+(1)/(k_(2))or (1)/(k)=(2)/(k_(2))+(1)/(k_(2))=(3)/(k_(2))` `because" "k_(2)=3k` `because" "k_(1)=(k_(2))/(2)=(3)/(2)k` |
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| 380. |
Answer the following questions: (a) The motion of a simple pendulum is approximately simple harmonic for small angle oscillation. For larger angles of oscilliation, a more involved analysis shows that `T` is greater than `2pisqrt((t)/(g))` Think of a qualitative argument to appreciate this result. (b) What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity ? |
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Answer» Correct Answer - (a) `sin theta lt theta` ; if the restoring force, `mg sin theta` is replaced by `mgtheta`, this amounts to effective reduction in `g` for large angels and hence an increase in time period `T` over that given by the formula `T = 2pisqrt((l)/(g))` where one assume `sin theta = theta`. (b) Gravity disappears for a man under free fall, so frequency is zero. |
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| 381. |
Which of the following statements is not true ? In the case of a simple pendulum for small amplitudes the period of oscillation isA. Directly proportional to square root of the length of the pendulumB. Inversely proportional to the square root of the acceleration due to gravityC. Dependent on the mass, size and material of the bobD. Independent of the amplitude |
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Answer» Correct Answer - C |
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| 382. |
A simple pendulum is set up in a trolley which moves to the right with an acceleration a on a horizontal plane. Then the thread of the pendulum in the mean position makes an angle `theta` with the verticalA. `(tan^(-1)) (a)/(g)` in the forward direactionB. `(tan^(-1)) (a)/(g)` in the backward direactionC. `(tan^(-1)) (g)/(a)` in the backward direactionD. `(tan^(-1)) (g)/(a)` in the forward direction |
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Answer» Correct Answer - B |
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| 383. |
To make the frequency double of an oscillator, we have toA. half the massB. quadruple the massC. double the massD. reduce the mass to one-fourth |
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Answer» Correct Answer - D (d) Frequency of oscillator, `n=(1)/(2pi)sqrt((K)/(m))` We get, `(n_(2))/(n_(1))=sqrt((m_(1))/(m_(2)))` `therefore 2=sqrt((m_(1))/(m_(2))` or `m_(2)=(1)/(4)m_(1)` |
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| 384. |
To make the frequency double of an oscillator, we have toA. Double the massB. Half the massC. Quadruple the massD. Reduce the mass to one-fourth |
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Answer» Correct Answer - D |
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| 385. |
What is constant in S.H.M.A. Restoring forceB. Kinetic energyC. Potential energyD. Periodic time |
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Answer» Correct Answer - D |
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| 386. |
A simple pendulum consisting of a ball of mass m tied to a thread of length l is made to swing on a circular arc of angle q in a vertical plane. At the end of this arc, another ball of mass m is placed at rest. The momentum transferred to this ball at rest by the swinging ball isA. zeroB. `mthetasqrt(g/l)`C. `(mtheta)/(l)sqrt(g/l)`D. `(mtheta)/(l)2pisqrt(g/l)` |
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Answer» Correct Answer - A |
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| 387. |
if a boby is released into a tunal dug across the diameter of earth, it executes simple harmonic motoin with time periodA. `T=2pi sqrt((R_(e))/(g)`B. `T=2pi sqrt((2R_(e))/(g)`C. `T=2pi sqrt((R_(e))/(2g)`D. ` T= 2` seconds |
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Answer» Correct Answer - A |
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| 388. |
Can we use the equation `v = u +` at in SHM or not ? |
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Answer» Correct Answer - A No, as acceleration in SHM `a = - omega^(2) x` is variable. |
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| 389. |
In the string mass system shown in the figure, the string is compressed by `x_(0) = (mg)/(2k)` from its natural length and block is relased from rest. Find the speed o the block when it passes through P (`mg//4k` distance from mean position) |
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Answer» `omega = sqrt((3k)/(m)), x = Asin(omegat + phi), v = Aomegacos(omegat+phi)` at `t = 0, x = 0 rArr phi = 0` `x = A sinomegat rArr (mg)/(4k) = (mg)/(2k)sin omegat rArr sinomegat = 1/2 rArr omegat = (pi)/(6) rArr t = (pi)/(6omega) = T/12` `v = Aomegacosomegat. V = (mg)/(2k)sqrt((3k)/(m)) cos[sqrt((3k)/(m)) (2pisqrt((m)/(3k))//12)] = gsqrt((9m)/(16k))` |
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| 390. |
Assertion : In ` x = A cos omega t`, the dot product of acceleration and velocity is positive for time interval `0 lt t lt (pi)/(2omega)`. Reason : Angle between them is `0^(@)` .A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true. |
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Answer» Correct Answer - A `(pi)/(2omega) = (pi)/(2(2pi//T)) = (T)/(4)`. In the given time particle moves from `x = A` to `x = 0`. |
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| 391. |
Acceleration of a particle in SHM at displacement `x=10 cm` (from the mean position is `a =-2.5 cm//s^(2)`). Find time period of oscillations. |
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Answer» Correct Answer - A::B::D Time period is given by `T = 2pisqrt(|(x)/(a)|)` Substituting the values we have, `T = 2pisqrt((10)/(2.5)) = (4pi)s` |
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| 392. |
`F-x` equation of a body in SHM is `F + 4x=0` Here, `F` is in newton and `x` in meter. Mass of the body is `1 kg`. Find time period of oscillations. |
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Answer» Correct Answer - A::B::D The given equation can be written as `F = -4x` Comparing this eqution with the standard equation of SHM or `F = -Kx` we get `K=4N//m` Now, `T=2pi sqrt((m)/(k))` `=2pi sqrt((1)/(4)) = (pi)s` |
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| 393. |
`F - x` equation of a body of mass `2kg` in SHM is `F + 8x = 0` Here, `F` is in newton and `x` in meter.Find time period of oscillations. |
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Answer» Correct Answer - A::C::D The given equation can be written as, `F = -8x` Comparing with the standard equation of SHM, `F = -Kx` we have, `K=8N//m` `:. T=2pi sqrt((m)/(K))` `=2pi sqrt((2)/(8)` `=(pi)s=3.14s` |
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| 394. |
`a - x` equation of a body in SHM is `a + 16 x = 0`. Here, `x` is in `cm` and a in `cm//s^(2)`. Find time period of oscillations. |
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Answer» Correct Answer - B::C `a = - 16x` Comparing with `a = - omega^(2)x rArr omega = 4 rad//s` Now, `T = (2pi)/(omega) = ((pi)/(2))sec` |
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| 395. |
Statement-1 : In S.H.M., the motion is ‘to and fro’ and periodic. Statement-2 : Velocity of the particle (v)`=omegasqrt(k^(2)-x^(2))`(where x is the displacement and k is amplitude)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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| 396. |
Given that the equation of motion of a mass is `x = 0.20 sin (3,0 t) m` . Find the velocity and acceleration of the mass when the object is `5 cm` from its equilibrium position. Repeat for `x = 0`. |
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Answer» Correct Answer - B::D `omega = 3rad//s` `A = 0.2m` At `x = 5cm` `v = +-omegasqrt(A^(2) - x^(2))` `= +- 3.0sqrt((0.2)^(2) - (0.05)^(2))` `= +- 0.58m//s` `a = - omega^(2)x = - (3)^(2)(0.05)` ` = - 0.45m//s` At `x = 0` `v = +-omega A` `= +- (3.0)(0.2)` `= +- 0.6 m//s` `a = - omega^(2)x = - (3)^(2)(0)` `= 0` |
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| 397. |
A small sleeve of mass `m=1//10kg` can move along the diameter of a horizontal disc, which slides without friction along a guide rode. The sleeve is tied to the end of the rod with the aid of a massless spring whose force constant `k=10 N//m`. When the spring is not under tension, the sleeve is at the oscillations of the sleeve when the disc rotates about its axis at the angular speed `omega` equal to a) 6 rad/s, b) 15 rad/s. |
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Answer» Correct Answer - `v= 1.27s^(-1)`,(b) v=0, the sleeve does not oscillate |
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| 398. |
Two particle undergoes `SHM` along parallel line with the same time period `(T)` and equal amplitude At a particular instant , one particle is at its extereme position while the other is at its mean position .They move in the same direction .They will cross each other after a further time A. `T//8`B. `3T//8`C. `T//6`D. `4T//3` |
| Answer» Correct Answer - B | |
| 399. |
Two particles undergo SHM along parallel lines with the same time period (T) and equal amplitudes. At particular instant, one particle is at its extreme position while the other is at its mean position. The move in the same direction. They will cross each other after a further time.A. `T//2`B. `3T//8`C. `T//6`D. `3T//4` |
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Answer» Correct Answer - B |
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| 400. |
A uniform thin cylindrical disk of mass M and radius R is attaached to two identical massless springs of spring constatn k which are fixed to the wall as shown in the figure. The springs are attached to the axle of the disk symmetrically on either side at a distance d from its centre. The axle is massless and both the springs and the axle are in horizontal plane. the unstretched length of each spring is L. The disk is initially at its equilibrium position with its centre of mass (CM) at a distance L from the wall. The disk rolls without slipping with velocity `vecV_0 = vacV_0hati.` The coefficinet of friction is `mu.` The centre of mass of the disk undergoes simple harmonic motion with angular frequency `omega` equal to -A. `sqrt((k)/(M))`B. `sqrt((2k)/(M))`C. `sqrt((2k)/(3M))`D. `sqrt((4k)/(3M))` |
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Answer» Correct Answer - D Total energy of system `E = (1)/(2)Mv^(2)(1+1/2) + 2 xx 1/2kx^(2)` `= 3/4 Mv^(2) + kx^(2)` `(dE)/(dt) = 0 rArr 3/4M(2v) (dv)/(dt) +2k xx (dx/dt) = 0` `rArr (dv)/(dt) + ((4k)/(3M))x = 0 rArr omega 2k xx ((dx)/(dt)) = 0` |
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