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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.
| 601. |
Acceleration of a particle, executing SHM, at it’s mean position isA. InfinityB. VariesC. MaximumD. Zero |
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Answer» Correct Answer - D |
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| 602. |
Which one of the following statement is true for the speed `v` and the acceleration a of a particle executing simple harmonic motion?A. when v is maxium , a is maxiumB. value if a is zero , whatever may be the value of vC. whaen v is zero , a is zeroD. when v is maxium , a is zero |
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Answer» Correct Answer - D |
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| 603. |
A particle moves along the x - axis according to `x = A[1 + sin omega t]`. What distance does is travel in time interval from `t = 0` to `t = 2.5pi//omega` ?A. `4A`B. `6 A`C. `5 A`D. `3 A` |
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Answer» Correct Answer - C `t = (2.5 pi)/(omega) = (2.5 pi)/(2pi//T)` `= (2.5)((T)/(2))` `= 5((T)/(4))` Particle starts from extreme position and in every `(T)/(4)` time it travels a distance `A`. So, in time `t = 5(T//4)` it will travel a distance `5A`. |
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| 604. |
A man is swinging on a swing made of `2` ropes of equal length `L` and in direction perpendicular to the plane of paper. The time period of the small oscillations about the mean position is A. `2pisqrt((L)/(2g))`B. `2pisqrt((sqrt3L)/(2g))`C. `2pisqrt((L)/(2sqrt3g))`D. `pisqrt((L)/(g))` |
| Answer» Correct Answer - B | |
| 605. |
A wire frame in the shape of an equilateral triangle is hinged at one vertes so that it can swing freely in a vertical plane, with the plane of the `Delta` always remaining vertical. The side of the frame is `1//sqrt(3)m`. The time period in secods of small oscillations of the frmae will be-A. `(pi)/(sqrt2)`B. `pisqrt2`C. `(pi)/(sqrt6)`D. `(pi)/(sqrt5)` |
| Answer» Correct Answer - D | |
| 606. |
The graphs in fugure show that a quantity `Y` varies with displacement `d` in a system undergoing simple harmonic motion. (I) , The unbalanced force acting on the system.A. `I`B. `II`C. `III`D. None |
| Answer» Correct Answer - D | |
| 607. |
The graphs in fugure show that a quantity `Y` varies with displacement `d` in a system undergoing simple harmonic motion. (I) , The timeA. `I`B. `II`C. `III`D. `IV` |
| Answer» Correct Answer - D | |
| 608. |
The length of a simple pendulum executing simple harmonic motion is increased by `21%`. The percentage increase in the time period of the pendulum of increased lingth is.A. `15%`B. `21%`C. `42%`D. `10%` |
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Answer» Correct Answer - D |
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| 609. |
If length of simple pendulum is increased by 6% then percentage change in the time-period will beA. `3%`B. `9%`C. `6%`D. `1//9 %` |
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Answer» Correct Answer - A `(DeltaT)/T=1/2 (Deltal)/l` `(DeltaT)/T=3%` |
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| 610. |
The graphs in fugure show that a quantity `Y` varies with displacement `d` in a system undergoing simple harmonic motion. (I) , Which graphs best represents the relationship obtained when `y` is The total energy of the systemA. `I`B. `II`C. `III`D. `IV` |
| Answer» Correct Answer - A | |
| 611. |
The maximum velocity of a simple harmonic motion represented by `y= 3 sin (100 t + pi/6 )` is given byA. 300B. `(3pi)/6`C. 100D. `(pi)/6` |
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Answer» Correct Answer - A |
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| 612. |
In the case of sustained force oscillations the amplitude of oscillationsA. decreases linearlyB. decreases sinusodiallyC. dereases exponentiallyD. always remains constant |
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Answer» Correct Answer - D (d) Amplitude always remains constant in case of sustained forced oscillations. |
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| 613. |
The diplacement , velocity and acceleration in a simple harmonic motion are related as theA. displacement , velocity and acceleration all act in the same directionbB. displacement and velocity act in the same direction but acceleration in the opposite direction.C. velocity and acceleration are parallel and both are perpendicular to the displacementD. displacement and acceleration are anti-parallel and both perpendicular to the velocity |
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Answer» Correct Answer - B (b) For SHM, `alpha=-omega^(2)y`, where `alpha` is acceleration, `omega` is angular velocity , y is displacement . Hence, it is clear that acceleration is directly proportional to displacement and is always opposite to displacement. |
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| 614. |
The diplacement of a particle varies with time according to the relation `y=a"sin"omegat+b " cos"omegat`.A. The motion is oscillatory but not SHMB. The motion is SHM with amplitude a+bC. The motion is SHM with amplitude `a^(2)+b^(2)`D. The motion is SHM with amplitude `sqrt(a^(2)+b^(2))` |
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Answer» Correct Answer - D (d) Accoding to the question , the displacement `y=a "sin"omegat+b "cos"omegat` Let `a=A"sin"phi " and "b=A"cos" phi` Now , `a^(2)+b^(2)=A^(2)"sin"^(2)phi+A^(2)"cos"^(2)phi` `=A^(2)impliesA=sqrt(a^(2)+b^(2))` `therefore y=A"sin"phi, "sin" omegat+A"cos"phi."cos"omegat` `=A"sin"(omegat+phi)` `implies(dy)/(dt)=Aomega"cos"(omegat+phi)` `implies(d^(2)y)/(dt^(2))=-Aomega^(2)"sin"(omegat+phi)=-Ayomega^(2)=(-Aomega^(2))y` `implies(d^(2)y)/(dt^(2))prop(-y)` Hence, it is an equation of SHM with amplitude, `A=sqrt(a^(2)+b^(2))`. |
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| 615. |
If `overset(vec)(F)` is force vector, `overset(vec)(v)` is velocity , `overset(vec)(a)` is acceleration vector and `overset(vec)(r)` vector is displacement vector `w.r.t.` mean position than which of the following quantities are always non-negative in a simple harmonic motion along a straight line ?A. `overset(vec)(F).overset(vec)(a)`B. `overset(vec)(v).overset(vec)(r)`C. `overset(vec)(a).overset(vec)(r)`D. `overset(vec)(F).overset(vec)(r)` |
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Answer» Correct Answer - A since all collision are elestic and total energy is conserved Time period of motion of right block is `t_(1) = (2L)/(v)`. Time period of motion of left block is `t_(2) = (1)/(2)2pisqrt((m)/(2k))` So. Total time period `= (2L)/(v) + (1)/(2) ` |
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| 616. |
How long after the beginning of motion is the displacment of a harmonically oscillation particle equal to one half its amplitude if the period is `24s` and perticle starts from rest.A. `12s`B. `2s`C. `4s`D. `6s` |
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Answer» Correct Answer - C `y = a cos omegat` `(a)/(2) = a cos omegat` `omegat = (pi)/(3)` `(2pi)/(24)t = (pi)/(3)` `t = 4 sec`. |
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| 617. |
The masses in figure slide on a frictionless table. `m_(1)` but not `m_(2)`, is fastened to the spring. If now `m_(1)` and `m_(2)` are pushed to the left , so that the spring is compressed a distance `d`, what will be the amplitude of the oscillation of `m_(1)` after the spring system is released ? |
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Answer» Correct Answer - A::B::D While returning to equilibrium position, `(1)/(2)kd^(2) = (1)/(2)(m_(1) + m_(2))v^(2)` `:. v = (sqrt((k_(1))/(m_(1) + m_(2))))d` Now, after mean position `m_(2)` is detached from `m_(1)` and keeps on moving with this constant velocity `v` towards right. Block `m_(1)` starts SHM with spring and this `v` becomes its maximum velocity at mean position. `:. v = omega A` `:. (sqrt((k)/(m_(1) + m_(2))))d = (sqrt((k)/(m_(1))))A` `:. A= (sqrt((m_(1))/(m_(1) +m_(2))))d`. |
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| 618. |
In the system shown in the figure the string, springs and pulley are light. The force constant of the two springs are `k_(1) = k " and " k_(2) = 2k`. Block of mass M is pulled vertically down from its equilibrium position and released. Calculate the angular frequency of oscillation. The top surface of the block (represented by line AB) always remains horizontal. |
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Answer» Correct Answer - `omega = sqrt((8k)/(3M))` |
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| 619. |
In simple harmonic motion, the ratio of acceleration of the particle to its displacement at any time is a measure ofA. spring constantB. angular frequencyC. `("angular frequency")^(2)`D. restoring force |
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Answer» Correct Answer - C `=-omega^(2)ximplies |(a)/(x)|=omega^(2)` |
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| 620. |
A particle starts SHM from the mean position. Its amplitude is A and time period is T. At the time when its speed is half of the maximum speed, its displacement y isA. `A/2`B. `A/sqrt2`C. `(Asqrt3)/2`D. `(2A)/sqrt3` |
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Answer» Correct Answer - C |
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| 621. |
For a particle performing `SHM`, equation of motion is given as `(d^(2))/(dt^(2)) + 4x = 0`. Find the time period |
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Answer» `(d^(2)x)/(dt^(2)) = -4x, omega^(2), omega = 2` Time speed, `T = (pi)/(omega) = pi` |
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| 622. |
Suppose an electron is attracted toward the origin by a force`(k)/(r )` where `k` is a constant and `r` is the distance of the electron from the origin .By applying Bohr model to this system the radius of the `n^(th)` orbital of the electron is found to be `r_(n)` and the kinetic energy of the electron to be `T_(n)` , Then which of the following is true ?A. `T_(n) prop 1/(n^(2)),r_(n) prop n^(2)`B. `T_(n)` independent of `n, r_(n) prop n`C. `T_(n) prop 1/n, r_(n) prop n`D. `T_(n) prop 1/n, r_(n) prop n^(2)` |
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Answer» Correct Answer - B `k/r=(mv^(2))/r implies mv^(2) =k` (independent or r) `n(h/(2pi))=mvrimplies r prop n` and `T=1/2 mv^(2)` is independent of n. |
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| 623. |
In forced oscillations , a particle oscillates simple harmonically with a frequency equal toA. frequency of driving forceB. natural frequency of bodyC. differnece of frequency of driving force and natrual frequencyD. mean of frequency of driving force and natural frequency |
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Answer» Correct Answer - A (a) In forced oscillations, particle oscillates with a frequency equal to the frequency of driving force. |
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| 624. |
A particle executing SHM while moving from one extremity is found at distance `x_(1),x_(2)` and `x_(3)` from the center at the end of three successive seconds. The time period of oscillatin is Here `theta=cos^(-1)((x_(1)+x_(3))/(2x_(2))`A. `(2pi)/theta`B. `pi/theta`C. `theta`D. `theta/(2theta)` |
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Answer» Correct Answer - A |
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| 625. |
Assertion : Resonance is special case of force vibration in which the nature frequency of vebration of the body is the same as the impressed frequency of external periodic force and the amplitude of force vibration is maximum Reason: The amplitude of forced vibrations of a bodyincrease with an increase in the frequency of the externally impressed perioic forceA. 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 - C |
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| 626. |
A block of mass m, attached to a spring of spring constant k, oscilltes on a smooth horizontal table. The other end of the spring is fixed to a wall. If it has speed v when the spring is at its naturla lenth, how far will it move on the table before coming to an instantaneous rest?A. `x= sqrt(m//k)`B. `x= 1/vsqrt(m//k)`C. `x= vsqrt(m//k)`D. `x= sqrt(mv//k)` |
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Answer» Correct Answer - C |
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| 627. |
A spring has force constant `k = 200 N//m` and its one end is fixed. There is a block of mass 2 kg attached to its other end and the system lies on a smooth horizontal table. The block is pulled so that the extension in the spring becomes 0.05 m. At this position the block is projected with a speed of `1 m//s` in the direction of increasing extension of the spring. Consider time t = 0 at the moment the block is projected and find (a) the extension (or compression) in the spring as a function of time. (b) the maximum extension in the spring and the time at which it occurs for the first time. (c) the time after which the speed of the block becomes maximum for the first time. Given: `sin^(-1) (0.446) = 0.46` radian |
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Answer» Correct Answer - `(a) x=0.112 sin (10t +046) m` (b) 0.112 m, 0.111 s (c ) t=0.268 s |
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| 628. |
The kinetic energy of a particle executing S.H.M. is 16 J when it is at its mean position. If the mass of the particle is 0.32 kg , then what is the maximum velocity of the particleA. 5 m/sB. 15 m/sC. 10 m/sD. 20 m/s |
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Answer» Correct Answer - C |
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| 629. |
Two spherical starts A and B emit black body radiation. The radius of A is 400 times that of B and A emits `10^(4)` times the power emitted from B. The ratio `(lambda_(A)//lambda_(B))` of their wavelengths `lambda_(A)` and `lambda_(B)` at which the peaks oc cur in their respective radiation curves is : |
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Answer» Correct Answer - 2 `P=e sigma AT^(-4)` `lambdaT=` constant `P_(A)/P_(B)=(e sigma A_(A) T_(A)^(4))/(e sigma A_(B)T_(B)^(4))=(A_(A) T_(A)^(4))/(A_(B) T_(B)^(4))=r_(A)^(2)/r_(B)^(2). lambda_(B)^(2)/lambda_(A)^(4)` `10^(4)=400^(2).(lambda_(B)/lambda_(A))^(4)` `10^(4)/(16xx10^(4))=(lambda_(B)/lambda_(A))^(4)` `lambda_(A)/lambda_(B)=(16/1)^(1/4)=2` |
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| 630. |
If `13.6 eV` energy is required to ionized the hydrogen atom then the energy required to ionize the hydrogen atom , then the energy required to remove an electron from `n = 2 ` isA. `10.2 eV`B. `0 eV`C. `3.4 eV`D. `6.8 eV` |
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Answer» Correct Answer - C Energy required to lonize a atom from `n^(th)` orbit is `= +13.6/n^(2) eV` `E_(2)=(+13.6)/2^(2) eV= +3.4 eV` |
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| 631. |
For a doubly ionised Li-atomA. Angular momentum of an electron in `3rd` orbit is `(3h)/(2pi)`B. energy of electron in `2nd` ecited state is `-13.6 eV`.C. Speed of electron in `3rd` orbit is `(c)/(137)`, where c is speed of lightD. Kinetic energy of electron is 2nd excited state is half of the magnitude of the potential energy. |
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Answer» Correct Answer - A::B::C::D (A) `L = (nh)/(2pi)` , (B) `E = (-13.6Z^(2))/(n^(2))` (C ) `v = (c )/(137)(Z)/(n)` , (D) `K = -(U)/(2)` |
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| 632. |
Consider the following energies :- 1. minimum energy needed to excite a hydrogen atom 2. energy needed to ionize a hydrogen atom 3. energy released in `.^(235)U` fission 4. energy released to remove a neutron from a `.^(12)C` nucleus Rank them in order of increasing value.A. `1,2,3,4`B. `1,3,2,4`C. `1,2,4,3`D. `2,1,4,3` |
| Answer» Correct Answer - C | |
| 633. |
If a simple harmonic motion is represented by `(d^(2)x)/(dt^(2)) + alphax = 0`, its time period is :A. `(2pi)/(alpha)`B. `(2pi)/(sqrt(alpha))`C. `2pialpha`D. `2pisqrt(alpha)` |
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Answer» Correct Answer - B `(d^(2)x)/(dt^(2)) = -alphax………(i)` We know `a = (d^(2)x)/(dt^(2)) = -omega^(2)x …….(ii)` From Eq. `(i)` and `(ii)`, we have `omega^(2) = alpha` `omega = sqrt(alpha)` or `(2pi)/(T) = sqrt(alpha) :. T = (2pi)/(sqrt(a))` |
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| 634. |
Two independent harmonic oscillators of equal mass are oscillating about the origin with angular frequencies `omega_(1)` and `omega_(2)` and have total energies `E_(1)` and `E_(2)` repsectively. The variations of their momenta `p` with positions `x` are shown in figures. If `(a)/(b) = n^(2)` and `(a)/(R) = n`, then the correc t equation(s) is (are) : A. `E_(1)omega_(1) = E_(2)omega_(2)`B. `(omega_(2))/(omega_(1)) = n^(2)`C. `omega_(1)omega_(2) = n^(2)`D. `(E_(1))/(omega_(1)) = (E_(2))/(omega_(2))` |
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Answer» Correct Answer - B::D For first oscillarator , For second oscillator `b = maomega_(1)` `(a)/(b) = (1)/(momega_(1)) = n^(2) , (1)/(momega_(2)) = 1` `(omega_(2))/(omega_(1)) = n^(2)` `E_(1) = (1)/(2)m omega_(1)^(2) a^(2) , E_(3) = (1)/(2)momega_(2)^(2) R^(2)` `(E_(1))/(E_(2)) = (omega_(1)^(2))/(omega_(2)^(2)) xx n^(2) = (omega_(1)^(2))/(omega_(2)^(2)) xx (omega_(2))/(omega_(1)) , (E_(1))/(E_(2)) = (omega_(1))/(omega_(2)) rArr (E_(1))/(omega_(1)) = (E_(2))/(omega_(2))` |
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| 635. |
A block of mass `0.9 kg` attached to a spring of force constant `k` is lying on a frictionless floor. The spring is compressed to `sqrt(2) cm` and the block is at a distance `1//sqrt(2) cm` from the wall as shown in the figure. When the block is released, it makes elastic collision with the wall and its period of motion is `0.2 sec`. Find the approximate value of `k` |
| Answer» Correct Answer - `100 Nm^(-1)` | |
| 636. |
The total energy of a particle in SHM is E. Its kinetic energy at half the amplitude from mean position will beA. E/2B. E/3C. E/4D. 3E/4 |
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Answer» Correct Answer - D KE`=(1)/(2)momega^(2)(A^(2)-x^(2))implies(1)/(2)momega(A^(2)-(A^(2))/(4))" "(becausex=(A)/(2))` `=(3)/(4)xx(1)/(2)momega^(2)A^(2)=(3)/(4)E` |
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| 637. |
A body of mass 10g executes SHM with amplitude `2 xx 10^(-2)` m and time period 2 s Calculate the energy of the particle. |
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Answer» Correct Answer - `1.974xx10^(-5)J` |
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| 638. |
Find the phase difference between two particles executing simple harmonic motion with the same frequency if they are found inn the states shown in the figures at four different points of time. |
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Answer» Correct Answer - (a)`pi`, (b)`(4pi)/(3)`, (c)`(5pi)/(6)`, (d) `(11pi)/(6)`, |
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| 639. |
A particle is moving with constant angular velocity along the circumference of a circle. Which of the following statements is trueA. The particle executes SHMB. The projection of the particle on any one of the diameters executes SHMC. The projection of the particle on any of the diameters executes SHMD. None of the above |
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Answer» Correct Answer - C When the particle moves along the cirumference of a circle with constant angular velocity, the prohection of the particle on any of the diameters executes SHM. |
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| 640. |
The phase of a particle executing simple harmonic motion is `pi/2` when it hasA. Maximum velocityB. Maximum accelerationC. Maximum energyD. Maximum displacement |
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Answer» Correct Answer - B::D |
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| 641. |
Which of the following expressions represent simple harmonic motionA. ` x= A sin (omegat+ delta)`B. ` x= B cos (omegat+ phi)`C. ` x= Atan (omegat+ phi)`D. ` x= A sin omegat cosomegat` |
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Answer» Correct Answer - A::B::D |
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| 642. |
A particle is moving with constant angular velocity along the circumference of a circle. Which of the following statements is trueA. The particle so moving executes S.H.M.B. The projection of the particle on any one of the diameters executes S.H.M.C. The projection of the particle on any of the diameters executes S.H.M.D. None of the above |
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Answer» Correct Answer - C |
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| 643. |
Which of the following equation does not represent a simple harmonic motionA. `y=asinomegat`B. `y=acosomegat`C. `y=asinomegat+bcosomegat`D. `y=a tan omegat` |
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Answer» Correct Answer - D For SHM acceleration`prop-` (displacement) |
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| 644. |
A clock with an iron pendulum keeps correct time at `20^(@)C`. How much time will it lose or gain in a day if the temperature changes to `40^(@)C`. Thermal coefficient of liner expansion `alpha = 0.000012 per^(@)C`.A. `10.3`s/dayB. 19s/dayC. `5.5`s/dayD. `6.8`s/day |
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Answer» Correct Answer - A `(Deltat)/(t)=(1)/(2)(DeltaL)/(2)=(1)/(2)alphaDeltatheta` `=(1)/(2)xx12xx10^(-6)xx(40^(@)-20^(@))=12xx10^(-5)` `Deltat=txx12xx10^(-5)=86400xx12xx10^(-5)=10.3s//day` |
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| 645. |
A clock with an iron pendulum keeps correct time at `20^(@)C`. How much time will it lose or gain in a day if the temperature changes to `40^(@)C`. Thermal coefficient of liner expansion `alpha = 0.000012 per^(@)C`.A. 10.3 second/ dayB. 20.6 seconds / daysC. 5 seconds/ dayD. 20 minutes /day |
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Answer» Correct Answer - A |
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| 646. |
The amplitude and the time period in a S.H.M. is 0.5 cm and 0.4 sec respectively. If the initial phase is `pi//2` radian, then the equation of S.H.M. will beA. `y=0.5 sin 5 pit`B. `y=0.5 sin 4pit`C. `y=0.5 sin2.5pit`D. `y=0.5 cos5pit` |
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Answer» Correct Answer - D |
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| 647. |
Two equations of two S.H.M. are `y=alpha sin (omegat-alpha) and y=b cos ( omega t -alpha)` . The phase difference between the two isA. `0^(@)`B. `alpha^(@)`C. `90^(@)`D. `180^(@)` |
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Answer» Correct Answer - C |
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| 648. |
A body oscillates with SHM according to the equation (in SHM unit ), `x=5"cos"(2pit+(pi)/(4))` . Its instantaneous displacement at t=1 s isA. `(sqrt(2))/(5)`mB. `(1)/(sqrt(3))`mC. `(1)/(2)` mD. `(5)/(sqrt(2))` m |
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Answer» Correct Answer - D (e) Given, `x=5"cos"(2pit+(pi)/(4))` (where, x is a displacement) `impliesx=5"cos"(2pi+(pi)/(4))" "("at " t=1 s)` `" "[because "cos"(360^(@)+theta)="cos"theta]` `implies x=5"cos"(pi)/(4)impliesx=5xx(1)/(sqrt(2))` `therefore x=(5)/(sqrt(2)) m` |
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| 649. |
A partilce is executive simple harmonic motion given by `x=5sin(4t-pi/6)` The velocity of the particle when its displacement is 3 units isA. `(2pi)/(3)` unitsB. `(5pi)/(6)` unitsC. `20` unitsD. `16` units |
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Answer» Correct Answer - D `x=5sin(4t-pi//6)` `x=Asin(omegat+phi)` Compaing (i) and (ii) `A=5, omega=4` `v=omegasqrt(A^2-x^2)` `=4sqrt(5^2-3^2)=16units` |
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| 650. |
A man of mass `60kg` is standing on a platform executing SHM in the vertical plane. The displacement from the mean position varies as `y = 0.5sin(2pift)`. The value of `f`, for which the man will feel weightlessness at the highest point, is (`y` in metre)A. `(g)/(4pi)`B. `4pig`C. `(sqrt(2)g)/(2pi)`D. `2pisqrt(2)g` |
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Answer» Correct Answer - C For welgthlessness `mg = momega^(2)a rArr g = (2pif)^(2) (0.5)` `rArr 2 pif = sqrt(2)g rArr f = (sqrt(2)g)/(2pi)` |
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