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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.
| 401. |
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 net external force acting on the disk when its centre of mass is at displacement x with respect to its equilibrium position is.A. `-kx`B. `-2kx`C. `-(2kx)/(3)`D. `-(4kx)/(3)` |
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Answer» Correct Answer - D For linear motion of disc `F_("net") = Ma = - 2kx + f` where `f =` frictional force For rolling motion `fR = - ((MR^(2))/(2))(alpha) = -((Ma)/2)R` `rArr f = -(Ma)/(2) = - (F_("ext"))/(2)` Therefore `F_(ext) = - 2kx - (F_("ext"))/(2) = -(4kx)/(3)` |
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| 402. |
Column I gives a list of possible set of parameters measured in some expermients. The varitions of the parameters in the form of graphs ar shown in Column II. Match the set of parameters given Column I with the graph given in Column II. Indicate your answer by darking the appropriate bubbles of the `4 xx 4` matrix given in the ORS. |
| Answer» Correct Answer - A::B::C::D | |
| 403. |
With respect to photoelectric experiment, match the entries of Column I with the enetires of Column II. |
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Answer» Correct Answer - `(A) rarr (P,R); (B) rarr (Q,S); (C) rarr (S); rarr (D) rarr (P,R)` From `eV_(0) = hv- phi` and `K_("max") - hv - phi `. If v increase keeping `phi` constant constant, the `V_(0)` and `K_(max)` both incease. If `phi` decrease keeping `v` constant, then `V_(0)` and `K_(max)` increase. If `I` incrases more photoelectrons would be librated, hence saturation photocurrent increases. If separation between cathode and anode is incrased, then there is no effect on `v_(0), K_(max)` or current. |
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| 404. |
The binding energy of nucei X and `Y` are `E_(1)` and `E_(2)`, respectively. Two atoms of X fuse to give one atom of `Y` and an energy `Q` is released. Then,A. `Q = 2E_(1) - E_(2)`B. `Q = E_(2) - 2E_(1)`C. `Q = 2E_(1) + E_(2)`D. `Q = 2E_(2) + E_(1)` |
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Answer» Correct Answer - B Energy released `=(BE)_("product")-(BE)_("reactant")` |
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| 405. |
The electron in hydrogen atom in a sample is in `n^(th)` excited state, then the number of differrent spectrum lines obtained in its emission spectrum will beA. `1+2+3+"……….."+(n-1)`B. `1+2+3+"…………."+(n)`C. `1+2+3+"…………."+(n+1)`D. `1xx2xx3xx"…………"xx(n-1)` |
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Answer» Correct Answer - B `n^(th)` excited state `=.^(n+1)C_(2) =(n(n+1))/(2) =Sigma n` |
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| 406. |
F-x and x-t graph of a principle in SHM are as shown in figure. Match the following. |
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Answer» Correct Answer - A::B::C |
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| 407. |
In the two block spring system, force constant of spring is k = 6N/m. Spring is stretched by 12 cm and then left. Match the following. |
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Answer» Correct Answer - A::B::C |
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| 408. |
F-x and x-t graph of a principle in SHM are as shown in figure. Match the following. |
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Answer» Correct Answer - `(A rarr q, B rarr r, C rarr s)` T=8 s `therefore omega=(2pi)/(T)=(pi)/(4) "rads"^(-1)` K=-slope of F-x graph `=10 Nm^(-1)` From, `T=2pisqrt((m)/(k))` We have, `8=2pisqrt((m)/(10))` `therefore m=(160)/(pi^(2))kg` `K_("max")=(1)/(2)momega^(2)A^(2)=(1)/(2)xx(160).(pi^(2))xx(pi^(2))/(16)xx(4xx10^(-2))^(2)` `=80xx10^(-3) J` |
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| 409. |
In the two block spring system, force constant of spring is k = 6N/m. Spring is stretched by 12 cm and then left. Match the following. |
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Answer» Correct Answer - `(A rarrq , B rarr s, C rarr p)` `omega=sqrt((k)/(mu))` Here, `mu`=reduced mass `=(m_(1)m_(2))/(m_(1)+m_(2))=(2)/(3)kg` `therefore omega=sqrt((6)/(2//3))=3 "rads"^(-1)` Amplitude 12 cm distance in inverse ratio of mass, `therefore A_(1)=8 cm` and `A_(2)=4 cm` Now, maximum kinetic energy `K_(1)=(1)/(2)m_(1)omegaA_(1)^(2)` `=(1)/(2)xx1xx3xx(8xx10^(-2))^(2)=9.6 mJ` `K_(2)=(1)/(2)xx2xx3xx(4xx10^(-2))^(2)=4.8 mJ` |
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| 410. |
In SHM match the following |
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Answer» Correct Answer - `(Ararrq, B rarr r, C rarr q)` Suppose `x=A"sin"omegat` Then, `v=(dx)/(dt)=omegaA"cos"omegat` and `a=(dv)/(dt)=-omega^(2)A"sin"omegat` |
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| 411. |
Out of the following functions representing motion of a particle which represents SHM ? I. `y = sin omega t - cos omega t` , II. `y = sin^(3) omega t` III. `y =5 cos ((3pi)/(4)-3 omega t)` , IV. `y = 1+omega t +omega^(2)t^(2)`A. A and BB. A and CC. A onlyD. A, B and C not D |
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Answer» Correct Answer - B (b) For a simple harmonic motion, `(d^()Y)/(dt^(2)) prop (-T)` Equations , `Y="sin"omegat-"cos"omegat` and `Y=5"cos"((3pi)/(4)-3omegat)` are satisfying this condition. Equations, `Y=1+omegat+omega^(2)t^(2)+omega^(3)t^(3)` is not periodic and `Y="sin"^(3)omegat` is periodic but not SHM. |
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| 412. |
The acceleration of a particle performing S.H.M. is ` 12 cm // sec^(2) ` cm at a distance of 3 cm form the mean position. Its period isA. 0.5 secB. 1.0 secC. 2.0 secD. 3.14 sec |
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Answer» Correct Answer - D |
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| 413. |
In the figure shown, the spring are connected to the rod at one end and at the midpoint. The rod is hinged at its lower end. Rotational `SHM` of the rod (Mass `m`, length `L`) will occur only if- A. `kgtmg//3L`B. `kgt2mg//3L`C. `kgt2mg//5L`D. `k=0` |
| Answer» Correct Answer - C | |
| 414. |
Equation of SHM is x=10sin10`pit`. Find the distance between the two points where speed is `50pi`cm/s. x is in cm and t is in seconds.A. 10cmB. 14 cmC. 17.32 cmD. 8.66 cm |
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Answer» Correct Answer - c |
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| 415. |
In a simple harmonic oscillator, at the mean positionA. kinetic energy is minimum, potential energy is maximumB. both kinetic and potential energies are maximumC. kinetic energy is maximum,potential energy is minimumD. both kinetic and potential energies are minimum |
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Answer» Correct Answer - C In a simple harmonic oscillator the potential energy is directly proportional to the square of displacement of the body from the mean position , at the mean postion the displacement is zero so the PE is zero but speed is maximum, hence KE is maximum. |
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| 416. |
If a spring has time period T, and is cut into (n) equal parts, then the time period of each part will be.A. `Tsqrt(n)`B. `(T)/(sqrt(n))`C. `nT`D. `T` |
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Answer» Correct Answer - B Time period of a mass loaded spring `T = 2pisqrt((m)/(k))` So `T prop (l)/(sqrt(k)) "….."(i)` Spring constant (k) is inversely proportional to the length of the spring i.e., `k prop 1/l` `(k_("complete spring"))/(k_("cut spring")) = ((1//l))/((l/(l//n))) = l/n` `rArr k_("cut spring") = n(k_("complete spring"))` `rArr (T_("cut spring"))/(T_("complete spring")) = sqrt((k_("complete spring"))/(k_("cut of spring"))) = (1)/(sqrt(n))` [From (i) `rArr T_("cut spring") = (T_("complete spring"))/(sqrt(n))` |
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| 417. |
The frequency of vertical oscillations of the three spring-mass system, shown in figure, is A. `(1)/(2pi)sqrt((3k)/(2m))`B. `(1)/(2pi)sqrt((2k)/(3m))`C. `(1)/(2pi)sqrt((2k)/(m))`D. `(1)/(2pi)sqrt((k)/(3m))` |
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Answer» Correct Answer - B `k_(eq)=((k+k)k)/((k+k)+k)=2k//3` `T=2pisqrt((m)/(k_(eq)))=2pisqrt((m)/(2k//3))=2pisqrt((3m)/(2k))` `f=1/T=(1)/(2pi)sqrt((2k)/(3m))` |
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| 418. |
STATEMENT-1 : In oscillatory motion, displacement of a body from equilibrium can be represented by `sin` or `cos` function. STATEMENT-2 : The body oscillates to and froabout its mean position.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
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Answer» Correct Answer - B |
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| 419. |
Passage XII)_ A particle of mass m is constrained to move along x-axis. A force F acts on the particle. F always points toward the position labelled E. For example, when the particle is to the left of E,Fpoints to the right. The magnitude of F is consant except at point E where it is zero. The systm is horizontal. F is the net force acting on the particle. the particle is displaced a distance A towards left from the equilibrium position E and released from rest at t=0. Find minimum time it will take to reach from x`=-A/2` to 0 A. `3/2sqrt((mA)/F)(sqrt(2)-1)`B. `sqrt((mA)/F)(sqrt(2-1))`C. `2sqrt((mA)/(F))(sqrt(2)-1)`D. None of these |
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Answer» Correct Answer - B |
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| 420. |
Two particles A and B are performing SHM along x and y-axis respectively with equal amplitude and frequency of `2 cm` and `1 Hz` respectively. Equilibrium positions of the particles A and B are at the coordinates `[3 cm, 0]` and `(0, 4 cm)` respectively. At `t = 0 ,B` is at its equilibrium position and moving towards the origin, while A is nearest to the origin and moving away from the origin- Minimum and maximum distance between A and B during the motion is-A. `sqrt(5) cm` and `sqrt(61) cm`B. `3 cm` and `7 cm`C. `1 cm` and `5 cm`D. `9 cm` and `16 cm` |
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Answer» Correct Answer - B Distance between A & B `= sqrt(x^(2) + y^(2))` `= sqrt((3-2 cos 2pit)^(2) + (4-2 sin 2pit)^(2))`, `= sqrt(9+4cos^(2)2pit - 12 cos2pit+16+4 sin^(2)2pit - 16sin2pit)` `= sqrt(29-20(3/5 cos 2pit + 4/5 sin 2pit))` `= sqrt(29-20sin(2pit + 37^(@)))` Maximum distance `= sqrt(29 + 20) = sqrt(49) = 7 cm` Minimum distance `= sqrt(29-20) = sqrt(9) = 3 cm` |
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| 421. |
Two particles A and B are performing SHM along x and y-axis respectively with equal amplitude and frequency of `2 cm` and `1 Hz` respectively. Equilibrium positions of the particles A and B are at the coordinates `[3 cm, 0]` and `(0, 4 cm)` respectively. At `t = 0 ,B` is at its equilibrium position and moving towards the origin, while A is nearest to the origin and moving away from the origin- Equation of motion of particle B can be written as-A. `y = (2 cm) cos 2 pit`B. `y = (4 cm) - (2 cm) cos 2pit`C. `y = (2 cm) sin 2 pit`D. `y = (4 cm) - (2 cm) sin 2pit` |
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Answer» Correct Answer - D As B is at its equilibrium position and moving towards negative exterme at `t = 0` so `y-4 = 0.2 sin (2pit + pi) rArr y= 4 -2 sin (2pit)` |
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| 422. |
A pariticle is performing `SHM` of amplitude `"A"` and time period `"T"`. Find the time taken by the period to go from `0` to `A//2`. |
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Answer» Let equation of `SHM` be `x = A sin omegat` when `x = 0`, `t = 0` when `x = A//2, A//2 = A sin omegat` or `sinomegat = 1//2, omegat = pi//6` `(2pi)/(T)t = pi//6t = T//12` Hence, time taken is `T//12`, where `T` is where `T` is time period of `SHM`. |
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| 423. |
Column-I gives different situtions in which bodes are floating in liquid. Column-II gives correspoding time periods of oscillation. A cube floating in liquid.A. `{:(,P,Q,R,S),((A),3,4,2,1):}`B. `{:(,P,Q,R,S),((B),1,3,4,2):}`C. `{:(,P,Q,R,S),((C),2,3,1,4):}`D. `{:(,P,Q,R,S),((D),4,3,2,1):}` |
| Answer» Correct Answer - A | |
| 424. |
x-tequation of a particle moving along x-axis is given as `x=A+A(1-cosomegat)`A. particle oscillates simple harmonically between `x=2A` and `x=A`B. velocity of particle is maximum at `x=2A`C. time taken by particles in travelling from x=A to x=3A is `pi/omega`D. time taken by particles in travelling from x=A to x=2A is `pi/(2omega)` |
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Answer» Correct Answer - B::C::D |
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| 425. |
Passage X) A 2kg block hangs without vibrating at the bottom end of a spring with a force constant of 400 N/m. The top end of the spring is attached to the ceiling of an elevator car. The car is rising with an upward acceleration of `5m/`s^(2)` when the accleration suddenly ceases at t=0 and the car moves upward with constant speed. (g=10m/`s^(2)`) What is the angular frequency of oscillation of the block after the acceleration ceases?A. `10sqrt(2)`rad/sB. `20 rad/s`C. `20sqrt(2)`rad/sD. `32 rad/s` |
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Answer» Correct Answer - A |
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| 426. |
Passage X) A 2kg block hangs without vibrating at the bottom end of a spring with a force constant of 400 N/m. The top end of the spring is attached to the ceiling of an elevator car. The car is rising with an upward acceleration of `5m/`s^(2)` when the accleration suddenly ceases at t=0 and the car moves upward with constant speed. (g=10m/`s^(2)`) The amplitude of the osciallations isA. 7.5 cmB. 5 cmC. 2.5 cmD. 1 cm |
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Answer» Correct Answer - C |
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| 427. |
Block `A` in the figure is released from the rest when the extension in the spring is `x_(0)`. The maximum downward displacement of the block will be `:` A. `(Mg)/(2k)-x_(0)`B. `(Mg)/(2k)+x_(0)`C. `(2Mg)/(k)-x_(0)`D. `(2Mg)/(k)+x_(0)` |
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Answer» Correct Answer - A For maximum displacement `Mg(x) = 1/2k(2x)^(2) rArr x = (Mg)/(2k) rArr x = (mg)/(2k) - x_(0)` |
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| 428. |
A constant force F is applied on a spring block system as shown in figure. The mass of the block is m and spring constant is k. The block is placed over a smooth surface. Initially the spring was unstretched. Choose the correct alternative(s) A. The block will executes SHMB. Amplitude of oscillation is `F/(2K)`C. Time period of oscillation is `2pisqrt(m/k)`D. The maximum speed of block is `sqrt(2Fx-kx^(2))/(m)` (Here, x=F/k). |
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Answer» Correct Answer - A::C::D |
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| 429. |
Velocity-time graph of a particle executing SHM is shown in figure. Select the correct alternative(s). A. At position 1, displacement of particles may be positive or negativeB. At position 2, displacement of particlees is negativeC. At position 3, acceleration of particle is positvieD. At position 4, acceleration of particle is positive |
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Answer» Correct Answer - B::C |
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| 430. |
Assertion : Bob is released from rest position A. Given `theta_(0)` very small. Angular velocity of bob about point O is maximum and `(sqrt((g)/(l)))theta_(0)` at point O. Reason : For small angular amplitudes , motion of bob is simple harmonicA. 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 - B (b) Maximum angular velocity `=omegatheta_(0)` `=(2pif)theta_(0)=(2pi)((1)/(2pi)sqrt((g)/(l)))theta_(0)` `=(sqrt((g)/(l)))theta_(0)` |
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| 431. |
The amplitude of vibration of a particle is given by `a_m=(a_0)//(aomega^2-bomega+c)`, where `a_0`, a, b and c are positive. The condition for a single resonant frequency isA. `b^2=4ac`B. `b^2gt4ac`C. `b^2=5ac`D. `b^2=7ac` |
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Answer» Correct Answer - A `aomega^2-bomega+c=0impliesomega=(b+-sqrt(b^2-4ac))/(2a)` For single frequency `b^2=4ac` |
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| 432. |
Two particles are in `SHM` in a straight line about same equilibrium position. Amplitude `A` and time period `T` of both the particles are equal. At time `t=0`, one particle is at displacement `y_(1)=+A` and the other at `y_(2)=-A//2`, and they are approaching towards each other. after what time they cross each other?A. `T//3`B. `T//4`C. `5T//6`D. `T//6` |
| Answer» Correct Answer - D | |
| 433. |
Two particles are executing SHM in a straight line. Amplitude A and the time period T of both the particles are equal. At time t=0, one particle is at displacement `x_(1)=+A` and the other `x_(2)=(-A/2)` and they are approaching towards each other. After what time they across each other? `T/4`A. `T/3`B. `T/4`C. `(5T)/6`D. `T/6` |
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Answer» Correct Answer - D |
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| 434. |
A simple harmonic oscillator of angular frequency `2 "rad" s^(-1)` is acted upon by an external force `F = sint N`. If the oscillator is at rest in its equilibrium position at `t = 0`, its position at later times is proportional to :-A. `sin t - 1/2 sin 2t`B. `sin t + 1/2 cos 2t`C. `sin t + 1/2 sin 2t`D. `cos t - 1/2 sin 2t` |
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Answer» Correct Answer - A `F = F sin t rArr a = F/m sin t rArr v = - (F)/(m) cos t` `x = - (F)/(m) sin t rArr X= vec(rx)_(1) + vec(rx)_(2) = A sin 2t - (F)/(m) sin t` |
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| 435. |
In the amplitude of vibration isA. 3 unitsB. 4 unitsC. 5 unitsD. None of these |
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Answer» Correct Answer - B `y-3=4 cos omegat` Comparing with x = A cos `omegat` `rArr` Amplitude of SHM, A = 4 units |
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| 436. |
Three masses of `500 g`, `300 g` and `100 g` are suspended at the end of a spring as shown and are in equilibrium. When the `500 g` mass is removed suddenly, the system oscillates with a period of `2 s`. When the `300 g` mass is also removed, it will oscillate with period `T`. Find the value of `T`. |
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Answer» Correct Answer - A::C `2 = 2pi sqrt((0.3 + 0.1))/(k))` ….(i) `T = 2pi sqrt ((0.1)/(k))` …(ii) Solving these two equation we get, `T = 1 s` |
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| 437. |
The displacement - time graph of a particle executing SHM is shown in figure. Which of the following statements is//are true ? A. The velocity is maximum at `t=T//2`B. The acceleration is maximum at `t=T`C. The force is zero at `t=3T//4`D. The potential energy equals the oscillation energy at `t=T//2`. |
| Answer» Correct Answer - B::C::D | |
| 438. |
A particle starts from a point `P` at a distance of `A//2` from the mean position `O &` travels towards left as shown in the figure. If the time period of `SHM`, executed about `O` is `T` and amplitude `A` then the equation of motion of particle is : A. `X=A sin ((2pi)/(T)t+(pi)/(6))`B. `X=A sin((2pi)/(T)t+(5pi)/(6))`C. `X=A cos((2pi)/(T)t+(pi)/(6))`D. `X=A cos ((2pi)/(T)t+(pi)/(3))` |
| Answer» Correct Answer - B::D | |
| 439. |
Find the frequency of oscillations of the system in the figure. The bar is rigid and light. [Hint: A spring of force constant `k_(1)` at a distance a is equivalent to a spring of force constant `k_(1)(a/b)o^(2)` at a distance b) |
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Answer» Correct Answer - `2nsqrt((m)/(k_(1)k_(2))[k_(1)+k_(2)((b)/(a))^(2)])` |
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| 440. |
A pipe in the form of a half ring is placed on a horizontal surface. If it is rotated through a small angle, and then released, assuming that it rolls without sliding determine the period of oscillations. The center of gravity of such a body is at distance `(2r)/pi)` from the center where r is the radius of the ring. Hint: `(vec_(c) = alpha rhat(i)` and `vec(a_(c)) = vec(a_(c)) +(veca_(G//c))_("tangent") +(veca_(G//c))_("normal")` |
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Answer» Correct Answer - `2nsqrt(((pi-2)r)/(g))` |
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| 441. |
A particle is subjected to two simple harmonic motions of sae time period in the same direction. The amplitude of the first motion is 3.0 cm and that of the second is 4.0 cm. Find the resultant amplitude if the phase difference between the motion is a. `0^@, b. 60^@, c. 90^@` |
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Answer» Correct Answer - A: 7cm B: 6.1cm C: 5cm The particle is subjected to two SHMs of same time period in the same direction. Given `r_1=3`cm, `r_2=4`cm and `phi`=phase difference Resultant amplitude `R=sqrt(r_1^2+r_2^2+2r_1r_2cos phi)` a. when `phi=0^@` `R=sqrt((3^2+4^2+2.3.4cos0^/2))` `=sqrt49=7cm` b. when `phi=60^@` `R=sqrt((3^2+4^2+2.3.4cos60^@))` `=sqrt37=6.1cm` c. when `phi=90^@` `R=sqrt((3^2+4^2+2.3.4cos90^@))` `=sqrt25=5cm` |
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| 442. |
A particle is subjected to two simple harmonic motions of the same frequency and direction. The amplitude of the first motion is `4.0 cm` and that of the second is `3.0 cm`. Find the resultant amplitude if the phase difference between the two motion is (a) `0^(@)` (b) `60^(@)` ( c) `90 ^(@)` (d) `180^(@)` |
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Answer» Correct Answer - A::B::C::D In such situation, amplitudes are added by vector method. `A_(R) = sqrt((4)^(2) + (3)^(2) + 2(4)(3)cos phi)` `= sqrt(25 + 24cos phi)` ...(i) Now, we can substitute different values of `phi` given in different parts in the question and can find the value of `A_(R)`. |
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| 443. |
A spring of force constant `K` is attached with a mass executes with `S.H.M` has a time period `T`. It is cut into three equal parts If the spring are attached to the load as shown in figure then effective spring constant is A. `(K)/(2)`B. `(K)/(3)`C. `(5K)/(3)`D. `(3K)/(2)` |
| Answer» Correct Answer - D | |
| 444. |
A spring of force constant `K` is attached with a mass executes with `S.H.M` has a time period `T`. It is cut into three equal parts The frequency of arrangement when the springs are connected in series to the parallel combination for the same load isA. same in both casesB. More in seriesC. More in parallelD. can not say |
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Answer» Correct Answer - C |
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| 445. |
A spring of force constant `K` is attached with a mass executes with `S.H.M` has a time period `T`. It is cut into three equal parts If the springs are connected in parallel then the time period of the combination for the same mass isA. `T`B. `(T)/(3)`C. `(T)/(9)`D. `(2T)/(3)` |
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Answer» Correct Answer - B |
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| 446. |
A block of mass `m`, when attached to a uniform ideal apring with force constant `k` and free length `L` executes SHM. The spring is then cut in two pieces, one with free length n `L` and other with free length `(1 - n)L`. The block is also divided in the same fraction. The smaller part of the block attached to longer part of the spring executes SHM with frequency `f_(1)` . The bigger part of the block attached to smaller part of the spring executes SHM with frequency `f_(2)`. The ratio `f_(1)//f_(2)` isA. `1`B. `(n)/(1 - n)`C. `(1 + n)/(n)`D. `(n)/(1 + n)` |
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Answer» Correct Answer - A `k prop (1)/("Length of spring")` `{:("Length","Force constant"),(L,k),(nL,k//n),((1-n),k//(1-n)):}` `m_(1) = nm` and `m_(2) = (1 - n)m` `(f_(1))/(f_(2)) = ((1)/(2pi)sqrt ((k//n)/((1 - n)m)))/((1)/(2pi)sqrt((k//(1 - n))/(nm))) = 1` |
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| 447. |
एक द्रव्यमानहीन स्प्रिंग की तनावहीन लम्बाई 4.9 सेमी है | इसका एक सिरा बँधा है तथा दूसरे पर एक छोटा गुटका लगा है (चित्र) | यह निकाय एक घर्षणहीन क्षैतिज तल पर रखा है | समय t = 0 पर गुटके को 0.2 मीटर खींचकर स्थिर अवस्था से छोड़ा जाता है, तब यह कोणीय आवृत्ति `omega = (pi)/(3)` रेडियन/सेकण्ड से सरल आवर्त गति करता है | ठीक उसी समय (t = 0) पर एक छोटा कंकड़ v चाल से बिन्दु P पर क्षैतिज से `45^(@)` के कोण पर प्रक्षपित किया जाता है, बिन्दु P को बिन्दु O से क्षैतिज दूरी 10 मीटर है | यदि t = 1 सेकण्ड पर कंकड़ गुटके पर गिरता है, तथा का मान है ( g = 10 मी/से`""^(2)`)A. `sqrt(50)m//s`B. `sqrt(51)m//s`C. `sqrt(52)m//s`D. `sqrt(53)m//s` |
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Answer» Correct Answer - A Time of flight for phojectile `T = (2usintheta)/(g) = 1` sec. `(2usin45)/(g) = 1` sec : `u = (g)/(sqrt(2)) : u = sqrt(50) m//s` |
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| 448. |
Phase space diagrams are useful tools in analysing all kond of dynamical problems. Theya re especially useful in studying the changes in motion as initial position and momentum are changed. Here we consider some simple dynamical system in one-dimelnsion. for such systeam, phase space is a plane in which position is plotted along horizontal axis and momentum is plotted along vertical axis. The phase space diagram is `x(t)` vs. `p(t)` curve in this plane. The arrow on the curve indicates the time flow. for example, the phase space diagram for a particle moving with constant velocity is a straight line as shown in the figure. We use the sign convention in which position or momentum. upwards (or to right) is positive and downwards (or to left) is negative. The phase space diagram for simple harmonic motion is a circle cen-tered at the origin.In the figure, the two circles represent the same oscillator but for different initial conditions, and `E_(1)` for `E_(2)` are the total mechanical energies respectively. Then A. `E_(1) = sqrt(2)E_(2)`B. `E_(1) = 2E_(2)`C. `E_(1) = 4 E_(2)`D. `E_(1) = 16 E_(2)` |
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Answer» Correct Answer - C In `1st` case amplitude of `SHM` is `a`. In `2nd` case amplitude of `SHM` is `2a` Total energy `= (1)/(2)K(amplitude)^(2)` `E_(1)/(2)2k(2a)^(2) , E_(1) = 4 E_(2)` Alternative : Linear momentum `P = mv` `= momegasqrt(A^(2) - x^(2))` `rArr P^(2) = m^(2)omega^(2) (A^(2) - x^(2))` `rArr P^(2) + m^(2)omega^(2)x^(2) = m^(2)omega^(2)A^(2)` Equation of circle (smaller) `P^(2) + x^(2) = a^(2) .......(iii)` Comparing `(i)` and `(ii)` Amplitude `A = 2a` and `(momega)^(2) = (1)/(m)` `(1)/(2)momega^(2)(A)^(2)` So energy `E_(1) = (1)/(2) momega^(2) (2a)^(2)` `(1)/(2)(1)/(m)xx(4a^(2))` `= (2a^(2))/(m)` Comparing `(i)` and `(iii)` `(momega)^(2) = 1 rArr momega^(2)A^(2) = (1)/(2) xx (1)/(m)a^(2) = (1)/(2)(a^(2))/(m^(2))(1a^(2))/(2 m)`. So, `(E_(1))/(E_(2)) = 4 rArr E_(1) = 4E_(2)` |
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| 449. |
idenigu correcy statement among the followingA. The greater the mass of a pendulum bob, the shorter is its frequency of oscillationB. A simple pendulum with a bob of mass M swing s with an angular amplitude of `40 ^(@)`, the tension in the string is less than `Mgcos 20^(@)`C. As the length of a simple pendulam is increased m the maximum velcoity of its bob during its oscillation will also decreasesD. The freactional change in the time period of a pendulum on changing the temperature is independent of the length of the pendulum |
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Answer» Correct Answer - C |
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| 450. |
The amplitude and the periodic time of a S.H.M. are 5 cm and 6 sec respectively. At a distance of 2.5 cm away from the mean position, the phase will beA. `(pi)/(3)`B. `(pi)/(4)`C. `(pi)/(6)`D. `(5pi)/(12)` |
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Answer» Correct Answer - C (c) The equation of motion is given as `y=5"sin"(2pit)/(6)` Here, `y=2.5 cm` `therefore 2.5=5 "sin"(2pit)/(6)` `implies(pi)/(6)=(2pit)/(6)` `implies t=(1)/(2)s` `therefore` The phase `=(2pit)/(6)=(2pi)/(6)xx(1)/(2)=(pi)/(6)` |
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