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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.
| 301. |
A particle of mass (m) is attached to a spring (of spring constant k) and has a narural angular frequency omega_(0). An external force `R(t)` proportional to cos omegat(omega!=omega)(0) is applied to the oscillator. The time displacement of the oscillator will be proprtional to.A. (a) `1/(m(omega_(0)^(2)+omega^(2))`B. (b) 1/(omega(omega_(0)^(2)-omega^(2))`C. (c ) `m/(omega_(0)^(2)-omega^(2)`D. (d) `m/(om_(0)^(2)+omega^(2))` |
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Answer» Correct Answer - B (b) Equation of desplacement is given by `x=A sin(omegat+phi) where `A=f_(0)/(m(omega_(0)^(2)-omega^(2)))`. |
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| 302. |
In forced oscillation of a particle the amplitude is maximum for a frequency `omega_(2)` of the force while the energy is maximum for a frequecyomega_(2) of the force, then . |
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Answer» Correct Answer - Both amplitued and energy of the particle can be maximum only in case of resonance. For resonacne to occur, `omega_(1) = omega_(2)` |
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| 303. |
In forced oscillation of a particle the amplitude is maximum for a frequency `omega_(2)` of the force while the energy is maximum for a frequecyomega_(2) of the force, then .A. (a) `omega_(1)ltomega_(2) when damping is small and `omega_(1)gtomega_(2)` when damping.B. (b) `omega_(1)gtomega_(2)`C. (c ) `omega_(1)=omega_(2)`D. (d) omega_(1)ltomwga_(2). |
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Answer» Correct Answer - C (c )The maximum of amplitude and energy is obtained when the frequecy is equal to the natural freqency (resonance condition) :. Omega_(1)=omega_(2)`. |
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| 304. |
A particle simultaneously participates in two mutually perpendicular oscillations `x = sin pi t` & `y = 2cos 2 pit`. Write the equation of trajectory of the particle. |
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Answer» Correct Answer - `2x^(2) + (y)/(2) = 1` `x = sinpit, y = 2 [1 - 2 sin^(2) pit]` `y = 2[1 - 2x^(2)]` or `2x^(2) + (y)/(2) = 1` Ans |
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| 305. |
In forcd oscillation of a particle, the amplitude is maximum for a frequency `omega_(1)` of the force, while the energy is maximum for a frequency `omega_(2)` of the force. What is the relation between `omega_(1)` and `omega_(2)` ? |
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Answer» Correct Answer - Both amplitued and energy of the particle can be maximum only in case of resonance. For resonacne to occur, `omega_(1) = omega_(2)` |
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| 306. |
According to a scientists, he applied a force `F = -cx^(1//3)` on a particle and the particle is performing SHM. No other force acted on the particle. He refuses to tell whether `c` is a constant or not. Assume that he had worked only with positive `x` then :A. as `x` increases `c` also increasesB. as `x` increases `c` also decreasesC. as `x` increases `c` remain constantD. the motion cannot be SHM |
| Answer» Correct Answer - A | |
| 307. |
Describe the motion of a particle acted upon by a force (i) `F= -2(x - 2)^(3)` (ii) `F= -2(x - 2)^(2)` (iii) `F= -2(x - 2)` |
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Answer» Correct Answer - A (i) `F= -2(x - 2)^(3)` `F = 0` at `x = 2` Force is along negative x-direction for `x > 2` and it is along positive x-direction for `x < 2`. Thus, the motion of the particle is oscillatory ( but not simple harmonic ) about `x = 2`. (ii) `F = 0` for `x = 2`, but force is always along negative x-direction for any value of `x` except at `x = 2`. Thus, the motion of the particle is rectilinear along negetive x-direction provided it is not kept at rest at `x = 2`. (iii) Let, us take `x - 2 = X`, then the given force can be written as, `F = -2X` This is the eqation of SHM. Hence, the particle oscillates simple harmonically about `X=0` or `x = 2`. |
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| 308. |
The block is allwed to fall, slowly from the position wher spring is in its natural length. Find, maximum extension in the string. |
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Answer» Since the block falls slowly from rest maximum extension occurs when `mg = Kx_(0)` `x_(0) = (mg)/(K)` is maximum extension, Ans.` (mg)/(K)` |
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| 309. |
The block is allwed to fall, slowly from the position wher spring is in its natural length. Find, maximum extension in the string. |
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Answer» Since the block falls slowly from rest maximum extension occurs when `mg = Kx_(0)` `x_(0) = (mg)/(K)` is maximum extension, Ans.` (mg)/(K)` |
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| 310. |
Repeat the previous exercise if the angle between each pair of springs is `120^@` initially. |
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Answer» Correct Answer - `Time period =T=(2pi)/omega=2pisqrt((2m)/(3k))` In this case if the particle m is pushed against C a by distance x. total resultant force acting on man m is given by `F=kx+(kx)/2=(3kx)/2` [Because net force by A and B ] ` sqrt((kx/2)^2+(kx/2)^2+2(kx/2)(kx/2)) cos120^@` `=kx/2` ` :. a=F/m=(2kx)/(2m)` `rarr a/x=(3k)/(2m)=w^2` `rarr omega=sqrt(3k)/(2m)` `:. Time period =T=(2pi)/omega=2pisqrt((2m)/(3k))` |
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| 311. |
If the period of oscillation of mass M suspended from a spring is one second, then the period of 4M will beA. 1 secB. 2 secC. 3 secD. 4 sec |
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Answer» Correct Answer - D |
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| 312. |
The potential energy of a simple harmonic oscillator when the particle is half way to its end point is (where, E is the total energy)A. `1/8 E`B. `1/4 E`C. `1/2 E`D. `2/3 E` |
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Answer» Correct Answer - B |
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| 313. |
For a simple pendulum the graph between length and time period will beA. HyperbolaB. ParabolaC. A curved lineD. A straight line |
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Answer» Correct Answer - B |
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| 314. |
A spring mass system is hanging from the ceiling of an elevator in equilbrium Elongation of spring is `l`. The elevator suddenly starts accelerating downwards with acceleration `g//3` find. (a) the frequency and (b) the amplitude of the resulting `SHM` |
| Answer» Correct Answer - (a) `(1)/(T)=(1)/(2pi)sqrt((g)/(L))`, (b) `(L)/(3)K` | |
| 315. |
Which of the following atoms has the lowest ionization potential ?A. `._(7)^(14)N`B. `._(55)^(133)Cs`C. `._(18)^(40)Ar`D. `._(8)^(16)O` |
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Answer» Correct Answer - B Ionization potential will be lowest for the atom in which the electrons are the farthest from the nucleus. So, the atom with the largest size will have the electron the farthest from the nuclei, hence to remove the electron from this atom will be easiest. so, the atom with least Ionization potential is `._(55)^(133)Cs`. |
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| 316. |
When a hydrogen atom is excited from ground state to first excited state thenA. Its kintic energy increases by `10.2 eV`B. Its kinetic energy decreases by `10.2 eV`C. It potential energy increases by `20.4 eV`.D. Its angular momentum increases by `1.05 xx 10^(34) J-s` |
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Answer» Correct Answer - B::C::D `|DeltaK| = 13.6[(1)/(1^(2)) - (1)/(2^(2))] = 10.2 eV` ltbr gt `|DeltaU| = 27.2 [(1)/(1^(2)) - (1)/(2^(2))] = 20.4eV` `DeltaL = (2h)/(2pi) - (h)/(2pi) = (h)/(2pi)` |
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| 317. |
If the binding energy of the electron in a hydrogen atom is `13.6 eV`, the energy required to remove the electron from the first excited state of `Li^(++)` isA. `30.6 eV`B. `13.6 eV`C. `3.4 eV`D. `122.4 eV` |
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Answer» Correct Answer - A Energy required to remove an electron from an orbit is `+ (13.6(Z)^(2))/n^(2) eV` So, to remove the electron from the first excited state of `Li^(2+)` is `E=(+13.6xx3^(2))/2^(2)=3.4xx9= +30.6 eV` |
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| 318. |
Two simple harmonic motions are given by `y_(1) = a sin [((pi)/(2))t + phi]` and `y_(2) = b sin [((2pi)/( 3))t + phi]`. The phase difference between these after `1 s` isA. zeroB. `pi//2`C. `pi//4`D. `pi//6` |
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Answer» Correct Answer - D At `t = 1 s` `phi_(1) = (pi)/(2) + phi` and `phi_(2) = (2pi)/(3) + phi` `:. Delta phi = phi_(2) - phi_(1) = (pi)/(6)` |
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| 319. |
Assertion : For a given simple harmonic motion displacement (from the mean position) and acceleration have a constant ratio. Reason : `T = 2pi sqrt(|("displacement")/("acceleration")|)`.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 `|(x)/(a)| = ((T)/(2pi))^(2) = (1)/(omega^(2))` = constant for a given SHM. |
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| 320. |
The equation of S.H.M. is`y=a sin (2 pint + a)`, then its phase at time t isA. `2 pint`B. aC. `2 pint+a`D. `2 pint` |
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Answer» Correct Answer - C |
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| 321. |
Assertion : We can call circular motion also as simple harmonic motion. Reason : Angular velocity in uniform circular motion and angular frequency in simple harmonic motion have the same meanings.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 - D If a particle `P` rotates in a circle with constant angle speed `omega` and we draw a perpendicular on and diameter. This perpendicular cuts the diameter at point `Q` then motion of `Q` is simple harmonic. But motion of `P` is circular. |
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| 322. |
A particle is oscillating according to the equation `X=7 cos 0.5 pit`, where t is in second. The point moves from the position of equilibrium to maximum displacement in timeA. 4.0 secB. 2.0 secC. 1.0 secD. 0.5 sec |
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Answer» Correct Answer - C |
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| 323. |
Find the equation fo simple harmonic motion of a particle whose amplitude is 0.04 and whose frequency is 50 Hz . The initial phase is `pi//3` . Assume that motion of particle is started from mean position. |
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Answer» From equation of SHM , `x=A "sin"(omegat+phi)` Here, `A=0.04 m,v=50 Hz , phi=(pi)/(3)` `therefore x=0.04 "sin"(2pi xx 50t+(pi)/(3)) " "(because omega=2piv)` or `x=0.04 "sin"(100pit+(pi)/(3))` |
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| 324. |
The phase (at a time t) of a particle in simple harmonic motion tellsA. Only the position of the particle at time tB. Only the direction of motion of the particle at time tC. Both the position and direction of motion of the particle at time tD. Neither the position of the particle nor its direction of motion at time t |
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Answer» Correct Answer - C |
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| 325. |
A simple harmonic oscillation has an amplitude `A` and time period `T`. The time required to travel from `x = A` to ` x= (A)/(2)` isA. `(T)/(6)`B. `(T)/(4)`C. `(T)/(3)`D. `(T)/(12)` |
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Answer» Correct Answer - A In equation `x = Acos omega t`, putting `x = (A)/(2)` we get `omega t = (pi)/(3)` `:. ((2pi)/(T))t = (pi)/(3)` or `t = (T)/(6)` |
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| 326. |
A simple harmonic oscillation has an amplitude `A` and time period `T`. The time required to travel from `x = A` to ` x= (A)/(2)` isA. T / 6B. T / 4C. T / 3D. T / 2 |
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Answer» Correct Answer - A |
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| 327. |
The diagram shows two oscillations. What is the phase difference betweenthe oscillations? A. `(pi)/(5) rad`B. `(2pi)/(5)rad`C. `(3pi)/(5)rad`D. `(4pi)/(5)rad` |
| Answer» Correct Answer - B | |
| 328. |
An object of mass `m` is attached to a spring. The restroing force of the spring is `F - lambdax^(3)`, where `x` is the displacement. The oscillation period depends on the mass, `l`mabd and oscillation amplitude. Suppose the object is initially at rest. If the initial displacement is `D` then its period is `tau`. If the initial displacement is `2D`, find the period.A. `8tau`B. `2tau`C. `tau`D. `tau//2` |
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Answer» Correct Answer - D Through dimensional analysis `[lambda] = [ML^(-2)T^(-2)]` `T prop m^(a)lambda^(b)A^(c)` `[T] = [M]^(a) [ML^(-2)T^(-2)]^(b)[L]^(c)` `a = 1/2, b = (-1)/(2), c = -1` `T = prop (1)/(A)` |
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| 329. |
A block of mass 0.2 kg is attached to a mass less spring of force constant 80 N/m as shown in figure. Find the period of oscillation. Take `g=10 m//s^(2)`. Neglect friction A. `pi` sB. `(pi)/(10)` sC. `(2pi)/(5)` sD. `(pi)/(2)` s |
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Answer» Correct Answer - B (b)`T=2pisqrt((m)/(k))=2pisqrt((0.2)/(80))=(pi)/(10)s` |
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| 330. |
A block of mass 0.2 kg is attached to a mass less spring of force constant 80 N/m as shown in figure. Find the period of oscillation. Take `g=10 m//s^(2)`. Neglect friction |
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Answer» Correct Answer - A::C::D `T = 2pi sqrt((m)/(k)) = 2pi sqrt ((0.2)/(80)) = (pi)/(10)s` ` = 0.314s` |
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| 331. |
A small ball of mass m is attached to the middle of a tightly stretched perfectly flexible wire AB of length 2 l (figure ). The ball is given a small lateral displacement in horizontal direction and released. The initial tension (T) in the wire is high and change in it due to small lateral displacement of the ball can be neglected. Prove that the ball will perform simple harmonic motion, and calculate the period. If there is a device which can change the tension in the wire at will , how will the time period change if tension in the wire is increased |
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Answer» Correct Answer - `2pisqrt((ml)/(2T))`, " Time period increases". |
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| 332. |
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 maximum value of `V_0` for whic the disk will roll without slipping is-A. `mug_(1)sqrt((M)/(k))`B. `mugsqrt((M)/(2k))`C. `mugsqrt((3M)/(k))`D. `mugsqrt((5M)/(2k))` |
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Answer» Correct Answer - C Using temperature conservation law `1/2Mv_(0^(2)) (1+1/2) = 2 xx (1)/(2)kx_(1^(2))` `2kx_(1) - f_("max") = Ma`& `f_(max) R = ((MR^(2))/(2))alpha` But `f_(max) = muMg` `rArr x_(1) = (3muMg)/(2K)` `rArr 3/4Mv_(0)^(2) = Kx_(1^(2)) = (1)/(K) ((9mu^(2)M^(2)g^(2))/(4))` `rArr v_(0) = mugsqrt((3M)/(K))` |
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| 333. |
A toy car of mass `m` is having two similar rubber ribbons attached to it as shown in the figure. The force constant of each rubber ribbon is `k` and surface is frictionless. The car is displaced from mean position by `x cm` and released. At the mean position the ribbons are undeformed. Vibration period is A. `2pisqrt((m(2k))/(k^(2))`B. `(1)/(2)pisqrt((m(2k))/(k^(2))`C. `2pisqrt((m)/(k^(2))`D. `2pisqrt((m)/(k+k))` |
| Answer» Correct Answer - C | |
| 334. |
A particle executing simple harmonic motion has angular frequency `6.28 s^(-1)` and amplitude `10 cm`. Find `(a)` the time period, `(b)` the maximum speed, `(c)` the maximum acceleration, `(d)` the speed when the displacement is `6 cm` from the mean position, `(e)` the speed at `t = 1//6 s` assuming that the motion starts from rest at `t = 0`. |
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Answer» a. Time period `=(2pi)/6.28s=1s` b. Maimum speed `=Aomega=(0.1m)(6.28m^-1)` `=0.628ms^-1` c. Maximum acceleration `=Aomega^2` ltbr.gt `=(0.1m)(6.28s^-1)^2` =4ms^-2` d. `v=omegasqrt(A^2-x^2)=(6.28s^-1)sqrt((10cm)^2-(6cm)^2)` `=50.2cms^-1` c. At t=0 the velocity is zero i.e. the particle is at an extreme. The equation for displacement may be written as `x=Acosomegat` The velocity is ` v=-Aomegasinomegat`. At `t=1/6s, v=-(0.1m)(6.28s^-1)sin(6.28/6)` `=(-0.628ms^-1)sinpi/3` `=-54.4cms^-1` |
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| 335. |
A spherical ball of mass m and radius r rolls without slipping on a rough concave surface of large radius R. It makes small oscillations about the lowest point. Find the time period. |
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Answer» Correct Answer - B Let the angular velocity of the system about the point is suspension at any time be `omega` `Sve=(R-r)omega` `Again ve=romega` `[where omega_1=`rotatioN/Al velocity of the sphere] `1-v_c/r=((R-r)/r)omega` ………..1 By energy method total energy in SHM is constant. `So, `mg(R-r)(1-costgheta)+/2mv_c^2+1/2Iomega^2` =constant `:. mg(R-r))1-costheta)+1/2m(R-r)^2omega^2`=constant `:. mg(R-r)(1-costheta)+1/2m(R-r)^2omega^2+1/2mr^2((R-r)/r)omega^2=constant`:.mg(R-r)(1-costheta)+1/2m(R-r)^2omega^2+1/2mr^2((R-r)/r)omega^2` =constant `rarr g(R-r)(1-costheta)+(R-r)omega^I2[1/2+1/2]` =costant Taking derivative `g(R-or)sintheta(dtheta)/(dt)=7/10 (R-r)^22omega (domega)/(dt)` `rarr gsintheta=2xx(7/10)(R-r)alpha` `implies gsin theta=(7/5)(R-r)alpha` `alpha=(5gsintheta)/(7(R-r))` `:. alpha/theta=omega^2` `=(5g)/(7(R-r))=constant So the motion is SHM again `omega=sqrt((5g)/(7(R-r)))` `rarr T=2pi sqrt((7(R-r))/(5g))` |
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| 336. |
A block of mass M connected to an ideal spring of force constant k, is placed on a smooth surface. The block is pushed to the left so as to compress the spring by a length A and then it is released. The block hits an elastic wall at a distance `(3A)/(2)` from its point of release. Assume the collision to be instantaneous. (a) Calculate the time required by the block to complete one oscillation (b) Draw the velocity - time graph for one oscillation of the block. (c) Find the value of k for which average force experienced by the wall due to repeated hitting of the block is `F_(0)`. |
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Answer» Correct Answer - (a)`(4pi)/(3)sqrt((M)/(k))`, (c ) `k=(4pi F_(0))/(3sqrt(3)A)` |
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| 337. |
Two ideal springs of same make (the springs differ in their lengths only) have been suspended from points A and B such that their free ends C and D are at same horizontal level. A massless rod PQ is attached to the ends of the springs. A block of mass m is attached to the rod at point R. The rod remains horizontal in equilibrium. Now the block is pulled down and released. It performs vertical oscillations with time period `T =2 pi sqrt((m)/(3k))` where k is the force constant of the longer spring. (a) Find the ratio of length RC and RD. (b) Find the difference in heights of point A and B if it is given that natural length of spring BD is L. |
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Answer» Correct Answer - `(a) (RC)/(RD)=(2)/(1) , (b) L`. |
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| 338. |
An object is attched to the bottom of a light vertical spring and set vibrating. The maximum speed of the object is `15cm//s` and the period is `628` milli seconds. The amplitude of the motion in centimetres isA. 3B. 2C. 1.5D. 1 |
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Answer» Correct Answer - C |
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| 339. |
If a body of mass 0.98 kg is made to oscillate on a spring of force constant 4.84 N/m the angular frequency of the body isA. 1.22 rad/sB. 2.22 rad/sC. 3.22 rad/sD. 4.22 rad/s |
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Answer» Correct Answer - B |
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| 340. |
When a mass m is attached to a spring, it normally extends by 0.2 m . The mass m is given a slight addition extension and released, then its time period will beA. `1/7 sec`B. 1 secC. `(2pi)/7 sec`D. `2/(3pi) sec` |
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Answer» Correct Answer - C |
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| 341. |
A spring mass system preforms `S.H.M` if the mass is doubled keeping amplitude same, then the total energy of `S.H.M` will become :A. doubleB. halfC. unchangedD. `4` times |
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Answer» Correct Answer - C |
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| 342. |
A mass at the end of a spring executes harmonic motion about an equilibrium position with an amplitude A.Its speed as it passes through the equilibrium position is V.If extended 2A and released, the speed of the mass passing through the equilibrium position will beA. `2V`B. `4V`C. `(V)/(2)`D. `(V)/(4)` |
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Answer» Correct Answer - A |
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| 343. |
A particle moves under the force `F(x) = (x^(2) - 6x)N`, where `x` is in metres. For small displacements from the origin what is the force constant in the simple harmonic motion approximation ? |
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Answer» For small values of `x`, the term `x^(2)` can be neglected. `:. F = -6x` Comparing with `F = -kx`, we get `k = 6N//m` |
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| 344. |
An alpha nucleus of energy `(1)/(2)m nu^(2)` bombards a heavy nucleus of charge `Ze` . Then the distance of closed approach for the alpha nucleus will be proportional toA. `v^(2)`B. `1//m`C. `1//v^(4)`D. `1//Ze` |
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Answer» Correct Answer - B Charge on `alpha`-particle `= 2e` Change on target nucles = Ze When the `alpha`-particle is projected towards the target nucleus, then at `r = r_(0)`, the `alpha`-particle comes to momentary rest. This position `r_(0)` from target nucleus is known as distance of closest approach. Applying law of conservation of energy, we get `(1)/(2) mv^(2) = (K (ze)(2e))/(r_(0)) rArr r_(0) prop (1)/(v^(2)), r_(0) Ze, r_(0) prop (1)/(m)` |
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| 345. |
A particle is subjected to two simple harmonic motions, one along the X-axis and the other on a line making an angle of `45^@` with the X-axis. The two motions are given by `x=x_0 sinomegat` and `s=s_0 sin omegat`. find the amplitude of the resultant motion. |
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Answer» Correct Answer - A::B The particle is subjected to two simple harmonic motions represented by `x=x_0sin2t` `s=s_0sinwt` and angle between two motion `theta=45^@` Resultant motions will be given by `R=sqrt(x^2+s^2+2xs.cos45^@)` `=sqrt({x_0^2sin^2 wt+s_0^2sin^2wt+2x_0s_0sin^2omegat)1/sqrt2)})` `=[x_0^2+s_0^2+sqrt(2x_0s_0)]^(1/2) sin omegat` Resultant amplitude `[x_0^2+s_0^2+sqrt(2x_0s_0)]^(1/2)` |
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| 346. |
A ring of radius `r` is suspended from a point on its circumference. Determine its angular frequency of small oscillations. |
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Answer» Correct Answer - B It is a physical pendulum, the time period of which is, `T = 2pi sqrt((I)/(mgl))` Here, `I` = moment of inertia of the ring about point of suspension ` = mr^(2) + mr^(2) = 2mr^(2)` and `l =` distance of point of suspension from center of gravity `= r` `:. T = 2pi sqrt ((2mr^(2))/(mgr))` `= 2pi sqrt ((2 r)/(g))` `:.` Angular frequency `omega = (2pi)/(T)` or `omega = sqrt ((g)/(2r))`. |
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| 347. |
Find the period of oscillation of a pendulum of length `l` if its point of suspension is (a) moving vertically up with acceleration `a` (b) moving vertically down with acceleration `a(lt g)` (c) failing freely under gravity moving horizontal with acceleration `a`. |
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Answer» Correct Answer - A::B::C::D (a) `g_(e) = g + a` (b) `g_(e) = g - a` ( c) `g_(e) = 0` (d) `g_(e) = sqrt(g^(2) + a^(2))` |
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| 348. |
Two solid spheres A and B of equal volumes but of different densities `d_A and d_B` are connected by a string. They are fully immersed in a fluid of density `d_F`. They get arranged into an equilibrium state as shown in the figure with a tension in the string. The arrangement is possible only if A. `d_(A)ltd_(F)`B. `d_(B)gtd_(F)`C. `d_(A)gtd_(F)`D. `d_(A)+d_(B)=2d_(F)` |
| Answer» Correct Answer - A::B::D | |
| 349. |
Two linear simple harmonic motions of equal amplitude and frequency are impressed on a particle along x and y axis respectively. The initial phase difference between them is `pi/2`. The resultant path followed by the particle isA. a circleB. a straight lineC. an ellipseD. a parabola |
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Answer» Correct Answer - A |
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| 350. |
One end of a spring of force constant `k` is fixed to a vertical wall and the other to a block of mass `m` resting on a smooth horizontal surface There is another and wall at a distance `x_(0)` from the block The spring is then compressed by `2x_(0)` and released The time taken to strike the wall is A. `1/6 pi sqrt(k/m)`B. `sqrt(k/m)`C. `(2pi)/6 sqrt(m/k)`D. `(pi)/4 sqrt(k/m)` |
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Answer» Correct Answer - C |
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