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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.
| 201. |
Assertion : The amplitude of an oscillation pendulum decreases gradually with time Reason : The frequency of the pendulum decrease with timeA. 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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| 202. |
A linear harmonic oscillator of force constant `2 xx 10^(6)N//m` and amplitude `0.01 m` has a total mechanical energy of `160 J`. ItsA. maximum potential energy is 160 JB. maximum potential energy is 100 JC. minimum potential energy is zeroD. minimum potential energy is 100 J |
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Answer» Correct Answer - A (a)`(1)/(2)kA^(2)=(1)/(2)xx2xx10^(6)xx(0.01)^(2)=100 J` At mean position , U=60 J and K=100 J At extreme positions , U=160 J and K=0 |
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| 203. |
A particle is subjected to two simple harmonic motions. `x_(1) = 4.0 sin (100pi t)` and `x_(2) = 3.0 sin(100pi t + (pi)/(3))` Find (a) the displacement at `t = 0` (b) the maximum speed of the particle and (c ) the maximum acceleration of the particle. |
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Answer» (a) At `t = 0, x_(1) = A_(1) sin omegat = 0` and `x_(2) = A_(2) sin (omegat + pi//3)` `= A_(2) sin (pi//3) = (A_(2)sqrt(3))/(2)` Thus, the resultant displacement at `t = 0` is `x = x_(1) + x_(2) = A_(2) sqrt(3)/(2)` (b) The resultant of the two motions is a simple harmonic motion of the same angular frequency `omega`. The amplitude of the resultant motion is `A = sqrt(A_(1)^(2) + A_(2)^(2) + 2A_(1)A_(2) cos(pi//3)), = sqrt(A_(1)^(2) + A_(2)^(2) + A_(1)A_(2))`. The maximum speed is `u_("mass") = Aomega = omegasqrt(A_(1)^(2) + A_(2)^(2) + A_(1)A_(2))` (c) The maximum acceleration is `a_(max) = Aomega^(2) = omega^(2) sqrt(A_(1)^(2) + A_(2)^(2) + A_(1)^(2))`. |
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| 204. |
Periodic Motion |
| Answer» Correct Answer - A motion which repeats itself over and over again after a regular interval of time is called a periodic motion. | |
| 205. |
A simple harmonic oscillator has a period of 0.01 sec and an amplitude of 0.2 m . The magnitude of the velocity in `m sec^(-1)` at the centre of oscillation isA. `20 pi`B. 100C. `40 pi`D. `100 pi` |
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Answer» Correct Answer - C |
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| 206. |
What is periodic motion ? |
| Answer» Correct Answer - A motion which repeats itself over and over again after a regular interval of time is called a periodic motion. | |
| 207. |
A child swinging on a swing in sitting position, stands up, then the time period of the swing will.A. IncreaseB. DecreaseC. Remain sameD. Increase if the child is long and decrease if the child is short |
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Answer» Correct Answer - B |
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| 208. |
A girl is swinging on a swing in the sitting position. How will the period of swing be affected if she stands up?A. The period will now be shorterB. The period will now longerC. The period will remain unchangedD. The period may become longer or shorter depending upon the height of girl |
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Answer» Correct Answer - A `TpropsqrtL` When girl stands up, L decreases |
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| 209. |
A particle executes S.H.M. with a period of 6 second and amplitude of 3 cm . Its maximum speed in cm / sec isA. `pi//2`B. `pi`C. `2pi`D. `3 pi` |
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Answer» Correct Answer - B |
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| 210. |
When a boy is playing on a swing in the sitting position , the time period of oscillations of the swing is T. If the boy stands up , the time period of oscillation of the spring will beA. more than TB. less than TC. equal to TD. Cannot be predicted |
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Answer» Correct Answer - B (b) As the boy stands up , the gravity of the pendulum is raised up , decreasing the effective length of the pendulum of the swing and hence the time period T decreases. |
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| 211. |
The periodic time of a simple pendulum of length 1 m and amplitude 2 cm is 5 seconds. If the amplitude is made 4 cm , its periodic time in seconds will beA. 2.5B. 5C. 10D. `5sqrt2` |
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Answer» Correct Answer - B |
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| 212. |
Two pendulums begin to swing simultaneously. If the ratio of the frequency of oscillations of the two is 7 : 8, then the ratio of lengths of the two pendulums will beA. `7:8`B. `8:7`C. `49:64`D. `64:49` |
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Answer» Correct Answer - D |
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| 213. |
A particle is executing simple harmonic motion with a period of T seconds and amplitude a metre . The shortest time it takes to reach a point `a/sqrt2` from its mean position in seconds isA. TB. T /4C. T/8D. T/16 |
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Answer» Correct Answer - C |
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| 214. |
particle is executing S.H.M. If its amplitude is 2 m and periodic time 2 seconds , then the maximum velocity of the particle will beA. `pi m//s`B. `sqrt2pi m//s`C. ` 2pi m//s`D. ` 4 pi m//s` |
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Answer» Correct Answer - C |
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| 215. |
Two pendulums differ in lengths by `22m`. They oscillate at the same place so that one of then makes 30 oscillations and the other makes 36 oscillations during the same time. The length `(` in `cm)` of the pendulum are `:` |
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Answer» `30xx2pisqrt((l_(1))/(g))=36xx2pisqrt((l_(2))/(g)) "or " (l_(1))/(l_(2))=(36xx36)/(30xx30)=(36)/(25)` Also, `l_(1)-l_(2)=22 cm (l_(1))/(l_(2))-1=(22)/(l_(2))` or `(22)/(l_(2))=(26)/(25)-1 " or "(l_(2))/(22)=(25)/(11) " or "l_(2)=50 cm,` `l_(1)=22+l_(2)=(22+50)cm=72 cm` |
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| 216. |
STATEMENT-1 : When two `SHMs` in the same direction with amplitude `A_(1)` and `A_(2)` are superimposed, the resultant amplitude will be between `|A_(1)-A_(2)|` and `(A_(1)+A_(2))` is `0ledeltaleI,delta` is phase difference between two `SHM`. STATEMENT-2 : At any instant energy will be conserved of system undergoing `SHM`.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 - A |
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| 217. |
Two particles execute SHMs of the same amplitude and frequency along the same straight line. They cross one another when going in opposite direction. What is the phase difference between them when their displacements are half of their amplitudes ?A. `0^(@)`B. `120^(@)`C. `180^(@)`D. `135^(@)` |
| Answer» Correct Answer - B | |
| 218. |
A particle of mass 10grams is executing simple harmonic motion with an amplitude of 0.5 m and periodic time of )`(pi //5)`seconds . The maximum value of the force acting on the particleA. 25 NB. 5 NC. 2.5 ND. 0.5 N |
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Answer» Correct Answer - D |
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| 219. |
A square plate of mass M and side length L is hinged at one of its vertex (A) and is free to rotate about it. Find the time period of small oscillations if (a) the plate performs oscillations in the vertical plane of the figure. (Axis is perpendicular to figure.) (b) the plate performs oscillations about a horizontal axis passing through A lying in the plane of the figure. |
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Answer» Correct Answer - (a) ` 2 pi sqrt((2sqrt(2))/(3)) (a)/(g)`, (b) ` 2pi sqrt((7)/(6sqrt(2))(a)/(g))` |
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| 220. |
Two particles A and B are describing SHM of same amplitude (a) and same frequency (f) along a common straight line. The mean positions of the two SHMs are also same but the particles have a constant phase difference between them. It is observed that during the course of motion the separation between A and B is always less than or equal to a. (a) Find the phase difference between the particles. (b) If distance between the two particles is plotted with time, with what frequency will the graph oscillate? |
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Answer» Correct Answer - `(a) phi_(1)-phi_(2)=(2pi)/(3) (b) 2f` |
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| 221. |
A body of mass 1 is executing simple harmonic motion. Its displacement y(cm) at t seconds is given by `y = 6 sin (100t + pi//4)` . Its maximum kinetic energy isA. 6 JB. 18 JC. 24 JD. 36 J |
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Answer» Correct Answer - B a=6 cm , `omega=100 "rad" s^(-1)` `K_("max")=(1)/(2)momega^(2)a^(2)=(1)/(2)xx 1 xx (100)^(2)xx (6xx 10^(-2))^(2)=18 J` |
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| 222. |
The displacement of a particle varies according to the relation `X=3 sin 100t+8 cos^(2)50t`. Which of the following `is//are` correct about this motion.A. the motion of the particle is not `S,H.M`.B. the amplitude of the `S.H.M.` of the particle is `5` unitsC. the amplitude of the resulting `S.H.M.` si `sqrt(73)` unitsD. the maximum displacement of the particle from the origin is `9` units. |
| Answer» Correct Answer - B::D | |
| 223. |
A simple pendulum oscillating with a small amplitude has a time period of `T = 1.0s` . A horizontal thin rod is now placed beneath the point of suspension at a distance equal to half the length of the pendulum. The string collides with the rod once in each oscillation and there is no loss of energy in such collisions. Find the new time period T of the pendulum |
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Answer» Correct Answer - `((sqrt(2)+1)/(2sqrt(2)))T` |
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| 224. |
A particle executes simple harmonic motion between `x = -A and x = + A`. The time taken for it to go from `0 to A//2 is T_1 and to go from A//2 to (A) is (T_2)`. Then.A. `T_(1)lt T_(2)`B. `T_(1)gtT_(2)`C. `T_(1)=T_(2)`D. `T_(1)=2T_(2)` |
| Answer» Correct Answer - A | |
| 225. |
A uniform stick of length `l` is mounted so as to rotate about a horizontal axis perpendicular to the stick and at a distance `d` from the centre of mass. The time period of small oscillation has a minimum value when `d//l` isA. `(1)/(sqrt(2)`B. `(1)/sqrt(12)`C. `(1)/sqrt(3)`D. `(1)/sqrt (6)` |
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Answer» Correct Answer - B `T = 2pi sqrt((I)/(mgd))` `= 2pi sqrt((ml^(2))/(12) + md^(2))/(mgd)` `= 2pi sqrt ((l^(2) + 12d^(2))/(12d))` `= 2pi sqrt (((l^(2))/(12d) + d))` `T` is minimum when first derivation of the quantity inside the root with respect to `d` is zero. `- (l^(2))/(12d^(2)) + 1 = 0` or `(d)/(l) = (1)/sqrt (12)` |
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| 226. |
A solid cube of side a and density `rho_(0)` floats on the surface of a liquid of density `rho`. If the cube is slightly pushed downward, then it oscillates simple harmonically with a period ofA. `2pisqrt((ld)/(rhog))`B. `2pisqrt((lrho)/(dg))`C. `2pisqrt((ld)/((rho-d)g))`D. `2pisqrt((lrho)/((rho-d)g))` |
| Answer» Correct Answer - A | |
| 227. |
A solid cube of side a and density `rho_(0)` floats on the surface of a liquid of density `rho`. If the cube is slightly pushed downward, then it oscillates simple harmonically with a period ofA. `2pisqrt((rho_(0))/(rho)(a)/(g))`B. `2pi sqrt((rho)/(rho_(0))(a)/(g))`C. `2pi sqrt ((a)/((1 - (rho)/(rho_(0)))g`D. `2pi sqrt ((a)/((1 + (rho)/(rho_(0)))g` |
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Answer» Correct Answer - A `T = 2pi sqrt((m)/(k))` Here `m = (rho_(0))(a^(3))` and `K = rho_(l) A g = rhoa^(2)g` `:. T = 2pi sqrt ((rho_(0)a)/(rhog))` . |
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| 228. |
A solid cube floats in water half immersed and h small vertical oscillations of time period `pi/5`s. Find the mass (in kg) (Take g=10m/`s^(2)`.A. 4B. 2C. 1D. 0.5 |
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Answer» Correct Answer - A (a) Since, the cube is half immersed. The density of cube should be half the density of water, i.e., `500 kgm^(-3)` `T=2pisqrt((m)/(rho_(u)Ag)) " "(k=rho_(u)Ag)` `(pi)/(5)=2pisqrt((a^(3)xx rho)/(rho_(w)xx a^(2) xx g))` `therefore (arho)/(grho_(w))=(1)/(100)` `therefore (a)/(20)=(1)/(100)` or a=(0.2)m Now, `m=a^(3)rho=4 kg" "`(`because` Mass =Density `xx` Volume) |
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| 229. |
In SHM , potential energy of a particle at mean position is `E_(1)` and kinetic enregy is `E_(2)` , thenA. `E_(1)=E_(2)`B. total potential energy at `x=(sqrt(3)A)/(2)` is `E_(1)+(3E_(2))/(4)`C. total kinetic energy at `x=(sqrt(3)A)/(2)` is `(3E_(2))/(4)`D. total kinetic energy at `x=(A)/(sqrt(2))` is `(E_(2))/(4)` |
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Answer» Correct Answer - B (b) `U=U_("mean")+(1)/(2)kX^(2)` Given , `U_("mean")=E_(1)` and `(1)/(2)kA^(2)=E_(2)` = maximum kinetic energy at mean position `therefore X=(sqrt(3)A)/(2), U=E_(1)+(1)/(2)k((sqrt(3)A)/(2))^(2)=E_(1)+(3)/(4)E_(2)` |
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| 230. |
The damping force on an oscillator is directly proportional to the velocity. The units of the constant to proportionality areA. `kgms^(-1)`B. `kgms^(-2)`C. `kgs^(-1)`D. `kgs` |
| Answer» Correct Answer - C | |
| 231. |
A car is driven at constant speed on a rough road. The shock absorbers vibrate vertically. As we know, if the displacement varies sinusoidally, the oscillation will be a simple harmonic. When there is matching between the disturbance and the oscillating shockers, we call it as resonance. Consider the mass of the car and its contents `M=1500 kg`with a man of mass `m=75 kg`, rising vertically `S=0.1 m` due to the peaks separated by `lambda=10 m` equally and the speed of the car being `v` and time period being `T`.Then: The frequency of oscillation of the car is proportional to:A. `sqrt((mg)/(MS))`B. `sqrt((mg)/(d))`C. `sqrt((3mg)/(2MS))`D. `sqrt((mg)/(Md))` |
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Answer» Correct Answer - A |
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| 232. |
A car is driven at constant speed on a rough road. The shock absorbers vibrate vertically. As we know, if the displacement varies sinusoidally, the oscillation will be a simple harmonic. When there is matching between the disturbance and the oscillating shockers, we call it as resonance. Consider the mass of the car and its contents `M=1500 kg`with a man of mass `m=75 kg`, rising vertically `S=0.1 m` due to the peaks separated by `lambda=10 m` equally and the speed of the car being `v` and time period being `T`.Then: Critical speed at maximum displacement is (in `ms^(-1)`):A. `2.56`B. `3.56`C. `5.26`D. `3.76` |
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Answer» Correct Answer - B |
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| 233. |
In the figure shown, a block A of mass m is rigidy attached to a light spring of stiffness k and suspended from a fixed support. Another block B of same mass is just placed on it and blocks are in equilibrium. Suddenly the block B is removed. Choose the correct options (s) afterward. A. Block A will start SHMB. Amplitude of oscillation of the block A is `(mg)/k)`C. Maximum speed acquired by the block A is `sqrt(m/(2k))`D. |
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Answer» Correct Answer - A::B::C |
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| 234. |
System shown in fig is in equilibrium and at rest. The spring and string are massless now the string is cut. The acceleration of mass 2m and m just after the string is cut will be: A. `1/6`sB. `1/2`sC. `1/3`sD. `2/3`s |
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Answer» Correct Answer - A::B |
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| 235. |
A hydrogen - like atom (described by the Bohr model) is observed to emit six wavelength , originating from all possible transitions between `- 0.85 eV and -0.544 eV` (including both these values ) (a) Find the atomic number of the atom (b) Calculate the smallest wavelength emitted in these transitions . (Take `hc = 1240 eV - nm`, ground state energy of hydrogen atom `= 13.6 eV)` |
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Answer» Correct Answer - (i) `Z = 3`, (ii) `4052.3 nm` (i) total 6 lines are emitted. Therefore `(n(n-1))/(2)=6 implies n=4` So, transition is taking place between `m^(th)` energy state and `(m+3)^(th)` energy state. `E_(m)= -0.85 eV` `implies -13.6 (Z^(2)/m^(2))= -0.85" "implies Z/m=0.25` …(i) Similarly `E_(m+3)= -0.544 eV` `implies -13.6 z^(2)/((m+3)^(2))= -0.544 implies z/((m+3))=0.2` ...(ii) Solving equation (i) and (ii) for z and m. We get `m=12` and `z=3` (ii) Smallest wavelength corresponds to maximum difference of energies which is obviously `E_(m+3)-E_(m)`. `:. DeltaE_("max")= -0.544-(-0.85)=0.306 eV` `:. lambda_("min")=(hc)/(DeltaE_("max"))=1240/0.306=4052.3 nm`. |
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| 236. |
The amplitude of a particle executing SHM about `O` is `10 cm`. ThenA. When the `K.E.` is `0.64` of its max. `K.E.` its displacement is `6cm` from `O`.B. When the displacement is `5 cm` from `O` its `K.E` is its max. `P.E.`C. Its total energy at any point is equal to its maximum `K.E.`D. Its velocity is half the maximumum velocity when its displacement is half the maximum displacement. |
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Answer» Correct Answer - A::B::C |
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| 237. |
A 100 g block is connected to a horizontal massless spring of force constant `25.6(N)/(m)` As shown in Fig. the block is free to oscillate on a horizontal frictionless surface. The block is displaced 3 cm from the equilibrium position and , at `t=0`, it is released from rest at `x=0` It executes simple harmonic motion with the postive x-direction indecated in Fig. The position time `(x-t)` graph of motion of the block is as shown in Fig. Q. When the block is at position B on the graph its.A. position and velocity both are negativeB. position and velocity both are positive.C. position is negative and velocity is positive.D. position is positive and velocity is negative. |
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Answer» Correct Answer - D At position B, x is positive and `(dx)/(dt)` is negative. |
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| 238. |
The period of oscillation of a simple pendulum of length (L) suspended from the roof of a vehicle which moves without friction down an inclined plane of inclination (prop), is given by.A. `2pisqrt((L)/(g cos alpha))`B. `2pisqrt((L)/(g sin alpha))`C. `2pisqrt((L)/(g))`D. `2pisqrt((L)/(g tan alpha))` |
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Answer» Correct Answer - A |
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| 239. |
A simple pendulum with a bob of mass ‘m’ oscillates from A to C and back to A such that PB is H . If the acceleration due to gravity is ‘g’ then the velcoity of the bob as it passes through B is A. mgHB. `sqrt(2gH)`C. 2ghD. zero |
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Answer» Correct Answer - B |
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| 240. |
The matallic bob of a simple pendulum has the relative density `rho`. The time period of this pendulum is `T` it the metallic bob is immersed in water the new time period is given byA. `T (rho-1)/ rho`B. `T rho/(rho-1) `C. `T sqrt((rho-1)/ rho)`D. `T sqrt(rho/(rho-1)) ` |
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Answer» Correct Answer - D |
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| 241. |
The period of a simple pendulum whose bob is hollow metallic sphere is `T`. The period is `T_(1)` when the bob is filled with sand, `T_(2)` where it is filled with mercury and `T_(3)` when it is half filled with mercury Which of the following is true? A. `T = T_(1) = T_(2)gtT_(3)`B. `T_(1) = T_(1) = T_(3)gtT`C. `Tgt TgtT_(1) = T_(2)`D. `T = T_(1) = T_(2)ltT_(3)` |
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Answer» Correct Answer - D |
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| 242. |
A pendulum has time period `T` in air when it is made to oscillate in water it acquired a time period `T = sqrt(2)T` The density of the pendulum bob is equal to (density) of water `= 1)`A. `sqrt2`B. 2C. `2sqrt2`D. none of these |
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Answer» Correct Answer - B |
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| 243. |
Assertion : When a simple pendulum is made to oscillate on the surface of moon , its time period increase Reason: Moon is much smaller compared to earthA. 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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| 244. |
A simple pendulum is vibrating in an evacuated chamber, it will oscillate withA. Increasing amplitudeB. Constant amplitudeC. Decreasing amplitudeD. First (c) then (a) |
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Answer» Correct Answer - B |
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| 245. |
The time period of a simple pendulum of length L as measured in an elevator descending with acceleration g / 3 isA. `2 pi sqrt((3L)/g)`B. `pi sqrt((3L)/(g)`C. `2 pi sqrt(((3L)/(2g)))`D. `2 pi sqrt((2L)/(3g)` |
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Answer» Correct Answer - C |
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| 246. |
An oscillation is superposition of three harmonic oscillations and decribed by the equation `x=A sin 2pi upsilon_(1)` where A changes with time according to `A=A_(0)(1+cos 2pi upsilon_(2)t)` with `A_0` to be constant. The frequencies of pure harmonic oscillations forming this oscillation areA. `gamma_(1):gamma_(2):(gamma_(1)-gamma_(2))`B. `gamma_(1):(gamma_(1)-gamma_(2)):(gamma_(1) + gamma_(2))`C. `gamma_(1):gamma_(2):(gamma_(2)-gamma_(1))`D. `gamma_(1):gamma_(2):(gamma_(1)+gamma_(2))` |
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Answer» Correct Answer - B `x = A_(0)(1+cos2pigamma_(2)t) sin (pigamma_(1))` `= A_(0)[sin 2pigamma_(1)t + cos 2pigamma_(2)t + sin 2pigamma_(1)t]` `= A_(0)[sin 2pigamma_(1)t + 1/2 sin (2pi(gamma_(1) + gamma_(2))t - 1/2 sin (2pi(gamma_(1) - gamma_(2)t)]` Required ratio `= v_(1) : (v_(1) - v_(2)) : (v_(1) + v_(2))` |
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| 247. |
The x-t graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the particle of `t=4//3s` is A. `(sqrt3)/(32)pi^2cm//s^2`B. `(-pi^2)/(32)cm//s^2`C. `(pi^2)/(32)cm//s^2`D. `-(sqrt3)/(32)pi^2cm//s^2` |
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Answer» Correct Answer - D Let `x=Asin(omegat+phi)` At `t=0`, `x=0`, `phi=0` `T=8s`, `omega=(2pi)/(T)=(2pi)/(8)=pi/4 rad//s`, `a=1cm` `x=A sin omegat=1xxsin((pi)/(4))t` At `t=4/3s`, `x=sin((pi)/(4))*4/3=sqrt3/2` `a=-omega^2x=-(pi/4)^2sqrt3/2=-(sqrt3pi^2)/(32)cm//s^2` |
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| 248. |
The (x - t) graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the particle at `t = 4//3 s` is .A. `(sqrt3)/(32)pi^(2)cm//s^(2)`B. `(-pi^(2))/(32)cm//s^(2)`C. `(pi^(2))/(32)cm//s^(2)`D. `-(sqrt3)/(32) cm//s^(2)` |
| Answer» Correct Answer - D | |
| 249. |
A block is kept on a horizontal table. The stable is undergoing simple harmonic motion of frequency `3Hz` in a horizontal plane. The coefficient of static friciton between block and the table surface is `0.72.` find the maximum amplitude of the table at which the block does not slip on the surface. |
| Answer» Correct Answer - `2 cm` | |
| 250. |
The `x-t` graph of a particle undergoing simple harmonic motion is shown in figure. Acceleration of particle at `t = 4//3 s` is A. `(sqrt(3))/(32)pi^(2)cm//s^(2)`B. `(-pi^(2))/(32)cm//s^(2)`C. `(pi^(2))/(32)cm//s^(2)`D. `-(sqrt(3))/(32)pi^(2)cm//s^(2)` |
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Answer» Correct Answer - D From graph `T = 8` second, `A = 1 cm, x = A sinomegat. = 1sin(2pi)/(8)t`. `a = -omega^(2)x = -((2pi)/(8))^(2) sin((2pi)/(8))t cm//s^(2)` At, `t = (4)/(3)` second, `a = -((2pi)/(8))^(2) sin"(pi)/(3) = - (sqrt(3) pi^(2))/(32) cm//s^(2)` |
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