

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.
1. |
Bats can ascertain distances, directions; nature and size of the obstacle without any eyes, explain how ? |
Answer» Bats emit ultrasonic waves of very small wavelength (high frequencies) and so high speed. The reflected waves from an obstacle in their path give them idea about the distance, direction, nature and size of the obstacle. |
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2. |
Bats can ascertain different directions, nature and size of obstacles without any eyes. Explain how? |
Answer» Bats emit ultrasonic waves of large frequencies. When these wave are reflected from the obstacles in their path, they give them the idea about the distance, direction, size and nature of the obstacle. |
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3. |
The maximum tension in the string of a simple pendulum is 1.2 times the minimum tenstion. If `0_(0)` is the angular amplitude, then `0_(0)` isA. `(1)/(2)`B. `(3)/(5)`C. `(3)/(4)`D. `(2)/(3)` |
Answer» Correct Answer - B `(l_(2))/(l_(1))=(g_(2))/(g_(1))=(1)/(6)` `l_(2)=(1)/(6)m` |
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4. |
A block of mass `M` is placed on a smooth table. Its two sides are attached to the fixed walls by means of collinear horizontal springs of spring constants `K_(1)` and `K_(2)` respectively `(K_(1) gt K_(2))` as shown in the figure. The block is made to oscillate horizontally along the line of two springs. The frequency of its oscillation is A. `(1)/(2pi) sqrt(((K_(1)K_(2))/((K_(1)+K_(2))M)))`B. `(1)/(2pi) sqrt(((M)/(K_(1)+K_(2))))`C. `(1)/(2pi) sqrt(((K_(1)+K_(2))/(M)))`D. `(1)/(2pi) sqrt(((K_(1)-K_(2))/(M)))` |
Answer» Correct Answer - C The springs are parallel |
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5. |
If an explosion takes place at the bottom of lake or sea, will the shock waves in water be longitudinal or transverse ? |
Answer» Explosion at the bottom of lake or sea create enormous increase in pressure of medium (water). A shock wave is thus a longitudinal wave travelling at a speed which is greater than that of ordinary wave. |
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6. |
Why the pitch of an organ pipe on a hot summer day is higher ? |
Answer» On a hot day, the velocity of sound will be more since (frequency proportional to velocity) the frequency of sound increases and hence its pitch increases. |
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7. |
The ralation between acceleration and displacement of four particles are given below which one of the particles is executing simple harmonic motion?A. `a_(x)=+2x`B. `a_(x)=+2x^(2)`C. `a_(x)=-2x^(2)`D. `a_(x)=-2x` |
Answer» Correct Answer - D For SHM, `a_(x)prop(-x)`. |
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8. |
Consider the situastion shown in figure. Show that if that blocks are displaced slightly in opposite directions and released, they will execute simple harmonic motion. Calculate the time period. A. `2pisqrt((m)/(k))`B. `2pisqrt((m)/(2k))`C. `2pisqrt((m)/(4k))`D. `2pisqrt((2m)/(k))` |
Answer» Correct Answer - B Reduced mass of the system `mu=((m)(m))/(m+m)=(m)/(2)` `therefore`Time period, `T=2pisqrt((mu)/(k))=2pisqrt((m)/(2k))` |
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9. |
Draperies and furniture often improve the acoustics of a room. Why? |
Answer» According to Sabine, acoustic of a room depends upon its time of reverberation which can be adjusted by placing such things in the room which may absorb the sound falling on them. As dampers and furniture in the room absorb the sound produced in the room, they help to control the time of reverberation of the room. Therefore the presence of draperies and furniture often improve the acoustics of the room. |
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10. |
Write down the difference between simple harmonic motion and angular simple harmonic motion. |
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Answer» Comparision of simple harmonic motion and angular harmonic motion
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11. |
Simple harmonic motion is the projection of uniform circular motion onA. x-axisB. y-axisC. reference circleD. any diameter of reference circle |
Answer» Correct Answer - D Simple harmonic motion is the projection of uniform circular motion on any diameter of reference circle. |
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12. |
STATEMENT -1 `:`The force acting on a particle moving alongx -axis is `F = - alpha ( x+ vt)`, where `alpha` is a constant. and STATEMENT-2 `:` To anobserver moving along x-axis with constant velocity v, it represents SHM. |
Answer» Correct Answer - 1 | |
13. |
When the amplitude of a particle executing S.H.M. is increased slightly its periodA. increasesB. remains unchangedC. decreasesD. may increase or decrease |
Answer» Correct Answer - B | |
14. |
STATEMENT-1 `:` The energy isincreased because the amplitude is increased . and STATEMENT-2 `:` The amplitude is increased because energy is increased. |
Answer» Correct Answer - 3 | |
15. |
A simple pendulum of length lis oscillating with a time period T `=1` minute . Match the columns. `{:(,"Column-I",,"Column-II"),((A),"Time period if the pendulum is osciallated inside liquid",(p),"More than T"),((B), "Time period if a constant force less than or equal to weight of bob is applied on the bob in vertically upward direction",(q),1 hr), ((C ), "Time period if the pendulum is oscillated in a moving lift",(r),"Inifinite"),((D),"Time period if its length becomes equal to radius of earth (R ) ", (s) , "Less than T "),(,,(t),T):}` |
Answer» Correct Answer - `A(p,q,r), B(p,q,r), C(p,q,r,s,t), D(p,q)` | |
16. |
STATEMENT-1 `:` An oscillatory motion is necessarily periodic. and STATEMENT -2 `:` A simple harmonic motion is necessarily osciallatory. |
Answer» Correct Answer - 4 | |
17. |
Assertion: The motion of the earth around the sun is perriodic but not oscillatory. Reason: Oscillatory motion is necessarily periodic but periodic motion is not necessarily oscillatory.A. If both assertion and reson are true and reason is the correct explanation of assertionB. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
Answer» Correct Answer - B The motion of earth around sun is not oscillatory as it not a to and from motion about a fixed point. |
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18. |
Assertion: The motion of a simple pendulum is simple harmoni for all angular displacement. Reason: Motion of simple pendulum is independent of the angular displacement.A. If both assertion and reson are true and reason is the correct explanation of assertionB. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
Answer» Correct Answer - D Motion of a simple pendulum is simple harmonic for small angular displacement only, not for all angular displacement. |
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19. |
The periodic time of a mass suspended by a spring (force constant k) is T. If the spring is cut in three equal pieces, what will be the force constant of each part? If the same mass be suspended from one piece, what will be the periodic time? |
Answer» Consider the spring be made of combination of three springs in series each of spring constant k. The effective spring constant K is given by: \(\frac{1}{K}=\frac{1}{k}+\frac{1}{k}+\frac{1}{k}=\frac{3}{k}\) Or K = \(\frac{k}{3}\) Or k = 3K ∴ Time period of vibration of a body attached to the end of this spring. T = \(2\pi\sqrt{\frac{m}{K}}\) = \(2\pi\sqrt{\frac{m}{(\frac{k}{3})}}\) = \(2\pi\sqrt{\frac{3m}{k}}\) …(i) When the spring is cut into three pieces, the spring constant = k. Time period of vibration of a body attached to the end of this spring. T1 = \(2\pi\sqrt{\frac{m}{k}}\) …(ii) From the equation (i) and (ii), \(\frac{T_1}{T}=\frac{1}{\sqrt3}\) Or T1 = \(\frac{T}{\sqrt 3}\) |
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20. |
Show that when a string fixed at its two ends vibrates in 1 loop, 2 loops, 3 loops, and 4 loops, the frequency are in the ratio 1:2:3:4. |
Answer» Length for each loop = \(\frac{\lambda}{2}\) Now, L = \(\frac{n\lambda}{2}\) λ = \(\frac{2L}{n}\) (1) But v = vλ or λ = \(\frac{v}{u}\) Putting in eqn, (1) \(\frac{v}{u}=\frac{2L}{n}\) v = \(\frac{n}{2L}\)u v = \(\frac{n}{2L}\sqrt{\frac{T}{\mu}}\) [∵ u = \(\sqrt{\frac{T}{\mu}}\) ] For n = 1, v1 = \(\frac{1}{2L}\sqrt{\frac{T}{m}}\) = v0 For n = 2, v2 = \(\frac{1}{2L}{\sqrt{\frac{T}{m}}}\) = 2v0 Therefore, u1: u2:u3: u4 = n1: n2: n3:n4 u1: u2:u3: u4 = 1: 2: 3: 4 |
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21. |
Define Maintained oscillation. |
Answer» The oscillation in which the loss of oscillator is compensated by the supplying energy from an external source are known as maintained oscillation. |
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22. |
What is the phase relationship between displacement, velocity and acceleration in SHM ? |
Answer» In SHM, The velocity leads the displacement by a phase π/2 radians and acceleration leads the velocity by a phase π/2 radians. |
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23. |
Can the motion of an artificial satellite around the earth be taken as SHM ? |
Answer» No, it is a circular and periodic motion but not SHM. |
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24. |
What would be the duration of the year if theA artificial satellite moving in a circular orbit around the earth has a total (kinetic + potential) energy E0. Its potential energy isA. 2E0B. E0C. 1.5E0D. -E0 |
Answer» Correct Answer - A |
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25. |
(1) periodic motion (2) oscillatory motion. Give two examples. |
Answer» (1) Periodic motion : A motion that repeats itself at definite intervals of time is said to be a periodic motion. Examples : The motion of the hands of a clock, the motion of the Earth around the Sun. (2) Oscillatory motion : A periodic motion in which a body moves back and forth over the same path, straight or curved, between alternate extremes is said to be an oscillatory motion. Examples : The motion of a taut string when plucked, the vibrations of the atoms in a molecule, the oscillations of a simple pendulum. [Note : The oscillatory motion of a particle is also called a harmonic motion when its position, velocity and acceleration can be expressed in terms of a periodic, sinusoidal functions-sine or cosine, of time. |
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26. |
Can the motion of an artificial satellite around earth be taken as S.H.M.? |
Answer» No, it is circular and periodic motion but not to and fro about a mean position which is essential for S.H.M., ∴ it cannot be taken S.H.M. |
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27. |
Which of the following functions of time represent (a) S.H.M., and (b) periodic but not simple harmonic motion? Give the period for each case: (i) sin ωt - cos ωt(ii) sin2 ωt. |
Answer» (i) sin ωt - cos ωt = \(\sqrt 2[sin\,\omega tcos\frac{\pi}{4}-\,cos\,\omega t\,sin\,\frac{\pi}{4}]\) = \(\sqrt 2\, sin\,(\omega t\,-\,\frac{\pi}{4})\) This function represents S.H.M., having period T = \(\frac{2\pi}{\omega}\) and initial phase = − \(\frac{\pi}{4}\)rad. (ii) sin2ωt = (1 – cos2ωt) The function is periodic, having a period T = \(\frac{2\pi}{2\omega}=\frac{\pi}{\omega}\) But does not represent S.H.M. |
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28. |
Can a motion be periodic but not oscillatory? If yes, given an example. If not, explain why? |
Answer» Yes, uniform circular motion is the example of it. |
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29. |
Define Oscillatory or Vibratory Motion. Give an examples. |
Answer» The motion in which a body moves to and fro or back and forth repeatedly about a fixed point in a definite interval of time. Oscillatory motion is also called as harmonic motion. Example : The motion of the pendulum of a wall clock. |
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30. |
What is meant by Periodic Motion ? |
Answer» A motion, which repeat itself over and over again after a regular interval of time is called a periodic motion and the fixed interval of time after which the motion is repeated is called period of the motion. Examples : Revolution of earth around the sun (period one year). |
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31. |
Which of the following functions of time represent (a) simple harmonic (b) periodic but not simple harmonic and (c ) non periodic motion? Give period for each case of periodic motion `(omega` is nay positive constant) (1) `Sin^(3) omegat` (2) `3cos (pi//4 -2 omegat)` (3) `cos omegat +cos 3 omegat + cos 5 omegat` (4) `e^(-omega^(2)t^(2))` (5) `1 + omegat +omega^(2) t^(2)` |
Answer» (1) `sin^(3) omegat` is a periodic functions but not simple harmonic because `a prop -y` conditionis not satisfied. Its time period is `T = (2pi)/(omega)` (2) `3cos (pi//4 -2omegat) = 3 cos (2 omegat - pi//4)` it is simple harmonic with a time period `T = (2pi)/(2omega) = (pi)/(omega)` (3) `Cos omega t + cos 3omegat +cos 5 omegat` is a periodic function but not simple harmonic. The time periods of each periodic function are `(2pi)/(omega),(2pi)/(3omega)` and `(2pi)/(5omega)`.Since `(2pi)/(omega)` is the multiple of the other two periods. The given function is periodic with time period `(2pi)/(omega)` (4) `e^(-omega^(2)t^(2))` is not periodic as `t` increases `e^(-omega^(2)t^(2))` decreasesa and tends to zero as `t rarr oo` (5) `1+ omegat + omega^(2) t^(2)` is not periodic, as function increases with time without repetition. |
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32. |
The acceleration due to gravity on the surface of the moon is `1.7ms^(-2)`. What is the time perioid of a simple pendulum on the surface of the moon, if its time period on the surface of earth is `3.5s ?` Take `g=9.8ms^(-2)` on the surface of the earth.A. 8.4 sB. 8.2 sC. 7.4 sD. 6.4 s |
Answer» Correct Answer - A Given, `g_(m)=1.7 ms^(-2),g_(e)=9.8 ms^(-2)` `T_(m) = ? and T = 3.5 s` As `T_(e)=2pi sqrt((l)/(g_(e)))` and `T_(m)=2pi sqrt((l)/(g_(m)))` `therefore" "(T_(m))/(T_(e))=sqrt(g_(e)/(g_(m))) or T_(m)=T_(e) sqrt((g_(e))/(g_(m)))` `rArr" "T_(m) = 3.5 sqrt((9.8)/(1.7))=8.4 s` |
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33. |
The acceleration due to gravity on the surface of the moon is `1.7ms^(-2)`. What is the time perioid of a simple pendulum on the surface of the moon, if its time period on the surface of earth is `3.5s ?` Take `g=9.8ms^(-2)` on the surface of the earth.A. 4.4 sB. 8.4 sC. 16.8 sD. |
Answer» Correct Answer - C For moon, `g_(m)=1.7ms^(-2)` for earth, `g_(e)=9.8ms^(-2),T_(e)3.5s` But, `T_(m)=2pisqrt((l)/(g_(m))) and T_(e)=2pisqrt((l)/(g_(e))) therefore(T_(m))/(T_(e))=sqrt((g_(e))/(g_(m)))` or `T_(m)=sqrt((g_(e))/(g_(m)))xxT_(e)=sqrt((9.8)/(1.7))xx3.5=8.4s` |
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34. |
The length of second’s pendulum on the surface of earth is nearly 1 m. Its length on the surface of moon should be [Given: acceleration due to gravity (g) on moon is 1/6 th of that on the earth’s surface](A) 1/6 m (B) 6 m (C) 1/36 m(D) 1/\(\sqrt{6}\) |
Answer» Correct option is: (A) 1/6 m |
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35. |
The length of a second's pendulum on the surface of earth is 1 m. What will be its length on the surface of the moon? |
Answer» T = 2π√{l/g} In both the cases, T is the same l ∝ g. On the moon, the value of acceleration due to gravity is 1/6th of that on the surface of earth. So length of the second's pendulum is 1/6m. |
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36. |
The length of a second’s pendulum on the surface of Earth is 1m. What will be the length of a second’s pendulum on the moon? |
Answer» lm = (1/6)lE =(1/6)m |
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37. |
The length of a seconds pendulum on the surface of earth is 100 cm. what will be the length of a seconds pendulum on the surface of moon? |
Answer» For a simple pendulum, time period (T) is given by; T = \(2\pi\sqrt{\frac{l}{g}}\) At Earth, T = \(2\pi\sqrt{\frac{l}{g}}\) (i) At Moon, T’=\(v2\pi\sqrt{\frac{l'}{g'}}\) (ii) Here, T = T’ and g’ = \(\frac{1}{6}g\) So, from equation (i) & (ii) \(\frac {l}{g}=\frac{l'}{g'}\) or,\(l'=\frac{g'}{g}\times l\) = \(\frac{1}{6}g\times\frac{1}{g}\times l\) = \(\frac{1}{6}\times l\) = \(\frac{1}{6}\)× 100 cm = 16.7 cm |
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38. |
Two identical blocks A and B, each of mass `m=3kg`, are connected with the help of an ideal spring and placed on a smooth horizontal surface as shown in Fig. Another identical blocks C moving velocity `v_0=0.6(m)/(s)` collides with A and sticks to it, as a result, the motion of system takes place in some way Based on this information answer the following questions: Q. Oscillation energy of the system i.e., part of the energy which is oscillation (changing) between potention and kinetic forms isA. `0.27 J`B. `0.09J`C. `0.18 J`D. `0.45 J` |
Answer» Correct Answer - A `B` accelerates and combined mass decelerates. The centre of mass of the system continues to move rightward with constant speed. While both the blocks oscillate about centre of mass. The velocity just after collision is: `v = (v_(0))/(2)` The velocity of centre of mass is: `V_(c.m) = (m_(1)v_(1)+m_(2)v_(2))/(m_(1)+m_(2)), T = 2pi sqrt((mu)/(k))`: reduced mass concept: Oscillation energy of the system is: `E = (1)/(2) (2m) v^(2)` Translational kinetic energy of the centre of mass of the system `= (1)/(2) (3m) v_(cm)^(2)` The remaining energy is oscillating between kinetic and potential energy during the motion of blocks: Oscillating energy `=` maximum elastic potential energy |
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39. |
Two identical blocks A and B, each of mass `m=3kg`, are connected with the help of an ideal spring and placed on a smooth horizontal surface as shown in Fig. Another identical blocks C moving velocity `v_0=0.6(m)/(s)` collides with A and sticks to it, as a result, the motion of system takes place in some way Based on this information answer the following questions: Q. The maximum compression of the spring isA. `3sqrt(30)mm`B. `3sqrt(20)mm`C. `3sqrt(10)mm`D. `3sqrt(50)mm` |
Answer» Correct Answer - A `B` accelerates and combined mass decelerates. The centre of mass of the system continues to move rightward with constant speed. While both the blocks oscillate about centre of mass. The velocity just after collision is: `v = (v_(0))/(2)` The velocity of centre of mass is: `V_(c.m) = (m_(1)v_(1)+m_(2)v_(2))/(m_(1)+m_(2)), T = 2pi sqrt((mu)/(k))`: reduced mass concept: Oscillation energy of the system is: `E = (1)/(2) (2m) v^(2)` Translational kinetic energy of the centre of mass of the system `= (1)/(2) (3m) v_(cm)^(2)` The remaining energy is oscillating between kinetic and potential energy during the motion of blocks: Oscillating energy `=` maximum elastic potential energy |
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40. |
Two blocks of masses `3kg` block is attached to a spring with a force constant, `k k = 900N//ma` which is compressed `2m` initially from its equilibrium position. When `3kg` mass is released, it strikes the `6kg` mass and the two stick togther in an inelastic collision. The velocities of a particle executing `S.H.M.` are `30cm//s` and `16 cm//s` when its displacements are `8cm` and `15cm` from the equilibrium position. then its amplitude of oscillation in cm is:A. `25`B. `21`C. `17`D. `13` |
Answer» Correct Answer - C `v^(2) = omega^(2) (A^(2)-x^(2))` For `x = 8cm, 30^(2) = omega^(2) (A^(2) - 8^(2)) …………..(1)` For `x = 15 cm, 16^(2) = omega^(2) (A^(2) -15^(2)) …………(2)` Solving (1) and (2), `16^(2)A^(2) - 16^(2) (8)^(2) = 30^(2)A^(2) - (30)^(2) - (15)^(2)` `A^(2) (30^(2)-16^(2)) = (30)^(2) (15)^(2) - (16)^(2) (8)^(2) rArr A = 17` |
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41. |
Find the period of the oscillations of the devices shown in figure if `m` is displaced slightly. |
Answer» Reduced mass of the system (1) `mu = ((mM)/(m+M))` `:. T_(1) = 2pi sqrt((mu)/(K)) = 2pi sqrt((mM)/((m+M)K))` (2) `mu = (mm)/(m+m) = (m)/(2)` and `K_(e) = K+K = 2K` `:. T_(2) = 2pi sqrt((mu)/(K_(e))) = 2pi sqrt(m)/(4K)` |
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42. |
A pendulum has a period `T` for small osillations. An obstacle is placed directly beneath the pivot, so that only the lowest one - quarter of the string can follow the pendulum bob when it swings to the left of its resting position. The pendulum is released from rest at a certain point. How long will it take to return to that point again ? In answering this question, you may assume that the angle between the moving string and the vertical stays small throughout the motion. A. `T//2`B. `3T//4`C. `T//4`D. `T` |
Answer» Correct Answer - B `t =(T)/(2)+(T^(1))/(2), T^(1) = 2pi sqrt((l)/(4g)) = (T)/(2)` |
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43. |
A pendulum has a period `T` for small osillations. An obstacle is placed directly beneath the pivot, so that only the lowest one - quarter of the string can follow the pendulum bob when it swings to the left of its resting position. The pendulum is released from rest at a certain point. How long will it take to return to that point again ? In answering this question, you may assume that the angle between the moving string and the vertical stays small throughout the motion. A. TB. `(T)/(2)`C. `(3T)/(4)`D. `(T)/(4)` |
Answer» Correct Answer - 3 | |
44. |
Define the terms 1. amplitude 2. period 3. frequency 4. phase related with a particle executes SHM. |
Answer» 1. Amplitude: Maximum displacement of the particle from the mean position is called amplitude. 2. Period : Time taken by the particle to complete one oscillation is called time period. (T) 3. Frequency : The number of oscillations completed by the particle in one second is called frequency (f) 4. Phase: Phase of a particle is defined as the fraction of the time period that has elapsed since the particle last passed through its mean position in the positive direction. |
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45. |
A second pendulum is attached to roof of a car that is moving with acceleration `10m//s^(2)` on a straight road. Its frequenct of oscillation isA. `(1)/(2)Hz`B. `1Hz`C. `(1)/(sqrt(2))Hz`D. `2^(-3//4)Hz` |
Answer» Correct Answer - D `f = (1)/(2pi) sqrt((sqrt(g^(2)+a^(2)))/(l))` |
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46. |
A simple pendulum is suspended from the roof of a train. If the train is moving with an acceleration `49 cm//s^(2)`. Then the angle of inclination of the string about the vertical will beA. `20^(@)`B. zeroC. `30^(@)`D. `3^(@)` |
Answer» Correct Answer - D `F=mgsintheta` `a=gsintheta` `therefore (49)/(980)=sintheta` `therefore theta=3^(@)` |
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47. |
A second pendulum is attached to roof of a car that is sliding down along a smooth inclined plane of inclination `60^(@)`. Its perios of oscillation isA. `2s`B. `sqrt(2)s`C. `2sqrt(2)s`D. `1s` |
Answer» Correct Answer - B `T = 2pi sqrt((l)/(g cos theta))` |
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48. |
The length of a weigtless spring increaes by 2cm when a weight of 1.0kg in the period of oscillation of the spring and its kinetic energy of oscillation. Take `g=10ms^(-2)`. |
Answer» Correct Answer - 0.28s, 2.5J |
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49. |
A uniform rod ofmass M and length L is hanging from its one end free to rotate in a veritcal plane.A small ball of equal mass is attached of the lowe end as shown. Time period of small oscillations of the rod is A. `2pi sqrt((2L)/(3g))`B. `2pi sqrt((4L)/(3g))`C. `2pi sqrt((8L)/(3g))`D. `2pi sqrt((L)/(g))` |
Answer» Correct Answer - 4 | |
50. |
The tota energy of a particle executing simple harmonic motion is `16J`. What will be total energy of particle if amplitude is halved and frequency is douboled?A. `1J`B. `64 J`C. `16J`D. `256J` |
Answer» Correct Answer - C `T.E. = (1)/(2) m omega^(2) A^(2)` |
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