InterviewSolution
Saved Bookmarks
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.
| 651. |
If the lift moves up and comes down with uniform speed, then the time period of pendulum in the liftA. increasesB. there will be no effectC. decreasesD. can not be predicted |
| Answer» Correct Answer - B | |
| 652. |
The period of oscillation of a simple pendulum of constant length at earth surface is T. Its period inside a mine is CA. greater than TB. less than TC. equal to TD. cannot be compared |
| Answer» Correct Answer - A | |
| 653. |
A person is swinging on a swing in the sitting position. How will be period of swing change, if the person stands up? |
|
Answer» The swinging person may be regarded as simple, pendulum. When the person stands up, 'l' decreases, Hence, T = 2π x √4 g decreases. |
|
| 654. |
A child swinging on a swing in sitting position, stands up, then the time period of the swing will.A. increasesB. decreasesC. remain sameD. first increase then decrease |
| Answer» Correct Answer - B | |
| 655. |
A girl is swinging a swing in a standing position. If the girl seat and swings, the period will beA. shorterB. longerC. sameD. first shorter and then longer |
| Answer» Correct Answer - B | |
| 656. |
Assertion `:` When a girl sitting on a swing stands up, the periodic time of the swing will increase. Reason `:` In standing position of the girl, the length of swing will increase.A. Statement-I is true, Statement-Ii is true, Statement-II is a correct explanation for Statement-I.B. Statement-I is true, Statement-II is true, Statement-II is NOT a correct explanantion for Statement-I.C. Statement-I is true, Statement-II is falseD. Statement-I is false, Statement-II is true |
| Answer» Correct Answer - D | |
| 657. |
A girl is in standing position in an oscillating swing. If the girl sits in the swing, the frequency of oscillationA. increasesB. decreasesC. does not changeD. becomes zero |
| Answer» Correct Answer - B | |
| 658. |
The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0m. If the piston moves with simple harmonic motion with an angular frequency of `200rev//`min., what is its maximum speed ? |
|
Answer» Stroke of piston =2 times the amplitude Let A = amplitude, stroke = 1m `:. rArr A = (1)/(2)m` Angular frequency, `omega = 200 rad//min`. `V_(max) = ?` We know that the maximum speed of the block when the amplitude is A, `V_(max) = omega A = 200 xx (1)/(2) = 100 m//min`. `=(100)/(60) = (5)/(3) ms^(-1) = 1.67 ms^(-1)` |
|
| 659. |
The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m. If the piston moves with simple harmonic motion with an angular frequency of 200 rad/ min, what is its maximum speed? |
|
Answer» Stroke = 1 m Amplitude = (troke/2) = 1/2 m ω = 200 rad / min Vmax = ωA = 200 × 1/2 = 100 m / min Vmax = 1.67 m/s. |
|
| 660. |
A spring of force constant `1200Nm^(-1)` is mounted on a horizontal table as shown in figure. A mass of 3.0kg is attached to the free end of the spring, pulled side ways to a distance of 2.0cm and released. Determing. (a) the frequency of oscillation of the mass. (b) the maximum acceleration of the mass. (c) the maximum speed of the mass. |
|
Answer» Here, `K = 1200 Nm^(-), m = 3.0 kg, a = 2.0 = 0.02m` (i) Frequency `v = (1)/(T) = (1)/(2pi) sqrt((k)/(m)) = (1)/(2 xx 3.14) sqrt((1200)/(3)) = 3.2 s^(-1)` (ii) Acceleration, `A = omega^(2) y = (k)/(m) y` Acceleration will be maximum when y is maximum i.e., `y = a` `:.` max. acceleration, `A_(max) = (ka)/(m) = (1200 xx 0.02)/(3) = 8 ms^(-2)` (iii) Max. speed of the mass will when it is passing through mean position `V_(max) = a omega = a sqrt((k)/(m)) = 0.02 xx sqrt((1200)/(3)) = 0.4 ms^(-1)` |
|
| 661. |
A spring of force constant `1200Nm^(-1)` is mounted on a horizontal table as shown in figure. A mass of 3.0kg is attached to the free end of the spring, pulled side ways to a distance of 2.0cm and released. Determing. (a) the frequency of oscillation of the mass. (b) the maximum acceleration of the mass. (c) the maximum speed of the mass. |
|
Answer» `K = 1200 N m^(-1) ,m = 3 kg A = 2cm = 0.02m` (a) Frequecny, `f = (1)/(2pi) sqrt((K)/(m)) = (1)/(6.28) sqrt((1200)/(3)) = 3.2 Hz` (b) Acceleration `a = omega^(2) y = (K)/(m)y` Acceleration will be maximum when `y` is maximum i.e. `y=A` `:.` Max.acceleration, `a_(max) = (KA)/(m) = (1200 xx 0.02)/(3) = 8ms^(-2)` (c) Maximum speed of the mass will be when it is passing through the mean position, given by `V_(max) = A omega =A sqrt((K)/(m)) = 0.02 xx sqrt((1200)/(3)) = 0.4 ms^(-1)` |
|
| 662. |
A mass of `2.0kg`is put on a that pan attached to a vertical spring fixed on the ground as shown in the figure The mass of the spring and the pen is negligible the mass executing a simple harmonic motion The spring constant is `200N//m` what should be the minimum amplitude of the motion so that the mass get detached from the pan? `(Taking g = 10m//s^(2))` A. `8.0 cm`B. `10.0 cm`C. Any value less than `12.0 cm`D. `4.0 cm` |
|
Answer» Correct Answer - B |
|
| 663. |
The time period of simple harmonic motionn depends uponA. amplitudeB. energyC. phase constantD. mass |
|
Answer» Correct Answer - D The time period of simple harmonic motion does not depend on amplitude, energy or the phase constant. |
|
| 664. |
A particle is subjected to two `SHMs x_(1) = A_(1) sin omegat` and `x_(2) = A_(2)sin (omegat +(pi)/(4))`. The resultant `SHM` will have an amplitude ofA. `(A_(1)+A_(2))/(2)`B. `sqrt(A_(1)^(2)+A_(2)^(2))`C. `sqrt(A_(1)^(2)+A_(2)^(2)+sqrt(2)A_(1)A_(2))`D. `A_(1)A_(2)` |
|
Answer» Correct Answer - C `A = sqrt(A_(1)^(2)+A_(2)^(2)+2A_(1)A_(2)cosphi)` |
|
| 665. |
Function `x=Asin^(2)omegat+Bcos^(2)omegat+Csinomegat cos omegat` represents simple harmonic motion,A. Any value of `A,B` and `C` (except `C = 0)`B. `A =- B, C = 2B`, amplitude `=|Bsqrt(2)|`C. `A =B, C = 0`D. `A =B,C = 2B`, amplitude `=|B|` |
|
Answer» Correct Answer - A::B::D Using `sin 2 omega =2 sin omegat cos omegat` and `cos 2omegat =1 -2 sin^(2) omegat =2 cos^(2)omegat -1`. We get, `x =(A)/(2) (1-cos 2omegat) +(B)/(2) (1_cos 2omegat) +(C )/(2)sin 2 omegat` For `A =0, B =0, x =(C )/(2)sin 2 omegat`, so choice (a) is correct for `A =-B` and `C=2B, x =B cos 2 omegat +B sin 2 omegat` Amplitude `=|Bsqrt(2)|` So choice (b) is correct For `A =B,C =0, X =A` Hence ,(c) is not correct option. For `A =B, C = 2B, X = B+B sin omegat` It also represents `SHM` will amplitude `B`. So choice (d) is correct. |
|
| 666. |
The displacement of a particle in SHM is given by `x=sinomegat+cosomegat`. Then its amplitude and initial displacement areA. `1, sqrt(2)` unitsB. `sqrt(2), 1` unitsC. `sqrt(2), sqrt(2)` unitsD. 1, 1 units |
|
Answer» Correct Answer - B `R=sqrt(A_(1)^(2)+A_(2)^(2))` `=sqrt(1^(2)+1^(2))=sqrt(2)` `x=sinomegaxx0+cosomegaxx0=0+1=1` |
|
| 667. |
The minimum phase difference between the two simple harmonic oscillations `x_(1)=(1//2)sinomegat+(sqrt(3)//2)cosomegat and` `x_(2)=(sqrt(3)//2)sinomegat+(1//2)cosomegat` isA. `pi//6`B. `pi//3`C. `pi//4`D. `pi//2` |
|
Answer» Correct Answer - A `phi_(1)=tan^(-1)(B//A)=tan^(-1)((sqrt(3)/(2))/((1)/(2)))=tan^(-1)(sqrt(3))=(pi)/(3)` `phi_(2)=tan^(-1)(B//A)=tan^(-1)(((1)/(2))/((sqrt(3))/(2)))=tan^(-1)((1)/(sqrt(3)))=(pi)/(6)` `deltaphi=phi_(1)-phi_(2)=(pi)/(3)-(pi)/(6)=(pi)/(6)` |
|
| 668. |
The equation of motion of a particle is x = a cos (α t )2.The motion is(a) periodic but not oscillatory.(b) periodic and oscillatory.(c) oscillatory but not periodic.(d) neither periodic nor oscillatory |
|
Answer» (c) oscillatory but not periodic. |
|
| 669. |
The displacement of a particle varies with time according to the relationy = a sin ωt + b cos ωt.(a) The motion is oscillatory but not S.H.M.(b) The motion is S.H.M. with amplitude a + b.(c) The motion is S.H.M. with amplitude a2 + b2.(d) The motion is S.H.M. with amplitude √a2 +b2 . |
|
Answer» (d) The motion is S.H.M. with amplitude √a2 +b2 |
|
| 670. |
Figure 14.2. shows the circular motion of a particle. The radius of the circle, the period, sense of revolution and the initial position are indicated on the figure. The simple harmonic motion of the x-projection of the radius vector of the rotating particle P is(a) x (t) = B sin (2πt/30).(b) x(t) = B cos (πt/15)(c) c(t) = B sin (πt/15 + π/2)(d) c(t) = B cos (πt/15 + π/2) |
|
Answer» (a) x (t) = Bsin(2πt/30) |
|
| 671. |
Does the change in frequency due to Doppler effected depend on:(i) Distance between source and observer?(ii) The fact that source is moving towards observer or observer is moving towards the source? |
|
Answer» (i) No, change in frequency due to Doppler’s effect has nothing to do with distance between source and listener. (ii) Yes, in case of sound waves. No, in case of light waves. |
|
| 672. |
Four pendulums A, B, C and D are suspended from the sameelastic support as shown in Fig. 14.1. A and C are of the same length, while B is smaller than A and D is larger than A. If A is given a transverse displacement,(a) D will vibrate with maximum amplitude.(b) C will vibrate with maximum amplitude.(c) B will vibrate with maximum amplitude.(d) All the four will oscillate with equal amplitude. |
|
Answer» (b) C will vibrate with maximum amplitude. |
|
| 673. |
Four pendulums A,B,C and D are suspended from the samme elastic support as shown in figure. A and C are of the same length, while B is smaller than A and D is larger than A. if A is given a transverse displacement,A. D will vibrate with maximum amplitudeB. C will vibrate with maximum amplitudeC. B will vibrate with maximum amplitudeD. All the four will oscillate with equal amplitude. |
|
Answer» Correct Answer - B Since length of pendulums A and C is same and `T=2pisqrt((L)/(g))`, hence their time period is same and they will have same frequency of vibration. Due to it, a resonance will take place and the pendulum C will vibrate with maximum amplitude. |
|
| 674. |
A pendulum is mounted on a cart rolling without friction down on an inclined surface of inclination θ with the horizontal. The period of the pendulum on an immobile cart is T. what will be the period of the pendulum on the cart when the cart rolls down the surface? |
|
Answer» When the cart is at rest, the effective acceleration due to gravity is g and when the cart is rolling down the inclined surface, the effective acceleration due to gravity involved perpendicular to plane is g cos θ. Time period of pendulum when art is immobile, T = \(2\pi\sqrt{\frac{l}{g}}\) Time period of cart when it moving down the plane, T’= \(2\pi\sqrt{\frac{l}{g\,cos\,\theta}}\) Or T' = \(\frac{T}{\sqrt {cos \theta}}\) |
|
| 675. |
When the displacement of a particle executing SHM is one-fourth of its amplitude, what fraction of the total energy is the kinetic energy?A. `(16)/(15)`B. `(15)/(16)`C. `(3)/(4)`D. `(4)/(3)` |
|
Answer» Correct Answer - B In SHM, Kinetic energy of the particle, `K=(1)/(2)momega^(2)(A^(2)-x^(2))` where m is the mass of particle, `omega` is its angular frequency, A is the amplitude of oscillation and x is its displacement. At `x=(A)/(4),K=(1)/(2)momega^(2)[A^(2)-((A)/(4))^(2)]=(1)/(2)((15)/(16)momega^(2)A^(@))` Energy of the particle, `E=(1)/(2)momega^(2)A^(2)` `therefore=(K)/(E)=((1)/(2)((15)/(16)momega^(2)A^(2)))/((1)/(2)momega^(2)A^(2))=(15)/(16)` |
|
| 676. |
Even after the breakup of one prong of tunning fork it produces a round of same frequency, then what is the use of having a tunning fork with two prongs ? |
|
Answer» Two prongs of a tunning fork set each other in resonant vitorations and help to maintain the vibrations for a longer time. |
|
| 677. |
Why is the sonometer box hollow and provided with holes ? |
|
Answer» When the stem of the a tunning fork gently pressed against the top of sonometer box, the air enclosed in box also vibrates and increases the intensity of sound. The holes bring the inside air in contact with the outside air and check the effect of elastic fatigue. |
|
| 678. |
Two independent harmonic oscillators of equal mass are oscillating about the origin with angular frequencies `(omega_1) and (omega_2) and have total energies (E_1 and E_2), respectively. The variations of their momenta (p) with positions (x) are shown (s) is (are). A. `E_(1)omega_(1) = E_(2)omega_(2)`B. `(omega_(2))/(omega_(1)) - n^(2)`C. `omega_(1)omega_(2) = n^(2)`D. `(E_(1))/(omega_(1)) = (E_(2))/(omega_(2))` |
|
Answer» Correct Answer - B::D For forst oscillator `E_(1) = (1)/(2) m omega_(1)^(2)a^(2)`, and `p = mv = m omega_(1)a = b rArr (a)/(b) = (1)/(m omega_(1)) …….(i)` for second oscillator `E_(2) = (1)/(2) m omega_(2)^(2) R^(2)`, and `m omega_(2) = 1………….(ii)` `((a)/(b)) = (omega_(2))/(omega_(1) =n^(2))` `(E_(1))/(omega_(1)^(2)a^(2))= (E_(2))/(omega_(2)^(2)R^(2)) rArr (E_(1))/(omega_(1)) = (E_(2))/(omega_(2))` |
|
| 679. |
A particle executes simple harmonic motion. At what point on its path is the acceleration maximum? |
|
Answer» Acceleration is maximum at a point of maximum displacement from the mean position. |
|
| 680. |
The phase difference between the displacement and acceleration of a particle execuliting simple harmonic motion isA. `pi, (pi)/(2)`B. `(pi)/(2),pi`C. `(pi)/(2),(pi)/(2)`D. `pi,pi` |
| Answer» Correct Answer - B | |
| 681. |
Statement I: If the amplitude of a simple harmonic oscillator is doubled, its total energy becomes four times. Statement II: The total energy is directly proportional to the square of the amplitude of vibration of the harmonic oscillator.A. Statement-I is true, Statement-Ii is true, Statement-II is a correct explanation for Statement-I.B. Statement-I is true, Statement-II is true, Statement-II is NOT a correct explanantion for Statement-I.C. Statement-I is true, Statement-II is falseD. Statement-I is false, Statement-II is true |
|
Answer» Correct Answer - A `E prop A^(2)` |
|
| 682. |
Where will a person hear maximum sound, at node or antinode? |
|
Answer» Perception of sound is due to pressure variations which is maximum at nodes. |
|
| 683. |
We can recognize our friends from their voices. Why? |
|
Answer» This is because quality of sound produced by them is different. |
|
| 684. |
Why is it difficult sometimes to recognize your friend’s voice on phone? |
|
Answer» This occurs due to poor quality of sound. |
|
| 685. |
A simple pendulum is set into vibrations. The bob of pendulum come to rest after some time due to A) Air frictionB) Moment of inertiaC) Weight of bobD) Combination of all above |
|
Answer» A) Air friction Explanations:
|
|
| 686. |
Define Non-mechanical waves. Give an examples. |
|
Answer» The waves which do not require medium for their propagation are called non-mechanical or electromagnetic waves. Examples : Light, heat (Infrared), radio waves, γ-rays. X-rays etc. |
|
| 687. |
Is superposition principle applicable to electromagnetic waves? |
|
Answer» Yes, it is applicable to electromagnetic waves. |
|
| 688. |
Which property is common in all types of mechanical waves? |
|
Answer» All mechanical waves travel in a medium, but the medium does not advance along with the wave. |
|
| 689. |
Is it possible to monitor the temperature of a wire by measuring its vibrational frequency. |
|
Answer» Yes. v = \(\frac{1}{2l}\sqrt{\frac{T}{m}}\) As temperature increase, length increases. Therefore frequency of vibration v decrease. Hence, changes in temperature can be monitored. |
|
| 690. |
Can a simple pendulum experiment be done inside a satellite? |
|
Answer» Since time period of a simple pendulum is :- T = 2π √1/g Since, inside a satellite, effective value of ‘g’ = O So, when g = O, T = α. Therefore, inside the satellite, the pendulum does not oscillate at all. So, it can not be preformed inside a satellite. |
|
| 691. |
Start gaining timeStart losing timeStill give correct timestop working |
|
Answer» Option " start losing time " will be correct answer Explanation Placing a magnet below the central position of an iron bob of a pendulum will enhance the effective force of restoration towards the equilibrium position as the magnetic force will act downward along with the gravitational pull. Hence time period will be shorter |
|
| 692. |
Assertion: The skill in swinging to greater heights lies in the synchronisation of the rhythm of pushing against the ground with the natural frequency of the swing. Reason: The phenomenon behind this is resonance.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 - A The phenomenon of increase in amplitude when the driving force is close to the natural frequency of the oscillator is called resonance. Swing is a good example of resonance. |
|
| 693. |
A body of mass 8 kg performs S.H.M. of amplitude 60 cm. The restoring force is 120 N, when the displacement is 60 cm. The time period isA. `0.628 s`B. `1.256 s`C. `1.884 s`D. ` 2.5 12s` |
| Answer» Correct Answer - 2 | |
| 694. |
The time period of oscillation of a particle that executes `SHM` is `1.2s`. The time starting from mean position at which its velocity will be half of its velocity at mean position isA. `2 A`B. `(sqrt(3))/(2)xxA`C. AD. `(A)/(2)` |
|
Answer» Correct Answer - B `V=(V_(m))/(2)` `omegasqrt(A^(2)-x^(2))=(Aomega)/(2)` `A^(2)-x^(2)=(A^(2))/(4)` `A^(2)-(A^(2))/(4)=x^(2)` `x=sqrt(3)/(2)A`. |
|
| 695. |
The displacement time graph of a particle executing S.H.M. is shown in Fig. 14.5. Which of the following statement is/are true?(a) The force is zero at t = 3T/4.(b) The acceleration is maximum at t 4T/4.(c) The velocity is maximum at t = T/4.(d) The P.E. is equal to K.E. of oscillation at t = T/2. |
|
Answer» (a) The force is zero at t = 3T/4. (b) The acceleration is maximum at t 4T/4. (c) The velocity is maximum at t = T/4. |
|
| 696. |
Statement-1 `:` During the oscillations of a simple pendulum, the direction of its acceleration at the mean position is directed towards the point of suspension and at extreme position is direction towards the mean position . Statement-2 `:` The directio of acceleration of a simple pendulum at an instant is decided by the tangential and radial components of weight at that instant.A. Statement-I is true, Statement-Ii is true, Statement-II is a correct explanation for Statement-I.B. Statement-I is true, Statement-II is true, Statement-II is NOT a correct explanantion for Statement-I.C. Statement-I is true, Statement-II is falseD. Statement-I is false, Statement-II is true |
|
Answer» Correct Answer - A Tangent component of weight `=mg sin theta` Radial component of weight `= mg cos theta` at mean position, radial present at exterme position, tangent present. |
|
| 697. |
Dimensions of force constant areA. `[L^(0)M^(1)T^(-2)]`B. `[L^(0)M^(-1)T^(-2)]`C. `[L^(1)M^(0)T^(-2)]`D. `[L MT^(-2)]` |
|
Answer» Correct Answer - A `K=(F)/(x)` `[L] =([L^(1)M^(1)T^(-2)])/([L^(1)])=[L^(0)M^(1)T^(-2)]` |
|
| 698. |
Statement-I : For a particle of mass `1kg`, executing `S.H.M.`, if slope of restoring force `vs` displacement graph is `=-1`, then the time period of oscillation with be `6.28s`. Statement-II : If `1kg` mass is replaced by `2 kg` mass and rest of the information remains same as in statement-1, then the time period of oscillation with remain `6.28s`.A. Statement-I is true, Statement-Ii is true, Statement-II is a correct explanation for Statement-I.B. Statement-I is true, Statement-II is true, Statement-II is NOT a correct explanantion for Statement-I.C. Statement-I is true, Statement-II is falseD. Statement-I is false, Statement-II is true |
|
Answer» Correct Answer - C As `F=- m omega^(2) y rArr` slope of `F -y` graph is `-m omega^(2) - 1 =- m omega^(2) =- omega^(2) =- omega^(2) rArr omega =1 T = 2pi = 6.28 s` If mass is changed but slope remains same, the time period will change. |
|
| 699. |
Displacement vs. time curve for a particle executing S.H.M. is shown in Fig. 14.4. Choose the correct statements.(a) Phase of the oscillator is same at t = 0 s and t = 2 s.(b) Phase of the oscillator is same at t = 2 s and t = 6 s.(c) Phase of the oscillator is same at t = 1 s and t = 7 s.(d) Phase of the oscillator is same at t = 1 s and t = 5 s. |
|
Answer» (b) Phase of the oscillator is same at t = 2 s and t = 6 s. (d) Phase of the oscillator is same at t = 1 s and t = 5 s. |
|
| 700. |
The length of a pendulum is halved. Its energy will beA. decreased to halfB. increased to 2 timesC. decreased to one fourthD. increased to 4 times |
|
Answer» Correct Answer - B `Eprop(1)/(l)` `(E_(2))/(E_(1))=(l_(1))/(l_(2))=2` `E_(2)=2E_(1)` |
|