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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.
| 151. |
A particle performing SHM starts from mean position. The phase of that particle is `pi//s` when it hasA. maximum displacementB. maximum velocityC. maximum energyD. maximum kinetic energy |
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Answer» Correct Answer - A maximum displacement |
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| 152. |
Give any three examples of e.m. waves. |
| Answer» Visible light, X-rays , Radio waves. | |
| 153. |
In a region of compression ,there is ……………….. in volume and a consequent …………………………. |
| Answer» temporary decrease, temporary increase in density | |
| 154. |
For propagation fo mechanical waves, the medium must possess ………………..,……………….,…………………. . |
| Answer» elasticity, inertia, minimum friction. | |
| 155. |
Change in temperature of the medium changesA. frequency of sound wavesB. amplitude of sound wavesC. wavelength of sound wavesD. loudness of sound waves |
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Answer» Correct Answer - C Change in temperature of the medium changes the velocity of sound waves and hence the wavelength of sound waves. This is because frequency `(v=(upsilon)/(lambda))` is fixed. |
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| 156. |
Which type of waves do not require a material medium for propagation? |
| Answer» Electromagnetic waves. | |
| 157. |
With propagation of longitudinal waves through a medium the quantity transmitted isA. matterB. energyC. energy and matterD. energy, matter and momentum |
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Answer» Correct Answer - B Propagation of longitudinal waves through a medium leads to transmission of energy through the medium. |
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| 158. |
The maximum acceleration of a simple harmonic oscillatore is `a_(0)` and maximum velocity is `v_(0)`. What is the displacement amplitude? |
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Answer» Let A be the amplitude and `omega` be the angular frequency of SHM. Then Maximum velocity, `v_(0)=omegaA or omega=v_(0)//A` Maximum acceleration, `a_(0)=omega^(2)A=(v_(0)^(2))/(A^(2))A=(v_(0)^(2))/(A)` `:.` Amplitude `A=(v_(0)^(2))/(a_(0))` |
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| 159. |
On an average a human heart is found to beat 75 times in a minute. Calculate its beat frequency and period. |
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Answer» The beat frequency of heart, `v=(75)/(60s)=1.25s^(-1)` The time period, `T=(1)/(v)=(1)/(1.25s^(-1))=0.8s` |
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| 160. |
Tha particle is executing SHM. The amplitude of motion is a. State those positions of the particle in terms of a when (i) the K.E. of the particle is zero (ii) P.E. is zero (iii) P.E. is one-fourth of the total energy (iv) P.E and K.E. are equal. |
| Answer» (i) `+-a` (ii)`+-a//2` (iv) `+-a//sqrt(2)` | |
| 161. |
The period of a particle in SHM is `8s`. At `t=0` it is at the mean position. The ratio of the distances travled by it in the first and the second second isA. `3.2:1`B. `4.2:1`C. `2.4:1`D. `1.6:` |
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Answer» Correct Answer - C `t_(1)=1s, t_(2)=2s` Now,`y_(1)=asinomegaxx1=asinomega` `y_(2)=asin omegaxx2-asinomega=asin2omega-asinomega` `:. (y_(2))/(y_(1))=(sin2omega-sinomega)/(sinomega)=(sin2omega)/(sinomega)-1` `=(2sinomegacosomega)/(sinomega)-1=2cosomega-1` `=2cos((2pi)/(T))-1=2cos((2pi)/(8))-1` or `(y_(2))/(y_(1))=2cos((pi)/(4))-1=2xx(1)/(sqrt(2))-1=sqrt(2)-1` `=1.414+1=2.414~~2.4:1` |
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| 162. |
A body weighing 10 kg is executing SHM, has a velocity of `6.0ms^(-1)`, after one second of starting from the mean position. If its time period is 6 second, find its K.E., P.E. and total energy. |
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Answer» Here, `m=10kg, V=6.0ms^(-1), t=1s, T=6s` K.E.=`(1)/(2)mV^(2)=(1)/(2)xx10xx6^(2)=180J` `V=aomegacosomegat=aomegacos((2pi)/(T))t` so `6=aomegacos (2pi)/(6)xx1=aomegaxx(1)/(2) or a omega=12` T.E.`=(1)/(2)momega^(2)a^(2)=(1)/(2)xx10xx(12)^(2)=720J :.` P.E.`=` T.E.-K.E. `=720-180=540J` |
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| 163. |
Time period of a simple pendulum is 2s and it can go to and fro from equilibrium position at a maximum distance of 6cm. If at the start of the motion the pendulum is in the position of maximum displacement towards the right of the equilibrium position, then write the displacement equation of the pendulum. |
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Answer» Here, `T=2s, A=6cm, ` `omega=(2pi)/(T)=(2pi)/(2pirad//s` Let `phi_(0)` be the initial phase of the pendulum, then displacement of particle executing SHM at time t is given by `y=Asin(omegat+phi_(0))=Asin((2pi)/(T)t+phi_(0))` `y=6sin((2pi)/(2)t+phi_(0))=6sin(pit+phi_(0))` when `t=0,y=6cm,` so `6=6sin(pixx0+phi_(0))=6sinphi_(0)` or `sinphi_(0)=1=sinpi//2or phi_(0)=(pi//2)rad` Hence displacement equation for the simple pendulum is `y=6sin(pit+pi//2)=6cospit` |
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| 164. |
Maximum velocity of a particle in SHM is `16cm^(-1)` . What is the average velocity during motion from one extreme position to other extreme position? |
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Answer» Given, Max. velocity `=aomega=16cms^(-1)` or `a(2pi)/(T)=16 or (a)/(T)=(1)/(2pi)` Average velocity, `v_(av)=(t otal displacement)/(t otal ti me)=(2a)/(T//2)=(4a)/(T)` `:.v_(av)=4xx(16)/(2pi)=(32)/(pi)cms^(-1)`. |
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| 165. |
Phase difference between two vibrating particles tells the …………………… in the vibrating states of the two particles at a given instant. |
| Answer» Correct Answer - lack of harmony | |
| 166. |
A transverse wave is represented by the equation `y=y_0sin.(2pi)/(lamda)(vt-x)` For what value of `lamda`, the maximum particle velocity equal to two times the wave velocity?A. `lambda=2piy_(0)`B. `lambda=(piy_(0))/(3)`C. `lambda=(piy_(0))/(2)`D. `lambda=piy_(0)` |
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Answer» Correct Answer - D Particle velocity `=(dy)/(dt)=[y_(0)cos((2pip)/(lambda))(upsilont-x)](2pi)/(lambda)upsilon` `((dy)/(dt))_(max)=y_(0)(2piupsilon)/(lambda)=2upsilon` `lambda=piy_(0)` |
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| 167. |
A wave, `y_( `(x, t)`)=0.03 sin pi(2t-0.01x)` tavels in a medium. Here, `x` is in metre. The instantaneous phase differenc ( in rad) between the two point separated by `25cm` isA. `pi//800`B. `pi//400`C. `pi//200`D. `pi//100` |
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Answer» Correct Answer - B `y=0.03sinpi(2t-0 .01x)` `=0.03sin(2pit-0.01pix)` …(i) Comparting it with the equation `y=rsin (omegat-(2pi)/(lambda)x)`, we have `(2pi)/(lambda)=0.01pi` or `lambda-(2)/(0.01)=200m` Phase diff, `Deltaphi=(2pi)/(lambda)Deltax=(2pi)/(200)xx(25)/(100)=(pi)/(400)rad` |
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| 168. |
The distance between two consecutive nodes in a stationary wave is 25cm. If the speed of the wave is `250ms^(-1)`, calculate the frequency. |
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Answer» Distance between two consecutive nodes is `(lambda)/(2)=25cm=0.25m,` `lambda=2xx0.25m =0.5m` `v=(upsilon)/(lambda)=(2500)/(0.5)=500Hz` |
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| 169. |
A force of `6.4N` streches a vertical spring by `0.1m` The mass that must be supended from the spring so that it oscilation with a time period of `pi//4` secondA. `2kg`B. `1kg`C. `1.5kg`D. `2.5kg` |
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Answer» Correct Answer - B Here, `F=6.4N, l=0.1m, T=pi//4s` Spring factor,`k=(6.4)/(0.1)64Nm^(-1)` Inertial factor`=` mass suspended `=`m In S.H.M. the time period is given by `T=2pisqrt((i n ertia fact o r)/(spri ng fact o r ))` `(pi)/(4)=2pisqrt((m)/(64))` or `(1)/(8)=sqrt((m)/(64))` squaring both sides `(1)/(64)=(m)/(64)` or `m=1kg` |
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| 170. |
The ratio of amplitude of two waves is `2:3`. What is the ration of intensities of these waves? |
| Answer» `(I_(1))/(I_(2))=(a^(2))/(b^(2))=(2^(2))/(3^(2))=(4)/(9)` | |
| 171. |
What type of graph you expect between speed of sound through a gas and pressure of the gas. |
| Answer» The graph will be straight line parallel to pressure exis, because at a given temp. there is no effect of change in pressure. | |
| 172. |
In a longitudinal wave, there is a state of maximum compression at a point at an instant. The frequency of the wave is 50Hz. After what time will the same point be in the state of maximum rarefaction ? |
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Answer» Time interval between a compression and adjoining rarefaction `(T)/(2)=(1)/(2n)=(1)/(2xx50)=(1)/(100)s` |
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| 173. |
How does velocity of sound in air change when temperature rises by `1^(@)C`? |
| Answer» Velocity of sound in air increases by `0.61m//s` when temperature rises by `1^(@)`C. | |
| 174. |
The amplitude of a simple harmonic oscillation is doubled. How does this affect (i) periodic time (ii) maximum velocity (iii) maximum acceleration and (iv) maximum energy |
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Answer» (i) As time period of oscillation os simple pendulum `T=2pisqrt((l)/(g))` is independent of amplitude of oscillation, hence time period of oscillation does no change with the change in the amplitude of oscillation. (ii) Max. velocity, `v_(0)=omegaA`. When amplitude A is doubled, then max. velocity becomed double. (iii) Max. acc., `a_(max)=omega^(2)A.` When amplitude A is doubled, then max, acceleration becomes double. (iv) Total energy `E=(1)/(2)m(2piv)^(2)A^(2)=2pi^(2)mv^(2)A^(2)` When amplitude A is doubled, then the energy of oscillator becomes four times. |
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| 175. |
Statement-1 `:` During the oscillation so f a simple pendulum, the direction of its acceleration at the mean position is directed towards the point of suspension and at exteme 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 -1 is true, Statement-2, Statement -2 is correct explanation of statement-1.B. Statement-1 is true, Statement-2 is true, Statement-2 is not a correct explanantion of Statement-1.C. Statement-1 is correct, Statement-2 is false.D. Statement-1 is false, Statement-2 is true. |
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Answer» Correct Answer - d When the length of pendulum makes an angle `theta` with the vertical direction, then the tangential component of weight `=mg sin theta` and radial component of weight `=mg cos theta. ` At extreme position, `theta` is maximum, so `mg sin theta ` is maximum, hence accelerated is directed towards mean position. At the mean position `theta=0`, so `mg sin theta ` is zero. The radial component of weight `=mg cos 0^(@)=mg` will provide the acceleration towards the point of suspension. |
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| 176. |
For an oscillating simple pendulum, is the tension in the string constant throughout the oscillation? If not, when it is (a) the least, (b) the greatest? |
| Answer» In simple pendulum, when bob is in deflection position, the tension in the string is `T=mgcostheta`. Since, the value of `theta` is differenet at different positions, hence tension in the string is not constant throughout the oscillation. (a) At end points, `theta` is maximum : the value of `costheta` is least, hence the value of tension in the string is least. (b) At th emean position, the value of `theta=0^(@)` and `cos0^(@)=1`, so the value of thension is greatest. | |
| 177. |
The tension in the string of a simple pnedulum isA. constantB. maximum in the extreme positionC. zero in the mean positionD. none of these |
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Answer» Correct Answer - D none of these |
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| 178. |
A spring mass system oscillates with a frequency v. If it is taken in an elavator slowly accelerating upward, the frequency willA. increaseB. decreaseC. remain sameD. become zero |
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Answer» Correct Answer - C remain same |
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| 179. |
A pipe 30.0 cm long is open at both ends. Which harmonic mode of the pipe resonates with a 1.1kHz source? Will resonance with the same source be observed if one end of the pipe is closed ? Take the speed of sound in air as `330ms^(-1)`. |
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Answer» Here, `L=30.0cm=0.3m` Let, nth harmonic of open pipe resonate with 1.1kHz source, i.e., `v_(n)=1.1kHz=110Hz` As, `v_(n) =(n upsilon)/(2L)` `:. n=(2Lv_(n))/(upsilon)=(2xx0.30xx1100)/(330)=2` i.e., 2nd harmonic resonates with open pipe. If one end of pipe is closed, its fundamental frequency `v_(1)=(upsilon)/(4L)=(330)/(4xx0.3)=275Hz.` As odd harmonics alone are produced in a closed organ pipe, therefore, possible freqencies are `3v_(1)=3xx275=825Hz, 5v_(1)=5xx275=1375Hz` and so on. As the source freqency is 1100 Hz, therefore, no resonance can occur when the pipe is closed at one end. |
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| 180. |
A series circuit consisting of a capacitor and a coil with active resistance is connected to a source of harmonic voltage whose frequency can be aried, keeping the voltage amplitdue are `n` times less than the resonance amplitude. Find `:` `(a)` the resonance frequency, `(b)` the quality factor of the circuit. |
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Answer» At resonance `omega_(0)L=(omega_(0)C)^(-1)` or `omega_(0)=(1)/(sqrt(LC))` `(I_(m))_(res)=(V_(m))/(R)`. Now `(V_(m))/( nR)=(V_(m))/sqrt(R^(2)+(omega_(1)L-(1)/(omega_(1)C))^(2))=(V_(m))/( sqrt(R^(2)+(omega_(2)L-(1)/( omega_(2)C))^(2)))` Then `omega_(1)L-(1)/( omega_(1)C)=sqrt(n^(2)-1)R` `omega_(2)L-(1)/( omega_(2)C)=sqrt(n^(2)-1)R` ( assuming `omega_(2) gt omega_(2))` or `omega_(1)-(omega_(0)^(2))/( omega_(1))=- omega_(2)+(omega_(0)^(2))/( omega_(2))=- sqrt (n^(2)-1 )(R)/(L)` or` omega_(1)+ omega_(2)=( omega_(0)^(2))/( omega_(1)omega_(2))(omega_(1)+ omega_(2))implies omega_(0)=sqrt( omega_(1) omega_(2))` and ` omega_(2) - omega_(1)= sqrt( n^(2)-1) (R)/(L)` `beta=(R)/(2L) = ( omega_(2)-omega_(1))/( 2 sqrt( n^(2)-1))` and `Q=sqrt( (omega_(0)^(2))/(4 beta^(2))-(1)/(4))=sqrt(((n^(2)-1)omega_(1)omega_(2))/( (omega_(2)-omega_(1))^(2))=(1)/(4))` |
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| 181. |
A circuit consisting of a capacitor with capacitance `C` and a resistance `R` connected in series was connected at the moment `t=0` to a source of ac voltage `V=V_(m) cos omegat` . Find the current in the circuit as a function of time `t`. |
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Answer» Here the equation is `(Q` is charge on the capacitor `)` `(Q)/(C)+R(dQ)/( dt)=V_(m) cos omegat` A solution subject to `Q=0 ` at `t=0` is of the form `(` as in the previous problem `)` `Q=Q_(m)[cos ( omegat-bar(varphi))-cos bar(varphi)e ^(-t//RC)]` Subtituting back `(Q_(m))/(C) cos ( omegat-bar(varphi))-omegaRQ_(m) sin ( omegat- bar( varphi))` `=V_( m) cos omega t ` `=V_(m){ cos bar(varphi)( omegat-bar(varphi))-sin bar(varphi) sin ( omegat-bar(varphi))}` so `Q_(m)=CV_(m) cos bar(varphi)` `omega RQ_(m)=V_(m) sin bar ( varphi)` This leads to `Q_(m)=(CV_(m))/( sqrt(1+ ( omegaRC)^(2))), tan bar(varphi)= omega RC` Hence `I=(dQ)/( dt)=(V_(m))/(sqrt(R^(2)+((1)/( omegaC))^(2)))[- sin omegat-bar(varphi)+( cos ^(2) bar(varphi))/( sin bar(varphi))e^(-t//RC)]` The solution given in the book satisfies `I=0` at `t=0`. Then `Q=0` at `t=0` but this will not satisfy the equation at `t=0`. Thus `I=0`, ( Equation will be satisfies with `I=0` only if `Q=0` at ` t=0)` With out `I, I(t=0) =(V_(m))/( R)` |
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| 182. |
When two waves of almost equal frequencies `n_(1)` and`n_(2)` reach at a point simultaneously, what is the time interval between successiv maxima ? |
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Answer» When two waves of almost equal frequencies `n_(1)` and `n_(2)` reach at a point simultaneously, beats are produced. No. of beats `//` sec `=(n_(1)-n_(2))` Time interval between two successive beats `=` time interval between two successive maxima `=(1)/((n_(1)-n_(2)))` |
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| 183. |
depict two circulat mothions. The radius of the circle, the period of revolution, the initial position and sense of revolution are indicated in the figures. Obtain the simple harmonic motinosof the x-projection of the radius vector of the rotating particle P in each case. |
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Answer» Refere to figure , `T=4s, ` intial pahse, `phi=45^(@)=(pi)/(4)rad` with the positive direction of x-axis. After a time t, it covers an angle `(2pi)/(T)t` in the anitclockwise direction and makes an angle `((2pi)/(T)t+(pi)/(4))` with the x-axis. The projectino of OP on the x-axis at time t is `x=acos((2ppit)/(T)+phi)=acos((2pit)/(4)+(pi)/(4))` Refer to figure `T=30s` inital phase, `phi=90^(@)=(pi)/(2)rad` with the x-axis. After a time t, it covers an angle `(2pit)/(T)` in the clockwise direction and makes an angle `((pi)/(2)-(2pit)/(T))` with the x-axis. Now, the projection of OP on the x-axis at time t is given by `x=bcos((pi)/(2)-(2pit)/(T))=bcos((2pit)/(T)-(pi)/(2))` `b cos((2pit)/(30)-(pi)/(2))=bcos((pit)/(15)-(pi)/(2))` |
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| 184. |
A tuning fork of unknown frequency gives 4beats with a tuning fork of frequency 310 Hz. It gives the same number of beats on filing. Find the unknown frequency. |
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Answer» Here, `m=4, n_(1)=310Hz, n_(2)=?` `n_(2)=n_(1)+-m=310+-4=314 or 306` On filing, frequency increases. Therefore, original frequency before filing `=306Hz` On filing, frequency increases to 314 giving four beats per second. |
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| 185. |
When two sets of progressive wave trains of the same type, having same amplitude an same time period, travelling with same speed along the same straight line in opposite directions superimpose, a new set of waves are formed. These are called stationary waves or standing waves. The resultant waves do not propagate in any direction, nor there is any transfer of energy in the medium. Read the above passage and answer the following questions: (i) Stationary waves are characterised by nodes and antinodes. What are these points ? (ii) What is the total energy associated with a stationary wave ? What values of life do you learn from this ? |
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Answer» (i) Nodes are the points in a medium in a stationary wave, which are permanantly at rest. Antinodes are the points in a medium in a stationary wave, the amplitude of vibration of which is maximum. (ii) The total energy associated with a stationary wave is twice the energy of each of incident and reflected wave. And there is no flow or transference of energy along the stationary wave. In stationary waves, disturbance is confined to a particular region. It does not travel on and on. In day to day life, it implies that when you confine your activities to a particular domain, all your energy is utilized in that domain and you become a great achiever. The idea points towards specialization in a particular field and becoming master of it, rather than diversifying and becoming jack of all trades, master of none. |
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| 186. |
A wave motion is a means of ………….. From one point to another without ………………… . |
| Answer» transferrring energy and momentum, without any actual transportation of matter between these points. | |
| 187. |
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). Its. |
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Answer» Here, `k=2xx10^(6)Nm^(-1)`, `a=2cm =0.02m` ltbtgt Total energy `=600J` As maximum K.E. `=`max. P.E. `=(1)/(2)ka^(2)` `=(1)/(2)xx(2xx10^(6))xx(0.02)^(2)=400J` In SHM P.E. is minimum at the mean position and K.E. is maximum at the mean position, whereas total meachanical energy is constant throughout. Therefore, at mean position total mechnaical energy `=max K.E.+min. P.E.` or `600=400+min. P.E.` or min. P.E.` =600-400=200J` |
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| 188. |
A linear harmonic oscillator of force ocnstant `2xx10^(6)N//m` and amplitude 0.01 m has a total mechanical energy of 160J. What is the maximum K.E. and minimum P.E.? |
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Answer» Here, maximum K.E. `=` max. PE `=(1)/(2)kr^(2)` `=(1)/(2)xx(2xx10^(6))xx(0.01)^(2)=100J` In SHM, P.E. is minimum at mean position and K.E. is maximum at mean position. But the total meachanical energy in SHM is constant. Therefore, at mean position., total mechanical energy `=` max. K.E. `+` minimum P.E. `160=100+` minimum P.E. So, minimum P.E. `=160-100=60J` |
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| 189. |
Define force constant of a sring. Give its SI unit and dimensional formula. |
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Answer» Force constant of a spring is defined as the force required to produce unit extension or compression in the spring i.e., `k=F//y`. The SI unit of k is `Nm ^(-1)` . Its dimensional formul `=MLT^(-2)//L=[M^(1)L^(0)T^(-2)]`. |
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| 190. |
Two tuning fork A and B give 5 beats `//` sec. A resounds with a closed column of air 15 cm long and B with an open column of air 30.5 cm long. Calculate their frequecies. Neglect end correction. |
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Answer» Freq. of A, `n_(1)=(upsilon)/(4l_(1))=(upsilon)/(4xx0.15)=(upsilon)/(0.6)` Freq. of B, `n_(2)=(upsilon)/(2l_(2))=(upsilon)/(2xx0.305)=(upsilon)/(0.61)` Beat frequency, `m=n_(1)-n_(2)` `5=(upsilon)/(0.6)-(upsilon)/(0.61)=((0.61-0.6)upsilon)/(0.6xx0.61)` `upsilon=(5xx0.6xx0.61)/(0.01)=183m//s` `:. n_(1)=(upsilon)/(0.6)=(183)/(0.6)=305Hz` `n_(2)=(upsilon)/(0.61)=(183)/(0.61)=300Hz` |
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| 191. |
Two tuning forks A and B give 9beats in 3 seconds. A resounds with a closed column of air 15cm long and B with an open column 30.5 cm long. Calculate their frequencies (neglect end correction). |
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Answer» For closed column, `l=15cm=0.15m` For open column, `l=30.5cm =0.305m` Let `v_(1)v_(2)`, be the frequencies of the tuning forks A and B respectively. As, tuning fork A resounds with a closed column `:. v_(2)=(upsilon)/(4l) =(upsilon)/(4xx0.15)=(upsilon)/(0.60)` ...(i) The fork B resounds with an open air column `:. v_(2)=(upsilon)/(2l)=(upsilon)/(4xx0.305)=(upsilon)/(0.61)` ...(ii) As, forks A and B produced` 9//3 =3` beats per second, therefore, `v_(1)-v_(2)=3` or `(upsilon)/(0.60)-(upsilon)/(0.61)=3` or `v(0.61-0.60)=3xx0.60xx0.61` or `upsilon=(3xx0.60xx0.61)/(0.01)=109.8ms^(-1)` `v_(1)=(upsilon)/(0.60)=(109.8)/(0.60)=183Hz` and `v_(2)=(upsilon)/(0.61)=(109.8)/(0.61)=180Hz` |
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| 192. |
Find the quality factor of the oscillator whose displacement amplitude decreases ` eta =2.0` times every `n=110` oscillation. |
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Answer» The displacement amplitude decrease `eta` times every `n` oscillations . Thus `(1)/(eta)=e^(-beta.(2pi)/(omega).n)` or ` (2pin beta)/(omega)=i n eta` or `(beta)/(omega)=(1n eta)/(2pi n ). ` `Q=( omega)/(2 beta)=( pi n )/( 1n eta)~~ 499` |
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| 193. |
Displacement versus time curve for a particle executing S.H.M. is shown in figure .Identidy the points marked at which (i) velocity of the oscillator is zero, (ii) speed of the oscillator is maximum. |
| Answer» We know that in SHM, the velocity is zero when particle is at the extreme position (i.e., displacement is maximum) and velocity is maximum when particle is at the mean position, i.e., displacement is zero. Thus velocity os oscillator is zero at locations A,C,E, and G. The speed of oscillator is maximum at locations B,DD,F and H. | |
| 194. |
A body is executing simple harmonic motion. At a displacement `x` its potential energy is `E_(1)` and at a displacement `y` its potential energy is `E_(2)` The potential energy `E` at displacement `(x+y)` isA. `E_(1)+E_(2)`B. `sqrt(E_(1)E_(2))`C. `sqrt(E_(1)^(2)+E_(2)^(2))`D. `E_(1)+E_(2)+2sqrt(E_(1)E_(2))` |
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Answer» Correct Answer - D `E_(1)=(1)/(2)momegat^(2)x^(2), E_(2)=(1)/(2)momega^(2)y^(2)` `:. sqrt(E_(1)E_(2))=[(1)/(2)m omega^(2)x^(2)xx(1)/(2)m omega^(2)y^(2)]^(1//2` `=(1)/(2)momega^(2)xy` `E=(1)/(2)momega^(2)(x+y)^(2)=(1)/(2)momega^(2)(x^(2)+y^(2)+2xy)` `(1)/(2)momega^(2)x^(2)+(1)/(2)momega^(2)y^(2)=2xx(1)/(2)momega^(2)xy` `=E_(1)+E_(2)+2sqrt(E_(1)E_(2))` |
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| 195. |
The displacement time graph of a particle executing S.H.M. is shown in figure. 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)` |
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Answer» Correct Answer - A::B::C For the given SHM, the displacement is given by `y=acos omegat` Velocity, `V=(dy)/(dt)=-aomega sin omegat=aomega sin (omegat+pi)` Force `=` mass `xx` acceleration `=-m a omega^(2) cos omega t` Force is zero, when `cos omega t =0 or omega t =(pi)/(2) or (3pi)/(2), i.e., (2pi)/(T)t=(pi)/(2) or (3pi)/(2)` If `(2pit)/(T)t=(pi)/(2),` then `t=(T)/(4)` If `(2pi)/(T)t=(3pi)/(2), ` then `t=(3T)/(4)s` (given) Acceleration is maximum if `cos omegat=1 or omega t=0 or 2pi or (2pi)/(T)t=2pi or t=T=(4T)/(4)s` Velocity is maximum if `sin(omegat+pi)=1 or omegat+pi=pi//2` or `omegat=(pi)/(2)-pi=-pi//2 or (2pi)/(T)t=-(pi)/(2) or t=-(T)/(4)s` `PE=(1)/(2)momega^(2)y^(2)=(1)/(2)momega^(2)a^(2)cos^(2)omegat` `KE=(1)/(2)momega^(2)a^(2)sin^(2)omegat` If `PE=KE ` then `cos^(2)omegat=sin^(2)omegat=sin^(2)omegat or cosomegat=sinomegat or tanomegat=1` or `omegat=(pi)/(4) or (2pi)/(T)t=(pi)/(4) or t=(T)/(8)s` |
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| 196. |
Displacement vs. time curve for a particle executing S.H.M. is shown in figure. Choose the correct statements A. Phase of the oscillator is same at `t=0s` and `t=2s`B. Phase of the oscillatore is same at `t=2s` and `t=6s`C. Phase of the oscillator is same at `t=1s` and `t=7s`D. Phase of the oscillatore is same at `t=1s` and `t=5s` |
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Answer» Correct Answer - B::D Phase is the state of a particle as regards its position and direction of motion w.r.t. mean position. In the given curve, phase is same when `t=1` and `t=5`s. Also phase is same when `t=2s` and `t=6s.` |
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| 197. |
Which of the following statements is`//` are true for a simple harmonic oscillator ?A. Force acting is directly proportional to displacement from the mean position and opposite to it.B. Motion is periodicC. Acceleration of the oscillator is constantD. The velocity is periodic |
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Answer» Correct Answer - A::B::D Knowledge based question. |
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| 198. |
Assertion : Resonance is special case of force vibration in which the nature frequency of vebration of the body is the same as the impressed frequency of external periodic force and the amplitude of force vibration is maximum Reason: The amplitude of forced vibrations of a bodyincrease with an increase in the frequency of the externally impressed perioic forceA. both, Assertion and Reason are true and Reason is the correct explanation of the Assertion.B. both, Assertion and Reason are true but Reason is not a correct explanation of the Assertion.C. Assertion is true but the Reason is false.D. both,Assertion and Reason are false. |
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Answer» Correct Answer - c Here Assertion is correct but Reason is wrong. As the amplitude of forced vibration decreases on changing the frequency of the impressed force on either side of the resonant frequency. |
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| 199. |
Find the displacement of a simple harmonic oscillator at which its P.E. is half of the maximum energy of the oscillator. |
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Answer» Let at an instatn the displacement of the particle from the mean position be y. Then its PE is `E_(P)=(1)/(2)momega^(2)y^(2)` Total energy, `E=(1)/(2)momega^(2)r^(2)` As per question, `PE=(1)/(2)xxTE` `:.(1)/(2)momega^(2)y^(2)=(1)/(2)xx(1)/(2)momega^(2)r^(2) or y^(2)=(r^(2))/(2) or y=+-(r)/(sqrt(2))` |
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| 200. |
At what displacement (i) the P.E. of a simple harmonic oscillator is maximum and minum (ii) the K.E. is maximum and minmum? |
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Answer» The P.E. of a particle executing SHM is given by, `U=(1)/(2)momega^(2)y^(2)` U is maximum when `y=a=` amplitude of vibration i.e., the particle is passing from the extremen position and is minimum when `y=0, ` i.e., the particle is passing from the mean position: The K.E. of a particle executing SHM is given by `K=(1)/(2)momega^(2)(a^(2)-y^(2))` K is maximum when `y=0`, i.e., the particle is passing from the mean position and K is minumum when `y=a` i.e., the particle is passing from the extremen position. |
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