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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.
| 101. |
The displacement from the position of equilibrium of a point 4 cm from a source of sinusoidal oscillations is half the amplitude at the moment `t=T//6` (T is the time perios). Assume that the source was at mean position at `t=0`. The wavelength of the running wave is :A. `0.96cm`B. `0.48cm`C. `0.24cm`D. `0.12cm` |
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Answer» Correct Answer - B Equation of wave is `lamda=Asin(omegat-kx)` `implies(A)/(2)=Asin((2pi)/(T)xx(T)/(6)-(2pi)/(lamda)xx4)` `implies(pi)/(6)=(pi)/(3)-(2pi)/(lamda)xx4implies(2pi)/(lamda)xx4=(pi)/(6)` `lamda=48cm` `lamda=0.48m` |
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| 102. |
Assertion : The speed of soun in solids is maximum though density is large. Reason : The coefficient of elasticity of solid is large.A. If both assertion and reason are true and reason is the correct expleanation of assertion.B. If both assertion and reason are true and reason is not the correct expleanation of assertion.C. If assertion is true but reason is false.D. If assertion and reason both are false. |
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Answer» Correct Answer - A The velocity of sound in solid is given by `v=sqrt((R)/(rho))`. Through`rho` is large for solids, but their coefficient of elasticity E is much larger (compared to that of liquids and gases). That is why `v` is maximum in case of solid. |
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| 103. |
Two parts of sonometer wire, divided by a movable knife edge differ in length by 2 cm and produce 1 beat/s when sounded together. Assume fundamental frequencies. If the total length of the wire is 100 cm. The frequencies. If the total length of the wire is 100 cm. The frequencies of the two parts of the wire areA. `51Hz`.`50Hz`B. `50.5Hz`,`49.5Hz`C. `25Hz`,`24Hz`D. `25.5Hz`,`24.5Hz` |
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Answer» Correct Answer - D Let the two parts are `L_1` and `L_2` then `L_1=49cm` and `L_2=51cm` For same tension `fprop(1)/(L)` `impliesf_1L_1=f_2L_2` `implies49f_1=51f_2` .(i) Also `f_1-f_2=1` .(ii) From (i) and (ii): `f_1=25.5Hz` `f_2=24.5` |
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| 104. |
Two sources of sound placed close to each other are wmitting progressive waves given by `y_1=4sin600pit` and `y_2=5sin608pit`. An observer located near these two sources of sound will hear:A. 4 beats per second with intensity ratio 81:1 between waxing and waningB. 4 beats per second with intensity ratio 25:16 between waxing and waningC. 8 beats per second with intensity ratio 25:16 between waxing and waningD. 8 beats per second with intensity ratio `81:1` between waxing and waning |
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Answer» Correct Answer - A `(I_1)/(I_2)=((A_1+A_2)^2)/((A_1-A_2)^2)=(81)/(1)` Beat frequency `=304-300=4Hz` |
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| 105. |
Two vibrating string of same length, same cross section area and stretched to same tension is made of meterials with densities `rho` and `2rho`. Each string is fixed at both ends If `v_0` represents the fundamental mode of vibration of the one made with density `rho` and `v_2` for another Then `v_1//v_2` is:A. `(1)/(2)`B. `2`C. `sqrt2`D. `(1)/(sqrt2)` |
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Answer» Correct Answer - C Velocity of sound is inversely proportional to the square rood of density of the medium. i.e., `V.rho=`constant `implies(V_1)/(V_2)=sqrt((rho_2)/(rho_1))=sqrt((2rho)/(rho))=sqrt2` |
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| 106. |
In the Q.89 the speed of the component progressive wave is:A. `20cm//sec`B. `6cm//sec`C. `120cm//sec`D. `40cm//sec` |
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Answer» Correct Answer - C `V=nlamda=20xx6=120cm//sec` |
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| 107. |
In the Q.89 the frequency of the component progressive wave isA. `20Hz`B. `40Hz`C. `((1)/(2))Hz`D. `((1)/(10))`Hz |
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Answer» Correct Answer - A `n=(1//T)=20Hz` |
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| 108. |
In the `5th` overtone of an open organ pipe, these are (N-stands for nodes and A- for antinodes)A. `2N,3A`B. `3N,4A`C. `4N,5A`D. `5N,4A` |
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Answer» Correct Answer - C In an open organ pipe, `5th` overtone corresponds to `4th` harmonic mode. Also in open pipe, number of nodes`=`order of mode of vibration and number of antinodes `=` (number of nodes `+1`). Here number of nodes `=4`, number of antinodes`=4+1=5`. |
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| 109. |
In the Q. 89 The separation between two consecutive antinodes is:A. `3 cm`B. `6 cmC. `12 cm`D. `20 cm` |
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Answer» Correct Answer - A Separation between two consecutive antinodes `=(lamda)/(2)=(6)/(2)=3cm` |
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