InterviewSolution
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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.
| 1. |
Two astronauts go the surface of the moon and would speak to each other. Explain how they could achieve this. |
| Answer» Absence of atmosphere on the surface of moon | |
| 2. |
When a disturbance is created in a medium, the particles in the medium perform ____A. non-periodic motionB. circular motionC. translatory motionD. periodic motion |
| Answer» Correct Answer - D | |
| 3. |
Assertion (A): Sound cannot be propagated on the moon. Reason (R) : Sound requires a medium to propagateA. A and R are correct and R is the correct explanation of AB. A and R are correct but R is not the correct explanation of AC. A is correct but R is not correctD. Both A and R are not correct |
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Answer» Correct Answer - A As there is no atmosphere on the moon, sound cannot propagate on the moon. Because, sound requires a medium to propagate |
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| 4. |
A sound wave of frequency 660 Hz incidents normally on a perfectly reflecting surface. If the speed of sound in air is `330 m s^(-1)`, what is the time taken by a particle in the medium to complete one vibration? Determine the wave length of the wave. |
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Answer» Frequency of the wave (n ) = 660 Hz Speed of the sound wave `(V) = 330 ms^(-1)` `:.` Wavelength `(lamda) (V)/(n) = (330)/(660) = (1)/(2) I` Time period, `T = (1)/(f) = (1)/(660) s` |
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| 5. |
Two pendula of the same length oscillate with amplitudes of 3 cm and 5 cm, respectively. The average time period noted for 3 trials for the first pendulum is found to be 2s, what will be the average time period for the second pendulum ? |
| Answer» The average time period for the second pendulum is also 2s. Because for a given length, the time period is independent of amplitude of oscillation | |
| 6. |
Both sound and light waves can be propagated throughA. vacuumB. airC. Both (a) and (b)D. None of the above |
| Answer» Correct Answer - B | |
| 7. |
Frequency of a sound wave produced by a vibrating body is 50 Hz. Find the wavelength of the sound wave produced. `(V_("air") = 330 ms^(-1))` |
| Answer» `V = n lamda rArr lamda = (V)/(n) = (330)/(50) = 6.6m` | |
| 8. |
Find the ratio of lengths of two pendula whose time periods are in the ratio `1 : 4` |
| Answer» `l_(1) : l_(1) = T_(1)^(2) : T_(2)^(2) = (12)/(4^(2)) = 1 : 16` | |
| 9. |
Find the ratio of time periods of two pendula whose lengths are in the ratio `1 : 4` |
| Answer» `(T_(1))/(T_(2)) = sqrt((l_(1))/(l_(2))) = sqrt((1)/(4)) = (1)/(2) rArr T_(1) : T_(2) = 1:2` | |
| 10. |
The length of the seconds pendulums is first increased by 10 cm and then decreased by 5 cm. If the time period is determined in each case, find their ratio (Take `g = 10 ms^(-2), pi^(2) ~= 10`) |
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Answer» (i) `T_(1) = 2pi sqrt((l + l^(1))/(g)), T_(2) = 2pi (l^(1) - l^(11))/(g)` (ii) `42 : 41` |
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| 11. |
Following are down the steps of an experiment to prove that medium is required for the propagation of sound waves. Arrange these steps in propagation of sound waves. Arrange these steps in proper sequence (a) Consider a glass jar with an outlet connected to a vacuum pump and a closed lid (b) Suspend an electric bell. (c) Evacuate the jar by using a vaccum pump (d) Switch on the electric bell (e) Observe that the intensity of the sound is less when the jar is evacuated (f) This proves that medium is required for the propagation of soundA. f e d a b cB. a b d e c fC. a b d c e fD. a b c d e f |
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Answer» Correct Answer - C Take a glass jar with closed lid and an outlet connected to a vaccum pump (a) Suspend an electric bell (b) and switch it on (d) Now evacuate the jar (c) and observe that intensity of sound is decreased when the jar is evacuated (e). This proves that a medium is required for the propagation of sound (f) |
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| 12. |
Which of the following statements is incorrect ?A. Sound travels faster in summer than in winterB. Sound travels in a straight lineC. Sound travels faster in vacuum than in airD. Sound travels in the form of longitudinal mechanical waves |
| Answer» Correct Answer - C | |
| 13. |
In a transverse wave, the time interval between 1st crest and 11th crest is 50s, then the time period of the wave is _____A. 5sB. 10 sC. 20 sD. 2 s |
| Answer» Correct Answer - A | |
| 14. |
If the time periods of two pendulums are 2s and 4s, then the ratio of their lengths is _____ |
| Answer» Correct Answer - `1 : 4` | |
| 15. |
Instrument (s) capable of producing musical notes is/are _____A. violaB. oboeC. celloD. All the above |
| Answer» Correct Answer - D | |
| 16. |
The loudness of sound of the bell in the bell jar experiment gradually decreases with _____A. decrease in the quantity of airB. increase in the quantity of airC. increase in the atmospheric pressureD. None of the above |
| Answer» Correct Answer - A | |
| 17. |
The bob of an oscillating simple pendulum arrives at one of the extreme positions 100 times in 200s. Then, the time period of the pendulum is ____A. `2.5`B. `2.0`C. `1.5`D. `1.0` |
| Answer» Correct Answer - B | |
| 18. |
If the period of a simple pendulum is 2s, then its frequency is `0.5 Hz` |
| Answer» Correct Answer - True | |
| 19. |
A disc siren with certain number of holes roates uniformly 270 times in one and half minute. The frequency of the note emitted is 48 Hz. If the velocity of sound in air is `336 m s^(-1)`, find its wavelength and number of holes on the disc. |
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Answer» (i) `lamda = (v)/(n)`, No. of holes `= ("frequency")/("No. of ratation in one sec")` (ii) 16 |
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| 20. |
The adjacent figure shows a displacement vs distance graph of a wave. If the velocity of the wave is `14 ms^(-1)`, calculate its (a) wavelength , (b) frequency and (c) amplitude of the wave |
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Answer» From the graph, (a) wavelength = 2m (b) frequency, `n (v)/(lamda) = (14)/(2) = 7Hz` and (c) amplitude of the wave `= 7 cm` |
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| 21. |
The wavelength of a sound wave travelling in a solid becomes one third when it propagates through air. If the frequency of the wave remains constant, then find the decreases in the velcoity of the wave |
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Answer» (i) `(V_(1))/(V_(2)) = (lamda_(1))/(lamda_(2)), V_(1) - V_(2)` (ii) `2//3` |
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| 22. |
The velocity of sound increases by 75% when it enters a liquid from air. Find the percentage increase in its wavelength |
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Answer» `V_(1) = V` `V_(2) = V + (75)/(100) V = 1.75V` `= (V_(1))/(V_(2)) = (lamda_(1))/(lamda_(2))` `(V)/(1.75 lamda_(1)) = (lamda_(1))/(lamda_(2))` `rArr lamda_(1) = 1.75 lamda_(1)` Increase in wavelength `= lamda_(2) - lamda_(1) = 1.75 lamda_(1) - lamda_(1) = 0.75 lamda_(1)` Percentage increase in wavelength `= (lamda_(2) xx lamda_(1))/(lamda_(1)) xx 100` `= (0.75 lamda_(1))/(lamda_(1)) xx 100 = 75%` |
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| 23. |
A stone is dropped from the top of a tower and it hits the groud at `t = 3.5s`. If the velocity of sound in air is `300 ms^(-1)`, find the time taken to hear the sound by a person on the top of the tower, from the instant the body is dropped `(g = 10 ms^(-2))` |
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Answer» (i) `S = (1)/(2) "gt"^(2), V = (d)/(t)` (ii) `3.75 s` |
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