Saved Bookmarks
This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
The correct statement among the following isA. sounds with frequency greater than 20 kHz are known as ultra-sonics .B. Dogs can hear ultrasonic sounds .C. In SONAR , ultra-sonics are used .D. All the above |
| Answer» Correct Answer - D | |
| 2. |
A man standing at a point on the line joining the feet of two cliffs fires a bullet . If he hears the `1^(st)` echo after 4 seconds and the next after 6 seconds , then what is the distance between the two cliffs ? (Take the velocity of sound in air as `330 m s^(-1)`) |
|
Answer» (i) `V = (2d)/(t)` (ii) Apply the formula d = Vt for both the cliffs. Then add the two distances to get the distance between the two cliffs. (iii) 1650 or 1.65 km . |
|
| 3. |
If wind blows in a direction opposite to the sound propagation , then the velocity of the soundA. increasesB. decreasesC. remains constantD. Cannot be determined |
|
Answer» Correct Answer - B If the wind blows in the opposite direction to the direction of propagation of sound , the velocity of sound direction. |
|
| 4. |
A source of sound and a listener are moving towards each other . The velocity of the source is `20 m s^(-1)` and that of the observer is `15 m s^(-1)` . If the velocity of sound is 340 m `s^(-1)` and its frequency is 640 Hz, find the apparent frequency of the sound . |
| Answer» Apparent frequency , n = `((V " "V)/(V" "V)) n = ((340 + 15)/(340 - 20)) 640 = 710` Hz | |
| 5. |
The velocity of sound in a gas is directly proportional to the square root of the temperature of the gas taken in degree celsius . |
| Answer» Correct Answer - False | |
| 6. |
A particle executing SHM completes 120`pi` vibrations in one minute . What is the frequency of this motion ?A. 5 HzB. 2`pi` HzC. 1 HzD. None of the above. |
|
Answer» Correct Answer - B Frequency = `(120 pi)/(60) = 2pi` Hz |
|
| 7. |
A string vibrating with a fundamental frequency of 8 Hz has tension `T_(1)`. This string vibrates with a fundamental frequency of 15 Hz when the tension is `T_(2)` . Find the fundamental frequency of the string when its tension is `T_(1) + T_(2)` . |
|
Answer» (i) Relation between frequency and tension of a stretched string . (ii) `n prop sqrtT` (iii) 17 Hz |
|
| 8. |
A source of longitudinal waves vibrates 320 times in two seconds. If the velocity of this wave in the air is `240 m s^(-1)` , find the wavelength of the wave . |
|
Answer» Velocity of wave , v = `240 m s^(-1)` Frequency of the wave , n = `(320)/(2) = 160` hertz Velocity of wave , `v = n lambda ` Wave length , `lambda = (v)/(n) = (240)/(160) = 1.5` m |
|
| 9. |
An air column enclosed in an open pipe is vibrating in its fundamental mode . The fundamental frequency is 30 Hz. If the velocity of sound air is `300 m s ^(-1)` , find the length of the pipe and frequency of 3rd overtone . |
|
Answer» Fundamental frequency of an open pipe = `(v)/(2l)` `implies 30 Hz = (300 m s^(-1))/(2l)` `implies l = (300)/(2 xx 30) = 5` m Frequency of third overtone = `4 xx n = 4 xx 30 = 120` Hz |
|
| 10. |
In simple harmonic motion , the acceleration of the body is inversely proportional to its displacement from the mean position . |
| Answer» Correct Answer - False | |
| 11. |
The following graph shows the displacement of the bob from its mean position versus time . The time period and the amplitude of the bob are : A. 4s , 5 cmB. 8s , 10 cmC. 4s , 10 cmD. 8s , 5 cm |
| Answer» Correct Answer - B | |
| 12. |
The special technique used in ships to calculate the depth of ocean beds isA. LASERB. SONARC. sonic boomD. reverberation |
| Answer» Correct Answer - B | |
| 13. |
A scooterist moves towards a vertical wall with a speed of 54 km `h^(-1)` . A person is standing on the ground and is behind the scooter , hears the sound . If the scooterist sounds the horn of frequency 400 Hz , calculate the apparent frequency sound heard by the person when (a) It is coming directly from the horn. (b) coming after reflection from the vertical wall. (Take speed of sound to be `330 m s^(-1)`) |
|
Answer» The speed of scooter = 54 km `h^(-1)` `= 54 xx 5//18 = 15 m s^(-1)` The frequency of the sound of horn = 400 Hz (a) The observer with reference to the ground is at rest `implies` Velocity of observer = `0 m s^(-1)` The source of moving away from observer . `therefore` The velocity of source = `15 ms^(-1)` The apparent frequency heard by the observer is `= n^(1) = (V - V_(o))/(V + V_(s)) n = (330 - 0)/(330 + 15) xx 400` `= 0.956 xx 400 = 382 . 4` Hz (b) The apparent frequency of sound received after from the wall is `= n' = ( 330)/(330 - 15) xx 400 = (330)/(315) xx 400 = 419` Hz [This is possible because the vertical wall reflects the sound without changing the frequency ]. |
|
| 14. |
A wound wave propagates in a medium which has the property/properties ofA. inertiaB. elasticityC. Both (a) and (b)D. Neither (a) and (b) |
|
Answer» Correct Answer - C The medium through which the sound waves are propagated must have both inertia and elasticity . |
|
| 15. |
When a sound wave passes from a highly polluted region to a pollution-free area , which of the following physical quantities remain unaltered ?A. amplitudeB. VelocityC. frequencyD. Wavelength |
|
Answer» Correct Answer - C Frequency of sound depends on the source . |
|
| 16. |
A rope of length of 2m is tied between two ends . If the speed of transverse waves propagating in the rope is `4 m s^(-1)` and the tension in the rope is 2 N , find the mass of the rope is C.G.S units . |
|
Answer» (i) `V = sqrt((T)/(m))` (ii) 250 g |
|
| 17. |
Two monoatomic gases of equal masses are in two different containers at S.T.P. if the ratio of velocities of sound in them is `1 :2` , then find the ratio of their volumes . |
|
Answer» Ratio of the velocities of sound in them = `1:2` we know that velocity of sound in a gas is inversely proportional to square root of molecular weight . `therefore V porp (1)/(sqrtM)` `M prop (1)/(V^(2))` `(M_(1))/(M_(2)) = ((V_(2))/(V_(1)))^(2) = ((2)/(1))^(2) = 4 :1` But PV = n.RT `V = nR((T)/(p))` `V prop (m)/(M) [ because T, p and R` are constant and n = `(m)/(M)`] `therefore V prop (1)/(M) [ therefore` m is also constant ] `therefore (V_(1))/(V_(2)) = (M_(2))/(M_(1)) = 1 :4` Alternate method : `V prop (1)/(sqrtd) ` or `V prop (1)/(sqrt((m)/(V)))` or `(V_(1))/(V_(2)) = sqrt((V_(1))/(V_(2)))` `therefore (V_(1)^(1))/(V_(2)^(1)) = (V_(1)^(2))/(V_(2)^(2)) = (1)/(2^(2)) = (1)/(4)` |
|
| 18. |
At S.T.P. the ratio of volumes occupied by 1 mole of each `O_(2)` and `CO_(2)` gases , respectively , is `"______"`A. `4 :1`B. ` 1: 4`C. `1: 2`D. `1 : 1` |
|
Answer» Correct Answer - D At S.T.P all gases of one mole occupies equal volume of 22.4 litre . `therefore` Ratio of their volumes is `1 :1`. |
|
| 19. |
The driver of a car approaching a cliff with a uniform velocity of 15 `m s^(-1)` sounds the horn and the echo is heard by the driver after 3 seconds . If the velocity of sound is `330 m s^(-1)` , calculate the distance between the cliff and the point where the horn was sounded ? Also calculate the distance between the cliff and the point where the echo is heard ? |
|
Answer» (i) 517.5 m , 472.5 m (ii) Equation for echo and velocity of sound (iii) Draw a rough linear figure depicting the positions of the car approaching the cliff when it sounded the horn , and when the echo is heard by its driver and the position of the cliff . From the figure find the distance travelled by the sound in the given time interval . For this consider the distance travelled by the car in the given time interval . Using the formula , velocity of sound = `("Distance travelled by the sound")/("time to heat the echo")` estimate the distance of the car from the cliff when it sounded the horn . |
|
| 20. |
A source of wave vibrates with a frequency 500 Hz. The wave travels 33m in `0.1`s . How far does the waves travels when the source executes 150 vibrations ? |
|
Answer» The frequency of the tuning fork = 500 Hz . The time taken for one wave to pass - `(1)/(500)` s `therefore` Therefore the time taken for 150 vibrations (waves ) `= (1)/(500) xx 150 = 0.3`s Hence the distance travelled by sound in 0.3 s is `= 0.3 xx 330` `= 99.0 m ` |
|
| 21. |
In a stationary wave , the phase difference between the particles in a given loop is `"_____"` . |
| Answer» Correct Answer - zero | |
| 22. |
Calculate the frequency of fifth harmonic of a closed organ pipe of length 50cm, if the velocity of sound in air is 330 m/s. |
| Answer» Correct Answer - 165 Hz | |
| 23. |
The velocity of sound is `"_____"` when the density of a gas is quadrupted , with the pressure remaining constant . |
| Answer» Correct Answer - halved | |
| 24. |
The velocity of sound in a gas is `30 m s^(-1)` at `27^(@)`C . What is the velocity of the sound in the same gas at `127^(@)C`?A. 20 `m s^(-1)`B. `30 m s^(-1)`C. `20 sqrt3 m s^(-1)`D. `60 m s^(-1)` |
| Answer» Correct Answer - C | |