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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. |
In a sonometer wire, the tension is maintained by suspending a 50.7 kg mass from the free end of the wire. The suspended mass has a volume of 0.0075 m 3. The fundamental frequency of the wire is 260 Hz . If the suspended mass is completely submerged in water, the fundamental frequency will become (take `g = 10 ms^(-2)`) [A. 220 HzB. 230 HzC. 240 HzD. 260 Hz |
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Answer» Correct Answer - C `n_2/n_1 = sqrt(T_2/T_1)` By archimedes principle `=T_1-T_2` `=vrhog` = loss of weight in water `therefore T_2=T_1-vrhog` `n_2/n_1 = sqrt((T_1-vrhog)/(T_1))` `=sqrt(((50.7-7.5xx10^(-3)xx10^3)g)/(50.7g))` `=sqrt((43.2)/(50.7)) = sqrt(144/169)` `n_2/n_1 = 12/13 therefore n_2 = 12/13 xx260 = 240Hz` |
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| 2. |
Fundamental frequency of pipe is 100 Hz and other two frequencies are 300 Hz and 500 Hz thenA. Pipe is open at both the endsB. Pipe is closed at both the endsC. One end open and another end is closedD. None of the above |
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Answer» Correct Answer - C `2n=n_2-n_1=500-300` `2n=200` n=100 The pipe is open at oen end and closed to other end. |
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| 3. |
A stretched wire of lenth 110 cm is divided into three segments whose frequencies are in ratio `1 : 2 : 3.` Their length must beA. 20 cm, 30 cm, 60 cm,B. 60 cm, 30 cm, 20 cmC. 60 cm, 20 cm, 30 cm,D. 30 cm, 60 cm, 20 cm |
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Answer» Correct Answer - b Given `l_(1)+l_(2)+l_(3)=110` cm and `n_(1)l_(1)=n_(2)l_(2)=n_(3)l_(3)` (`because` T and d are same) `n_(1):n_(2):n_(3)::1 : 2: 3` `because n_(1)/n_(2)=1/2=l_(2)/l_(1)rArr l_(2)=l_(1)/2` and `because n_(1)/n_(3)=1/3=l_(3)/l_(1)rArr l_(3)=l_(1)/3` `therefore l_(1) +l_(1)/2+l_(1)/3=110` `rArr l_(1) = 60 ` cm So, ` l_(1) = 60 ` cm, `l_(2) = 30` cm, ` l_(3) = 20` cm |
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| 4. |
The integral multiple of fundamental frequencies . Which are actully present in the sound isA. HarmonicsB. ResonanceC. OvertonesD. Forced vibration |
| Answer» Correct Answer - C | |
| 5. |
When a string is divided into three segments of lengths `l_(1),l_(2) and l_(3)` the fundamental frequencies of these three segments are `v_(1), v_(2) and v_(3)` respectively. The original fundamental frequency (v) of the string isA. `sqrt(v)= sqrt(v_(1))+sqrt(v_(2))+sqrt(v_(3))`B. `v= v_(1) + v_(2)+v_(3)`C. `1/v=1/v_(1) + 1/v_(2)+1/v_(3)`D. `1/sqrt(v)= 1/sqrt(v_(1))+1/sqrt(v_(2))+1/sqrt(v_(3))` |
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Answer» Correct Answer - c The fundamental frequency of string `v=1/(2l)sqrt(T/m)` `therefore v_(1)l_(1)=v_(2)l_(2)=v_(3)l_(3)=k ... (i) ` Form Eq. (i), `l_(1)=k/v_(1), l_(2)=k/v_(2), l_(3)=k/v_(3)` Original length, `l=k/v` Here, `l=l_(1)+L_(2)+l_(3)` `k/v=k/v_(1)+k/v_(2)+k/v_(3)` `rArr 1/v=1/v_(1)+1/v_(2)+1/v_(3)` |
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| 6. |
When the string of the sonometer of length l between the bridges vibrates in the first overtone, then the antinode formed from one end of the string will beA. `lamda/2`B. `lamda/4 " and " (3lamda)/(4)`C. `lamda/6, (3lamda)/(6) " and " (5lamda)/(6)`D. `lamda/8, (3lamda)/(8) " and " (7lamda)/(8)` |
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Answer» Correct Answer - B In the first overtone second harmonics i.e. second mode of vibration |
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| 7. |
The length of the stretched sonometer wire is 1m, if the fundamental frequency are in the ratio 1:2 . Where should be the bridge be placed to divide the wire two segments ?A. `1/3 : 2/3`B. `2/3:1/3`C. `3/2:3`D. `2:3` |
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Answer» Correct Answer - B `n_1/n_2 = l_2/l_1 = 1/2 therefore l_1=2l_2` `l_1+l_2 = 1 " hence " 2l_2 +l_2 =1 therefore l_2=1/3` `2l_2 +l_2 = 1` `l_2=1/3 m therefore l_1=2/3m` |
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| 8. |
A sonometer wire of length 100cm has fundamental frequency of 320Hz. Then the velocity of the transverse waves is ,A. 640 m/sB. 160 m/sC. 320 m/sD. 480 m/s |
| Answer» Correct Answer - A | |
| 9. |
A sonometer wire , `100 cm` in length has fundamental frequency of `330 Hz`. The velocity of propagation of tranverse waves along the wire isA. 330 m/sB. 660 m/sC. 990 m/sD. 115 m/s |
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Answer» Correct Answer - B `v=nlamda " as " lamda = 2l = 2 xx 100 = 220` `therefore v=2xx330=660m//s` |
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| 10. |
One metre long sonometer wire is stretched with a force of 4kg wt, another wire of same material and diamter is arranged along a side. The second wire is stretched with a force of 16kg wt. if the length of the second wire is in its second harmonic is the same as fifth harmonic of the first wire, then the length of the second wire will beA. 40cmB. 40cmC. 80cmD. 70cm |
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Answer» Correct Answer - C `n_1 prop (sqrtT_1)/(l_1) " and " n_2 prop (sqrtT_2)/(l_2)` `2n_2=5n_1` `therefore (2xx4)/(l_2) = (5xx2)/(1)` `therefore l_2= 0.8m` |
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| 11. |
A sonometer wire carries a brass weight of density 8 gm/cc at its free end and has a fundamental frequency 320 Hz, when the brass weight is completely immersed in water , then new frequency will beA. 340HzB. 320HzC. 280HzD. 300Hz |
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Answer» Correct Answer - D `n_1 prop sqrt(T_1) " and " n_2 prop sqrt(T_2)` Loss of weight `=T_1-T_2 therefore Vrhog = T_1-T_2` `T_2=T_1 - vrhog` `n_2/n_1 = sqrt(T_2/T_1) " and " n_2/n_1 = sqrt((rho-rho_1)/(rho))` |
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| 12. |
The frequency of vibration of 20cm length of a sonometer wire (linear density is 0.0294 gm/cm) under a tension 3kg wt is (g=`980cm//s^2`)A. 750HzB. 250HzC. 500HzD. 125Hz |
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Answer» Correct Answer - B `n=(1)/(2l) sqrt(T/m) = (1)/(2xx20) sqrt((3000xx980)/(0.294))` `=1/40xxsqrt((294xx10^4)/(294xx10^(-4)))` `=1/40 sqrt(108) = 104/40` `=(10000)/(40)` `=250Hz` |
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| 13. |
In open organ pipe, first overtone produced is of such frequency that length of the pipe is equal toA. `lamda//4`B. `lamda//3`C. `lamda//2`D. `lamda` |
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Answer» Correct Answer - D `l_2/l_1 = sqrt(T_2/T_1)` `=sqrt(196/100) = 13/10` `(l_2-l_1) / (l_1) = (13-10)/(10) = 3/10 = 0.3` `(Deltal)/(l) = 30%` |
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| 14. |
A sonometer wire of length `l_(1)`is in reasonance with a frequency 250 Hz. If the length of wire is increasesed then 2 beats per second are heard. What is ratio of the lengths of the wire ?A. 125:124B. 250:313C. `5:3`D. `41:57` |
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Answer» Correct Answer - A `n_1/n_2 = l_2/l_1` `250/248 = l_2/l_1` `125/124 = l_2/l_1` |
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| 15. |
If the length of an open organ pipe is 33.3cm , then the frequency of fifth overtone will beA. 3000HzB. 2500HzC. 1500HzD. 1250Hz |
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Answer» Correct Answer - A `l_0 = 33.3 xx10^(-2) m` `n=(v)/(2l) = (33.3)/(2xx33.3) xx10^(-2)` `n_5 = 6n = 6xx500 = 3000Hz` |
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| 16. |
Flute is an example ofA. A pipe open at one end and closed to other endB. Pipe open at both the endsC. A pipe closed at one endD. Neither a and b |
| Answer» Correct Answer - B | |
| 17. |
If oil of density higher than that of water is used in place of water in a resonance tube its frequency will beA. DecreaseB. IncreaseC. Remain constantD. Can not be predicted |
| Answer» Correct Answer - C | |
| 18. |
For a resonance tube, the air columns for the first and the second resonance differ in length by `31.5 cm`. The wavelength of the wave isA. 31.5cmB. 63.0cmC. 126.0cmD. 252.0cm |
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Answer» Correct Answer - B The first resonating length is `l_1+e=lamda/4` The second resonating length is `l_2 +e= (3lamda)/(4)` `l_2-l_1 = lamda//2 = 2xx 31.5=63cm` |
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| 19. |
A cylinderical tube open at both ends, has a fundamental frequency f in air. The tube is dipped vertically in water so that half of it is in water. The fundamental frequency of air column is nowA. n/2B. nC. 2nD. 4n |
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Answer» Correct Answer - B `n_1=n " " l_2=l/2` `n_2/n_1 = (v)/(4l//2) (2l)/(v) = 1` `n_2 = n_1 =n` Since when the tube is dipped in water it acts as open at one end closed to the other end. |
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| 20. |
The two organ pipes of same length and material, but of different radii. The loss of sound will beA. More from a wider pipeB. More from a narrow pipeC. Same for bothD. None of these |
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Answer» Correct Answer - A Wider pipe has larger area of cross section. Therefore , loss of sound is more from a wider pipe. |
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| 21. |
Two closed organ pipes have lengths L and `L+X`. When two pipes are sounded together , the beat frequency isA. `(YV)/(4l(l+y))`B. `(YV)/(2l(l+y))`C. `(YV)/(2l^2(l+y))`D. `(2l^2)/(YV)` |
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Answer» Correct Answer - A `n=n_1-n_2` `=(v)/(4l_1) - (v)/(4l_2)` `=v/4[1/l - (1)/(l+y)] = v/4[(l+y-l)/(l(l+y))]= (vy)/(4l(l+y))` |
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| 22. |
Water is slowly drained out in a cylindrical tube then the fundamental frequency of vibration of the column will beA. Continuously decreasesB. Continuously increasesC. Remain sameD. First increases, later on decreases |
| Answer» Correct Answer - A | |
| 23. |
A closed pipe has certain frequency. Now its length is halved. Considering the end correction, its frequency will now becomeA. DoubleB. More than doubleC. Less than doubleD. Four time |
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Answer» Correct Answer - C `n_1 = (v)/(4(l+e))` and `n_2 = (v)/(4(l/2+e))= (2v)/(4(l+2e))` Thus `n_2` is less than double of `n_1` |
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| 24. |
If oil of density higher than that of water is used in place of water in a resonance tube its frequency will beA. DecreaseB. IncreaseC. Remain the sameD. Can not say |
| Answer» Correct Answer - C | |
| 25. |
Two organ pipes give 4 beat/s when sounded together at `27^@C`, then the number of beat/s at `127^@C` will beA. 4.8B. 4.6C. 4D. 5 |
| Answer» Correct Answer - B | |
| 26. |
The value of end correction for the pipe open at both the ends is a (If the d is the inner diameter of the tube and r is the radius of the tube .)A. e=0.6dB. e=1.2rC. Both a and bD. Neither a nor b |
| Answer» Correct Answer - C | |
| 27. |
The value of end correction for an open organ pipe of radius r is _________ .A. 0.3rB. 0.6rC. 0.9rD. 1.2r |
| Answer» Correct Answer - D | |
| 28. |
If the end correction of closed organ pipe is 0.3 cm then the diameter of the tube will beA. 0.05mB. 0.5mC. 1cmD. 3cm |
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Answer» Correct Answer - C e=0.3d 0.3=0.3d d=1cm |
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| 29. |
The end correction of the open pipe isA. 0.3dB. 0.6dC. `(l_2-3l_1)//2`D. Both a and b |
| Answer» Correct Answer - B | |
| 30. |
If the length of the closed pipe whose fundamental frequency is equal to that of open pipe of length 60cm will beA. 20cmB. 24cmC. 28cmD. 30cm |
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Answer» Correct Answer - D `n_0=n_c` `(v)/(2L_0) = (v)/(4L_c)` `L_c= (L_0)/(2) = 30cm` |
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| 31. |
An organ pipe of length 0.4m is open at both ends . The speed of sound in air is `340ms^(-1)` . The fundamental frequency isA. 400HzB. 425HzC. 450HzD. 475Hz |
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Answer» Correct Answer - B `n=(v)/(4l) = (340)/(4xx0.4)` `=3400/8 = 425Hz` |
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| 32. |
If the end correction of an open tube is 0.3 cm, then the diameter of the tube will beA. 1cmB. 0.3cmC. 0.5cmD. 3cm |
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Answer» Correct Answer - C Pipe open at both the ends have and correction is `e=0.6d` `d=(0.3)/(0.6) = 0.5cm` |
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| 33. |
An open organ pipe of length is 50 cm, if the velocity of sound in air is 320m/s , then the frequency of its fundamental note will beA. 160HzB. 480HzC. 320HzD. 640Hz |
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Answer» Correct Answer - C `n=(v)/(2l) = (320)/(50xx2xx10^(-2))-320Hz` |
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| 34. |
The fundamental frequency of an open organ pipe is 300Hz. The first overtone of this has same frequency as that of first overtone of a closed organ pipe. If speed of sound is 330m/s, then the length of closed of organ pipe will beA. 41cmB. 37cmC. 31cmD. 80cm |
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Answer» Correct Answer - A (The frequency of first overtone of open pipe ) = (The frequency of first overtone of a closed pipe) `2n_0 = 3n_c` `2xx300=3xxn_c` `therefore n_c=200Hz` `therefore n_c=(v)/(4l_c)` `therefore l_c =(v)/(4n_c) = (330)/(4xx200) = 41.25cm` |
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| 35. |
The fundamental frequency of an open and closed organ pipe of the same length is in the ratioA. `2:1`B. `1//2:2`C. `1:2`D. `2:1//2` |
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Answer» Correct Answer - A `n_0/n_c = (v)/(2l_0) xx(4l_c)/(v) = 2` `n_0/n_c= 2/1` |
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| 36. |
The fundamental frequency of a closed organ pipe of length `20 cm` is equal to the second overtone of an organ pipe open at both the ends. The length of organ pipe open at both the ends isA. 80 cmB. 100 ccmC. 120 cmD. 140 cm |
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Answer» Correct Answer - c The fundamental frequencies of closed organ pipe are given as, `v_(c)=v/(4l)` `because` Second overtone of an open organ pipe `v_(2)=(3v)/(2l_(1))` Where `l_(1)=` length of the open pipe `because v_(c)=v_(2)` `rArr v/(4l)=(3v)/(2l_(1))rArr l_(1)=6l=6xx20` `therefore l_(1) = 120` cm |
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| 37. |
A fork of frequency 256Hz resonates with a closed organ pipe of length 25.4cm. If the length of the pipe is increased by 2mm then the number of beat/s will beA. 4B. 1C. 2D. 3 |
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Answer» Correct Answer - C `Deltan)/(n) = (Deltal)/(l)` `therefore Delta n =2` |
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| 38. |
Two tuning forks A and B gives 6 beat/s. The fork A resonates with a closed end air column of length 15cm and the fork B with an open air column of length 30.5cm . If the air in both the cases vibrates in the air fundamental modes, then the frequencies of fork A and B will beA. 354 and 360HzB. 366 and 360HzC. 360 and 354 HzD. 360 and |
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Answer» Correct Answer - B Let `n_A "and " n_B` denotes the frequency of tuning fork A and B . Then `n_A = (v)/(4(15)) = v/60 " and " n_B = (v)/(2(30.5)) = v/61` As `n_A gt n_B " " therefore n_A -n_B = 6` `v/60 - v/61 = 6 therefore v[(61-60)/(60xx61)]=6` `n_A = v/60 = (6xx60xx61)/(60) = 366` `n_B=v/61=(6xx60xx61)/(61) = 360Hz` |
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| 39. |
An open organ pipe has fundamental frequency 100 Hz. What frequency will be produced if its one end is closed ?A. 100, 200, 300B. 50, 150, 250C. 50, 100, 200, 300D. 50, 100, 150 |
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Answer» Correct Answer - B When one end is closed `n_1 = 1xxn_c = 50Hz` `n_2 = 3n_c = 3xx50=150Hz` `n_3=5n_c=5xx50=250Hz` |
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| 40. |
A tuning fork of frequency n resonates with a closed organ pipe when the length of the shortest air columns are a and b respectively, then the speed of sound in air will beA. n(a-b)B. `(n(b-a))/(2)`C. `2n(b+a)`D. 2n(b-a) |
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Answer» Correct Answer - D The first resonating length is `lamda/4 = a+e` The second resonating length is `(3lamda)/(4) = b+e` `b-a = lamda/2 therefore lamda = 2(b-a)` As `v=nlamda =2n(b-a)` |
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| 41. |
A steel wire of length 1m is stretched between two rigid supports. The wire is vibrating in its fundamental mode with a frequency 100 Hz. The maximum acceleration at the mid point of the wire is 986 `m//s^2` . Then the amplitude of vibration at the midpoint isA. 25 cmB. 0.5 cmC. 0.25 cmD. 50 cm |
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Answer» Correct Answer - C `a_m = omega^2A therefore A=(a_m)/(omega^2)` = `(986)/(4pi^2xx100xx100) = (9.86xx100)/(4xxpi^2xx 100xx100)` =0.25cm |
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| 42. |
the frequency of a sonometer wire is 10 Hz. When the weight producing th tensions are completely immersed in water the frequency becomes 80 Hz and on immersing the weight in a certain liquid the frequency becomes 60 Hz. The specific gravity of the liquid isA. `1.42`B. `1.77`C. `0.36`D. `1.82` |
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Answer» Correct Answer - b Case I `n=1/(2l) sqrt(T_(L)/(pir^(2)p))` `100 = 1/(2l)sqrt((Vd_(1)g)/(pir^(2)p))` ...(i) Case II `80=1/(2l) sqrt((Vg(d_(1)-d))/(pir^(2)p))` ...(ii) Case III `60=1/(2l) sqrt((Vg(d_(1)-d_(2)))/(pir^(2)p))` ...(iii) where, `d_(1)=` Density of weight `d_(2)=` Density of water `d_(2)=` Density of second liquid From Eqs. (ii) and (i), we get ltbtgt `80/100 = sqrt((d_(1)-d)/d_(1))rArr 4/5 = sqrt(1-d/d_(1))` `rArr 16/25 = 1 - d/d_(1) rArr therefore d_(1)/d=25/9` ...(iv) From Eqs. (iii) and (iv), we get `80/60= sqrt((d_(1)-d)/(d_(1)-d_(2)))rArr 4/3 = sqrt((d_(1)/d-1)/(d_(1)/d-d_(2)/d))` `rArr 16/9 = (25/9-1)/(25/9-d_(2)/d)` Putting `d_(1)/d-25/9` from Eqs. (iv) `rArr 16/9=(16//9)/(25/9-d_(2)/d)rArr 25/9 - d_(2)/d=1 rArrd_(2)/d=16/9` Thus, The specific gravity of the liquid, `d_(2)/d=16/9=1.77` |
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| 43. |
Two wires are fixed on a sonometer with their tensions are in the ratio 8:1, the length are in the ratio 36,35, the diameters in the ratio 4:1 and densities in the ratio 1:2. If the note of higher pitch has a frequency of 360Hz, then the frequency of other string will beA. 370HzB. 345C. 350HzD. 425 |
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Answer» Correct Answer - C `n_1/n_2 = sqrt (T_1/T_2xxrho_2/rho_1) xx l_2/l_1 xx r_2/l_1` |
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| 44. |
Two wires of same material of length l and 2l vibrate with frequencies 100 Hz and 150 Hz respectively. The ratio of their tensions isA. `2:3`B. `3:2`C. `1:9`D. `1:3` |
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Answer» Correct Answer - C `n_1 (1)/(2l_1)sqrt(T_1/m) " and " n_2 = (1)/(2l_2) sqrt(T_2/m)` `n_1/n_2 = l_2/l_1 sqrt(T_1/T_2) = (2l)/(l) sqrt(T_1/T_2)` `100/150 = 2 sqrt(T_1/T_2)` `2/6 = sqrt(T_1/T_2) therefore 1/9 = T_1/T_2` |
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| 45. |
Air is blown at the mouth of a tube of length 25 cm and diameter equal to 2 cm open at both ends. If velocity of sound in air is `330 ms^(-1)`, the sound emitted will have all the frequencies in the groupA. 330, 990, 1690 HzB. 302, 664, 1320 HzC. 660, 1320, 1980 HzD. 660, 100, 3300 Hz |
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Answer» Correct Answer - c Fundamental frequency , `n=v/(2l)=330/(2xx0.25)=660` Frequency of overtones are `2n, 3n, 4n, ...=1320, 1980, 2640` Hz |
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| 46. |
The stirng stretched by tension T and length L vibrates in resonance with a tuning fork of frequency n. the tension in the stretched string is increased by 69% and length of string reduced by 35%. Then, the frequency of vibrating string isA. nB. 1.5nC. 2nD. n/2 |
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Answer» Correct Answer - C `n_2/n_1 = l_1/l_2 sqrt(T_2/T_1)` `=(l_1)/(65/100l_1) xx sqrt(169/100) = 100/65 xx13/10` `=130/65 = 2` `n_2=2n` |
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| 47. |
A string of mass 0.2 kg/m and length l= 0.6 m is fixed at both ends and streteched such that it has a tension of 80 N. the string vibrates in 3 segments with maximum amplitude of 0.5 cm. the maximum transevers velcotiy amplitude isA. `9.43 ms^(-1)`B. `3.14 ms^(-1)`C. `1.57 ms^(-1)`D. `6.28 ms^(-1)` |
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Answer» Correct Answer - c As the string is vibrating in three segments, Therefore, `l=(3lambda)/2 rArr lambda = (2l)/3=(2(0.6))/3=0.4` m Now, `v=sqrt(T/m) rArr v= sqrt(80/2.0)=20 ms^(-1)` `n= v/lambda = 20/0.4 = 50` Hz Amplitude of particle belocity `=(dy/dt)_(max)=(a_(max))omega="a"_(max )(2pin)` `=(0.5xx10^(-2))xx2pixx50=1.57 ms^(-1)` |
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| 48. |
The fundamental frequency of a sonometer wire increases by 5Hz, if the tension is increased by 21% . The fundamental frequency of the sonometer wire is a Hz isA. 45B. 50C. 100D. 55 |
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Answer» Correct Answer - B `n_2/n_1 = sqrt(T_2/T_1) = sqrt(121/100) = 1.1` `(n+5)/(n) = 1.1` 1.1n - n = 5 0.1n = 5 n = 50 |
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| 49. |
The length of pipe 4cm in diameter and open at both ends when it is set into vibrations by a tuning fork of frequency 288Hz. (v=332m/s)A. 85.23cmB. 75.23cmC. 65.23cmD. 55.23cm |
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Answer» Correct Answer - D `v=2n(l+e)` `l+e=(v)/(2n)=(332)/(2xx288)` `l+e=166/288=0.576` `l=0.576-0.3d` `=0.576-0.3xx4xx10^(-2)` `=0.576-0.0012` =0.5526 =55.26cm |
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| 50. |
A tuning fork makes `256` vibrations per second in air. When the speed of sound is `330 m//s`, the wavelength of the note emitted is :A. `1.29` mB. `2.1` mC. `0.5` mD. `0.9` m |
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Answer» Correct Answer - a Wave motion is a disturbance travelling through a medium. Wave velocity `v = nlambda` This relation holds for transverse as well as longitudinal waves. Here, `v(s) = 330ms^(-1), n = 256 ` So `lambda = v_(s)/n=330/256=1.29 m` |
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