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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.
| 151. |
Two strings of the same material and the same area of cross-section are used in an wxperiment. One is loaded with 12 kg and the other with 3 kg The fundamental fequency of the first string is equal to the first overtone of the second string. If the length of the second string is 100 cm, then the length of the first string isA. 300 cmB. 200 cmC. 100 cmD. 50 cm |
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Answer» Correct Answer - c The frequency of the string `n = 1/(2l) sqrt(T/m)=1/(2l) sqrt((Mg)/(pir^(2)d))(because T=Mg, m = pi r^(2)d)` `rArr n_(1) =1/(2l_(1)) sqrt((Mg)/(pir^(2)d))and n_(2) =2/(2l_(2)) sqrt((M_(2)g)/(pir^(2)d))` From the question`n_(1) n_(2) rArr l_(2)/l_(1)=2sqrt(M_(2)/M_(1)` `rArr 100/l_(1) = 2sqrt(3/12)=2sqrt(1/4)=2xx1/2rArrl_(1)=100` cm |
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| 152. |
The equation of stationary wave along a stretched string is given by `y=5 sin ((pix)/(3)) cos 40 pi t `, where x and y are in cm and t in second. The separation between two adjacent nodes isA. 6 cmB. 4 cmC. 3 cmD. `1.5` cm |
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Answer» Correct Answer - c The standad equation of statonary wave is `y=2 A sin kx. Cos omegat` …(i) where, a is amplitude, `lambda` is wavelength. Given, `y=5sin frac{pix}{3} cos 40 pit` ...(ii) Comparing Eqs. (i) with (ii), we get `k=pi/3 rArr (2pi)/lambda = pi/3 [because lambda=3 cm]` Distance between adjacent nodes `lambda/2=6/2=3` cm `therefore lambda/2 = 3` cm |
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| 153. |
A wire under a tension vibrates with fundamental frequency of 256 Hz . What should be the fundamental frequency of the wire if its length is half, thickness twice and one - fourth of the tension ?A. 128 HzB. 64 HzC. 200 HzD. 96 Hz |
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Answer» Correct Answer - A `n_2/n_1=l_1/l_2d_1/d_2 sqrt(T_2/T_1)` `=2/1xx1/2xxsqrt(1/4) = 1/2` `n_2 = (n_1)/(2) = 256/2 = 128Hz` |
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| 154. |
A segment of wire vibrates with fundamental frequency of 40 Hz under a tension of 9 kg-wt. Then, tension at which the fundamental frequency of the same wire becomes 900 Hz isA. 36 kg-wtB. 27 kg-wtC. 18 kg-wtD. 72 kg-wt |
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Answer» Correct Answer - a Fundamental frequency, `n=1/(2l)sqrt(T/m)and nprop sqrt(T)` `therefore n_(1)/n_(2)=sqrt(T_(1)/T_(2))rArr 450/900=sqrt(9/T_(2))` `T_(2)=36` kg-wt |
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| 155. |
A wire under tension vibrates with a frequency of 450Hz. What would be the fundamental frequency, if the wire were half as long, twice as thick and under one fourth tension ?A. 225HzB. 190HzC. 247HzD. 174Hz |
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Answer» Correct Answer - A `n_2/n_1 = l_1/l_2 xx d_1/d_2 xx sqrt(T_2/T_1)` `=2xx1/2xxsqrt(1/4) = 1/2` `n_2 = n_1/2 = 450/2` =225Hz |
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| 156. |
A metal wire of linear mass density of `9.8g//m` is stretched with a tension of `10 kg-wt` between two rigid support `1meter` apart. The wire passes at its middle point between the poles of a permanent magnet, and it vibrates in resonance when carrying an alternating current of frequency `n`. the frequency `n` of the alternating source isA. 50 HzB. 100 HzC. 200 HzD. 25 Hz |
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Answer» Correct Answer - a The wire will vibrate with the same frequency as that of source. This can be considered as an example of forced vibration. `T=10xx9.8 N=98 ` n and `m=9.8 xx 10^(-3)`kg `m^(-1)` Frequency of wire `f=1/(2L) sqrt((T/m))=1/(2xx1) sqrt((98)/(9.8xx10^(-3)))` = 50 Hz |
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| 157. |
The speed of a wave in a certain medium is `960 m//s`. If 3600 waves pass over a certain point of the medium in 1 min, the wavelength isA. 6mB. 32mC. 16mD. 8m |
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Answer» Correct Answer - C `lamda = v/n = 960/60 = 16m` |
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| 158. |
What tension should be there in a string of length 0.8m and of mass `5xx10^(-3) kg`, if it vibrates with fundamental frequency of 100 Hz?A. 100NB. 160NC. 140ND. 120N |
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Answer» Correct Answer - B `n=(1)/(2l) sqrt(T/m) therefore n^2 = (1)/(4l^2) T/m` `therefore T= 4n^2l^2m` `=(4xx10^(4)xx0.6xx5xx10^(-3))/(0.8) = 160N` |
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| 159. |
A string vibrates with a frequency of 200Hz. Its length is doubled and its tension is altered till it begins to vibrate with a frequency of 300Hz. What is the ratio of new tension to the original tension ?A. `9:01`B. `1:09`C. `3:01`D. `1:03` |
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Answer» Correct Answer - a `v = 1/(2l)sqrt(T/m) rArr v prop sqrt(T)/l` `T_(2)/T_(1)=[v_(2)/v_(1)]^(2)[l_(2)/l_(1)]^(2)=[300/200]^(2)[(2l)/l]^(2)=9/1` |
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| 160. |
A string is hanging from a rigid support. A transverse pulse is excited at its free end. The speed at which the pulse travels a distance x is proportional toA. xB. `1/x`C. `1/sqrt(x)`D. `sqrt(x)` |
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Answer» Correct Answer - d Speed of transverse pulse, `v=sqrt(T/mu)` where, `mu` is mass of string per unit length. For a transverse pulse, `T=muxg` `therefore v=sqrt((muxg)/mu)rArr v prop sqrt(x)` |
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| 161. |
A stone in hung in air from a wire which is stretched over a sonometer . The bridges of the sonometer are `40 cm` apart when the wire is in unison with a tuning fork of frequency `256 Hz`. When the stone is completely immersed in water , the length between the bridges is `22 cm` for re - establishing unison . The specific gravity of the material of the stone isA. 0.767B. 1.38C. 0.983D. 9.23 |
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Answer» Correct Answer - B `T_=Kl_1^2 " and " T_2=Kl_2^2` By definition of specific gravity `therefore (T_1)/(T_1-T_2) = (l_1^2)/(l_1^2-l_2^2) = ((40)^2)/((40)^2-(22)^2)=1.38` |
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| 162. |
A pipe `30.0` cm long is opened at both ends. Which Harmonic mode of the pipe will be at `1.1` kHz frequency also find fundamental frequency, if one end of the pipe is colsed. Take the speed of sound in air as 330 `"ms"^(-1)`A. 2nd harmonic and 275 HzB. 1st harmonic and 260 HzC. 3rd harmonic and 260 HzD. Fundamental harmonic and 240 Hz |
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Answer» Correct Answer - a Here, `L = 30.0` cm `=0.3` cm Let, nth harmonic of open pipe is at a frequency of 1 kHz, i.e. `v_(n) = 1.1" kHz"= 1100` Hz As, `v_(n) = (nv)/(2L)` `therefore n =(2Lv_(n))/v=(2xx0.30xx1100)/330=2` i.e. 2nd harmonic of the open pipe will present. If one end of pipe is closed, its fundamental frequency `n=0" in " v_(0)=(2n+1)v/(4L)` ` v_(1)=v/(4L)=330/(4xx0.3)=275` Hz |
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| 163. |
The string of monochord , vibrates 100 times a second. Its length is doubled and its tension is altered unitl it makes 150 vib/s . What is the ratio of the new tension to the string increases by a force ?A. `1:9`B. `9:1`C. `1:3`D. `3:1` |
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Answer» Correct Answer - B `T_1/T_1 =(n_2/n_1)^2xx(l_2/l_1)^2 = 2.25xx4 = 9.1` |
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| 164. |
A string of length 10.0 m and mass 1.25kg stretched with a tension of 50N. If a transverse pulse is created at one end of the string, how long does it take to reach the other end ?A. 0.5sB. 1.0sC. 1.5sD. 2.0s |
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Answer» Correct Answer - A `V=sqrt(T/m) = sqrt((50)/(0.125)) = sqrt((5xx10^4)/(125))` `V=10^2/5 = 100/5 = 20` m/s `t=d.v = 10/20 = 0.5s |
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| 165. |
A wire is under a tension of 32N and length between the two bridges in 1m. If the length of the sample wire is 10m and its mass is 2 g, then the fundamental frequency of the wire will beA. 400HzB. 200HzC. 100HzD. 800Hz |
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Answer» Correct Answer - B `n=(1)/(2l) sqrt(T/m) = (1)/(2xx1) sqrt((32)/(2xx10^(-4))` `=1/2xx4xx10^2` `=400/2 = 200` |
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| 166. |
A pipe is open at both ends and 40 cm of its length is in resonance with an external frequency `1.1` kHz. If the speed of sound is `330ms^(-1)`, which harmonic is in resonance?A. FirstB. SecondC. ThirdD. none |
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Answer» Correct Answer - d Fundamental frequencyu of open pipe First harmonic `=n_(1)=v/(2l)=(330)/(2xx0.4)=412.5` Hz Frequency of nth harmonic of vibration `v_(n)=n(v//2l)` Given `v_(n)=1.1kHz = 1100 Hz` `rArr nxx (412.5)=1100 rArr n = 2.666` `therefore` No harmonic is in resonance |
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| 167. |
in an experiment it was found that string vibrates in n loops when a mass M is placed on the pan. What mass should be placed on the pan to make it vibrate in 2n loops with same frequency ? ( neglect the mass of pan )A. `M//4`B. 4 MC. 2 MD. `M//2` |
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Answer» Correct Answer - b As the string vibrates in n loops therefore, `l=(nlambda)/2` therefore, v would become `1/2` time. As `v prop sqrt(T)` Threrfore to make v half- time T must be made `1/4` time i.e. M/4. |
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| 168. |
A mass of 10m long wire is 100gm. If a tension of 100N is applied, what is the time taken by transverse wave to travel from one end to the other end of the wire ?A. 0.1 sB. 0.3 sC. 0.2 sD. 0.4 s |
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Answer» Correct Answer - A `V=d/t therefore t=d/V = 10xxsqrt((10xx10^(-3))/(100)) = 0.1` sec |
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| 169. |
While measuring the speed of sound by performing a resonance column experiment, a student gets the first resonance condition at a column length of `18 cm` during winter. Repeating the same experiment during summer, she measures the column length to be `x cm` for the second resonance. ThenA. `18gtx`B. `xgt54`C. `54gtxgt36`D. `36gtxgt18` |
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Answer» Correct Answer - d Here, `l_(1)= 18` cm `f=v_(1)/(4l_(1))=(3v_(2))/(4l_(2)), ` where `l_(2)=x` according to given situation and also `v_(1)ltv_(2)` as during summer temperature would be higher. `(3v_(2))/(4l_(2))=v_(1)/(4l_(1))rArr l_(2) = 3l_(1) xxv_(2)/v_(1)` `rArr x=54 xx` (A quantity greater than 1) So, `xgt 54` |
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| 170. |
In an experiment , it was found that a string vibrated in 5 loops when 100gm was placed in the pan. What must be placed in the pan to make the string vibrate in 10 loops ?A. 100gmB. 50gmC. 75gmD. 25gm |
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Answer» Correct Answer - D `T_1P_1^2 = T_2P_2^2` `100xx25 = T_2xx100` `T_2=25g` |
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| 171. |
In a resonance tube experiment, the first resonating length of air column is 0.2m and second resonating length of air column is 0.62 m, then the inner diameter of the tube isA. 2.33cmB. 3.33cmC. 1.33cmD. 04.44cm |
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Answer» Correct Answer - B `e=(l_2-3l_1)/(2) = (0.62-3xx0.2)/(2) = (0.02)/(0.2)` `e=0.01 therefore 0.3d=0.01` `therefore d=(0.01)/(0.3) = 0.03333m` |
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| 172. |
The condition for first resonance in a resonating air column closed at one end isA. l+e=`lamda//2`B. `l+2e = lamda//2`C. `l+e = lamda//4`D. `l+2e=lamda//4` |
| Answer» Correct Answer - C | |
| 173. |
In a resonance tube experiment ,a tuning fork resonates with an air column produces first resonating length of 12.5cm and second resonating length of 40cm . Then the end correction isA. 1cmB. 1.25cmC. 3cmD. 0.12cm |
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Answer» Correct Answer - B `e=(l_2-3l_1)/(2) = (40-37.5)/(2) = (2.5)/(2)` `=1.25cm` |
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| 174. |
In a resonance tube, using a tuning fork of frequency `325 Hz`, the first two resonance lengths are observed at `25.4 cm` and `77.4 cm`. The speed of sound in air isA. 338 `ms^(-1)`B. 328 `ms^(-1)`C. 330 `ms^(-1)`D. 320 `ms^(-1)` |
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Answer» Correct Answer - a Velocity of sound in air, `v=2n(l_(2)-l_(1))=2xx 325(77.4-25.4)` `rArr v(650xx52)/100=338ms^(-1)` |
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| 175. |
In a resonance tube experiment, two resonating lengths are taken , becauseA. One reading is not suffcientB. Average is to be calculated sC. End correction is eliminatedD. To minimise the error |
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Answer» Correct Answer - C Two resonance positions are taken to eliminate end correction |
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| 176. |
The length and diameter of a metal wire is doubled the fundamental frequency of vibration will change from n to (tension being kept constant and material of both the wires is same)A. n/4B. n/8C. n/12D. n/16 |
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Answer» Correct Answer - A `n_2/n_1 = (l_1d_1)/(l_2/d_2) sqrt((T)/(rhopi)(rhopi)/(T)) = 1/2xx1/2` `n_2 = 1/4n_1 = n/4` |
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| 177. |
If an addition of 75 kg wt to the vibrating string rises the pitch is an octave of the original pitch, then the tension in the string isA. 50 kg wtB. 100 gk wtC. 75 kg wtD. 25 kg wt |
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Answer» Correct Answer - D `n_2/n_1 therefore (2n)/(n) = sqrt((T_1 +75)/(T_1))` `therefore 3T_1 = 75 therefore T_1 = 25` kg wt |
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| 178. |
In a resonance pipe the first and second resonance are obtained at depths 22.7 cm and 70.2 respectively. What will be the end correction?A. `1 .05` cmB. `115.5` cmC. `92.5` cmD. `113. 5` cm |
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Answer» Correct Answer - a For and correction `x, (l_(1)+x)/(l_(1)+ x ) = (3lambda //4)/(lambda//4) =3` `rArr x= (l_(2)-3l_(1))/2=(70.2-3xx 22.7)/2` ` = 1.05` cm |
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| 179. |
In a resonance tube experiment, two successive resonances are heard at 15 cm 48 cm. End correction will beA. 1.5cmB. 3cmC. 2.5cmD. 1cm |
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Answer» Correct Answer - A `rho=(l_2-3l_1)/(2) = (48-3xx15)/)(2)` `rho = (48-45)/(2) = 3/2 = 1/5cm` |
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| 180. |
A pipe closed at one end open at the other end, resonates with sound waves of frequency 135 Hz and also 165 Hz, but not with any wave of frequency intermediate between these two. Then, the frequency of the fundamental note isA. 30HzB. 15HzC. 60HzD. 7.5Hz |
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Answer» Correct Answer - B `n_1-n_2` = 165 - 135 = 30 The difference between two successive harmonics = 2 fundamental frequency `30 therefore 2n therefore n = 15Hz` |
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| 181. |
A pipe closed at one end open at the other end, resonates with sound waves of frequency 135 Hz and also 165 Hz, but not with any wave of frequency intermediate between these two. Then, the frequency of the fundamental note isA. 30 HzB. 15 HzC. 60 HzD. 7.5 Hz |
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Answer» Correct Answer - b In a closed pipe, resonance frequency, `v= (2r-1)//4l=135 and 165.` The lowest frequency must be highest common factor of 135 and 165, which is 15 Hz. |
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| 182. |
A tuning fork of frequency 340Hz vibrated above a cylindrical hallow tube closed at one end. The height of the tube is 120cm . Water is slowly poured in it. What is the minimum height of water required for resonance ?A. 95cmB. 25cmC. 45cmD. 105cm |
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Answer» Correct Answer - C `l_1 = (v)/(4n) = (340)/(4xx340) = 25cm` `l_2=(3v)/(4n) = 3xx0.025 = 75cm` `l_3=5l_1 = 5xx0.25 = 125cm` The height of the tube is 120cm , the maximum height of the air column for resonance is 75 cm so the minimum height of the water is 120-75=45cm. |
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| 183. |
For a certain organ pipe , three successive resonance frequencies are observed at 255 Hz, 425Hz, 595Hz respectively, then the pipe isA. A pipe closed at one end with fundamental frequency 85 HzB. Closed at one end with fundamental frequency 255HzC. Open at both the ends with fundamental frequency 85HzD. Open at both the ends with fundamental frequency 255Hz |
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Answer» Correct Answer - A `n_3-n_2 = 595 - 425 = 170` `n_2-n_1 = 425 - 255 = 170` `n_2 -n_1 = 2n = 170 " " therefore n=85` `n_2=3n=3xx85=255` `n_3=5n=5xx85=425` odd multiples of fundamental frequency |
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| 184. |
Stationary waves are setup in an air column. Velocity of sound in air is `330 ms^(-1)` and frequency is `165 Hz`. The distance between two successive nodes isA. 2mB. 1mC. 0.5mD. 4m |
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Answer» Correct Answer - B The distance between antinode to antinode is `=lamda /2 = 2/2 = 1m.` |
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| 185. |
For a certain organ pipe , three successive resonance frequencies are observed at 170 Hz, 255Hz, 340Hz respectively, then the pipe isA. Open end pipe with fundamental frequency 85HzB. Open end pipe with fundamental frequency 170HzC. Closed at one end pipe with fundamental frequency 85HzD. Closed at one end pipe with fundamental frequency 170 Hz |
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Answer» Correct Answer - A The successive frequency are 170, 255 and 340 Hz `n_2 - n_1 = 255 - 170 = 85 Hz` and `n_3 - n_2 = 340 - 255 = 85 Hz` `170= 85xx2` `255=85xx3` `340=85xx4` Hence it is a open organ pipe with fundamental frequency equal to 85 Hz. |
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| 186. |
Two successive resonance frequencies in an open organ pipe are 1944 Hz and 2592 Hz. Find the length of the tube. The speed of sound in air is `324 m s^-1`.A. 25cmB. 100cmC. 50cmD. 12.5cm |
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Answer» Correct Answer - A `n_2-n_1 = 2592 - 1944 = 648` `n=(v)/(2l)` `therefore l=(v)/(2n) = (324)/(2xx648)` 0.25cm |
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| 187. |
The frequency of open organ pipe isA. `(v)/(2l+0.4d)`B. `(v)/(2l+0.6d)`C. `(v)/(2l+0.8d)`D. `(v)/(2l+0.6d)` |
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Answer» Correct Answer - B `n=(v)/(2l+0.6d)` |
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| 188. |
Two closed organ pipe A and B have the same length. A is wider than B. They resonate in the fundamental mode at frequencies `V_(A) and V_(B)` respectively , thenA. `n_(A) = n_(B)`B. `n_(A) gt n_(B)`C. `n_(A) lt n_(B)`D. Either (b) or ( c) depending on the ratio of their diameters |
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Answer» Correct Answer - c In closed organ pipe. First resonance occurs at `lambda//4.` So, in fundamental mode of vibration of organ pipe `lambda/4=(l+0.3d)` where, `0.3` d is necessary end correction. Frequency of vibration, `n=v/lambda = v/(4(l+0.3d))` As / is same, wide pipe A will resonate at a lower Frequency, I.e. `n_(A) lt n_(B).` |
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| 189. |
Velocity of a particle in stationary wave is maximum atA. AntinodesB. NodeC. Mid way between node and antinodeD. Can not be predicted |
| Answer» Correct Answer - A | |
| 190. |
The amplitude of a stationary waves isA. ConstantB. Varies periodically with distance xC. Varies periodically with both x and tD. Varies periodically with time t |
| Answer» Correct Answer - B | |
| 191. |
The equation of a stationary wave is given by `y=6sin(pi)//(x)cos40pit` Where `y` and `x` are given in cm and time `t` in second. Then the amplitude of progressive wave isA. 3cmB. 6cmC. 12cmD. 2cm |
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Answer» Correct Answer - A `2A = 6 therefore A=3cm` |
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| 192. |
Which characteristic of musical sound determines the shrillness or graveness of the sound?A. QualityB. LoudnessC. PitchD. None of these |
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Answer» Correct Answer - c Pitch is the characteristic of sound that depends upon the frequency. So it determines the shrillness or graveness of the sound. |
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| 193. |
A glass tube of length 1m is filled with water . The water can be drained out slowly at the bottom of the tube . If a vibrating tuning fork of frequency 250Hz is brought at the upper end of the tube and the velocity of sound is 330 m/s , then the total number of resonance obtained will beA. 4B. 3C. 2D. 1 |
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Answer» Correct Answer - D `l_1=(v)/(4n) = (330)/(4xx250) = 330/1000` `l_1 = 0.33m` =33cm `l_2=3xxl_1 = 0.3x33=0.99cm` Number of resonances observed are 2 |
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| 194. |
An open pipe is suddenly closed at one end with the result that the frequency of third harmonic of the closed pipe is found to be higher by `100 Hz` then the fundamental frequency of the open pipe. The fundamental frequency of the open pipe isA. 200 HzB. 480 HzC. 240 HzD. 300 Hz |
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Answer» Correct Answer - a Frequency of third harmonic of closed pipe, `n_(1)(3v)/(4l)` Fundamental frequenccy of open pipe `n_(2)=v/(2l)` As `n_(1)-n_(2) = 100` `rArr v/(4l)=100` Fundamental frequency, `v/(2l)=200` Hz |
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| 195. |
The amplitude of S.H.M. at antinodes of the problem No. 51 isA. 0.01 mB. `(sqrt(3)//2)xx0.02m`C. 0.02mD. `0.02xx1.714` |
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Answer» Correct Answer - C `R_("max") = 2A = 2xx0.1=0.02` |
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| 196. |
In a resonance tube experiment, the first resonance is obtained when the level of water in the tube is at 26cm from the open ened. Then the next resonance may be obtained , when the level of water will beA. 32cmB. 48cmC. 47cmD. 50cm |
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Answer» Correct Answer - D `e=(l_2-3l_1)/(2) gt 0 therefore l_2 gt 3l_1` `therefore l_2 gt 3xx16=48 " " l_2 gt 48` i.e. 50cm |
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| 197. |
A travelling wave represented by `y=Asin (omegat-kx)` is superimposed on another wave represented by `y=Asin(omegat+kx).` The resultant isA. A standing wave having node at `x=(n+1/2)lambda/2,` `n=0, 1, 2`B. A wave travelling along + x-directionC. A wave travelling along - x-directionD. A standing wave having nodes at `x=(nlambda)/2,n=0,1,2` |
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Answer» Correct Answer - a `y=y_(1)+ y_(2) = a sin (omegat-kx)+a sin (omega t+kx)` `y= 2a sin omegat cos kx` Clearly it is equation of standing wave for position of nodes `y = 0 ` i.e `x=(2n+1)lambda/4` `= (n+1/2)lambda/2,n=0,1,2,3` |
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| 198. |
At resonance, the amplitude of forced oscillations isA. MinimumB. Equal to amplitude of periodic forceC. MaximumD. Can not be predicted |
| Answer» Correct Answer - C | |
| 199. |
In a stationary wave (A) All the particles of the medium vibrate in phase (B) All the antinodes vibrate in phase © All alternate antinodes vibrate in phase (D) All the particles between consecutive nodes are in phaseA. A and BB. C and DC. A and DD. B and C |
| Answer» Correct Answer - B | |
| 200. |
The distance between any two successive nodes or antinodes in stationary waves isA. `lamda`B. `lamda/4`C. `lamda/2`D. `lamda/8` |
| Answer» Correct Answer - C | |