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hDSwe4zz1mcwvD0MbthSZ | physics | waves | basic-of-waves-and-progressive-waves | The end correction of a resonance column is 1 cm. If the shortest length resonating with the tunning fork is 10 cm, the next resonating length should be : | [{"identifier": "A", "content": "28 cm"}, {"identifier": "B", "content": "32 cm"}, {"identifier": "C", "content": "36 cm"}, {"identifier": "D", "content": "40 c"}] | ["B"] | null | Given, End correction (e) = 1 cm
<br><br>For first resonance,
<br><br>$${\lambda \over 4} = {l_1} + e$$ = 10 + 1 = 11 cm
<br><br>For second resonance,
<br><br>$${3\lambda \over 4} = {l_2} + e$$
<br><br>$$ \Rightarrow $$ $${l_2}$$ = 3 $$ \times $$ 11 - 1 = 32 cm | mcq | jee-main-2018-online-16th-april-morning-slot | 13,150 |
B9xnNpMaJ5UPp6lIpimJT | physics | waves | basic-of-waves-and-progressive-waves | 5 beats / econd are heard when a tuning fork is sounded with a sonometer wire under tension, when the length of the sonometer wire is either 0.95 m or 1 m. The frequency of the fork will be : | [{"identifier": "A", "content": "195 Hz"}, {"identifier": "B", "content": "150 Hz"}, {"identifier": "C", "content": "300 Hz"}, {"identifier": "D", "content": "251 Hz"}] | ["A"] | null | <p>Length of wire is L<sub>1</sub> = 0.95 m; L<sub>2</sub> = 1 m. Number of beats per second heard = 5</p>
<p>Let frequency of fork be f. Therefore,</p>
<p>$${v \over {2{L_1}}} - f = 5 \Rightarrow {v \over {2{L_1}}} = 5 + f$$ ...... (1)</p>
<p>and $$f - {v \over {2{L_2}}} = 5 \Rightarrow {v \over {2{L_2}}} = f - 5$$ .... (2)</p>
<p>Dividing Eq. (1) by Eq. (2), we get</p>
<p>$${{{v \over {2{L_1}}}} \over {{v \over {2{L_2}}}}} = {{5 + f} \over {f - 5}} \Rightarrow {{{L_2}} \over {{L_1}}} = {{f + 5} \over {f - 5}}$$</p>
<p>$$ \Rightarrow {1 \over {0.95}} = {{f + 5} \over {f - 5}} \Rightarrow f - 5 = 0.95f + 4.75$$</p>
<p>$$ \Rightarrow f - 0.95f = 5 + 4.75 \Rightarrow 0.05f = 9.75 \Rightarrow f = {{9.75} \over {0.05}}$$</p>
<p>$$ \Rightarrow f = 195$$ Hz</p> | mcq | jee-main-2018-online-15th-april-evening-slot | 13,151 |
bErMi1ftWwu3xRhj6kKR8 | physics | waves | basic-of-waves-and-progressive-waves | A tuning fork vibrates with frequency $$256$$ $$Hz$$ and gives one beat per second with the third normal mode of vibration of an open pipe. What is the length of the pipe ? (Speed of sound in air is $$340\,m{s^{ - 1}}$$) | [{"identifier": "A", "content": "$$220$$ $$cm$$"}, {"identifier": "B", "content": "$$190$$ $$cm$$"}, {"identifier": "C", "content": "$$180$$ $$cm$$"}, {"identifier": "D", "content": "$$200$$ $$cm$$"}] | ["D"] | null | The tuning fork vibrates with frequency 256 Hz and give one beat per second So, the organ pipe will have frequency (256 $$ \pm $$ 1) Hr.
<br><br>For open organ pipe,
<br><br>Frequency n = $${{N\upsilon } \over {2\ell }}$$
<br><br>Here n = 255 Hz
<br><br>N = 3
<br><br>$$\upsilon $$ = 340 m/s
<br><br>$$\therefore\,\,\,\,$$ 255 = $${{3 \times 340} \over {2 \times \ell }}$$
<br><br>$$ \Rightarrow $$$$\,\,\,\,$$ $$\ell $$ = $${{3 \times 340} \over {2 \times 255}} = 2\,m$$
<br><br>$$\therefore\,\,\,\,$$ $$\ell $$ = 2m or 200 cm | mcq | jee-main-2018-online-15th-april-morning-slot | 13,152 |
CMBFz9IyO51LtRUL | physics | waves | basic-of-waves-and-progressive-waves | A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The
density of granite is 2.7 $$\times$$ 10<sup>3</sup> kg/m<sup>3</sup> and its Young’s modulus is 9.27 $$\times$$ 10<sup>10</sup> Pa. What will be the fundamental frequency of the longitudinal vibrations ? | [{"identifier": "A", "content": "7.5 kHz"}, {"identifier": "B", "content": "5 kHz"}, {"identifier": "C", "content": "2.5 kHz"}, {"identifier": "D", "content": "10 kHz"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263660/exam_images/b4tr3ruvqzcwhfoma1sa.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2018 (Offline) Physics - Waves Question 108 English Explanation">
<br><br>As rod length = 60 cm
<br><br>$$\therefore\,\,\,$$ $${\lambda \over 2}$$ = 60
<br><br>$$ \Rightarrow $$$$\,\,\,$$ $$\lambda $$ = 120 cm = 1.2 m
<br><br>In solid, velocity of wave,
<br><br>V = $$\sqrt {{Y \over \rho }} $$
<br><br>= $$\sqrt {{{9.27 \times {{10}^{10}}} \over {2.7 \times {{10}^3}}}} $$
<br><br>= 5.85 $$ \times $$ 10<sup>3</sup> m/sec.
<br><br>As we know,
<br><br>v = f $$\lambda $$
<br><br>$$\therefore\,\,\,$$ f = $${v \over \lambda }$$
<br><br>= $${{5.85 \times {{10}^3}} \over {1.2}}$$
<br><br>= 4.88 $$ \times $$ 10<sup>3</sup> Hz
<br><br>$$ \simeq $$ 5 kHz | mcq | jee-main-2018-offline | 13,153 |
ishxRlvOA3XL98ow4g1KA | physics | waves | basic-of-waves-and-progressive-waves | A heavy ball of mass M is suspendeed from the ceiling of a car by a light string of mass m (m < < M). When the car is at rest, the speed of transverse waves in the string is 60 ms<sup>$$-$$1</sup>. When the car has acceleration a, the wave-speed increases to 60.5 ms<sup>$$-$$1</sup>. The value of a, in terms of gravitational acceleration g, is closest to : | [{"identifier": "A", "content": "$${g \\over {30}}$$"}, {"identifier": "B", "content": "$${g \\over 5}$$"}, {"identifier": "C", "content": "$${g \\over 10}$$"}, {"identifier": "D", "content": "$${g \\over 20}$$"}] | ["B"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1ljexxhav/257f9e0f-db8d-4565-a4ac-a642545d47e9/6393ee70-1545-11ee-a085-a368e58ec1a5/file-6y3zli1ljexxhaw.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1ljexxhav/257f9e0f-db8d-4565-a4ac-a642545d47e9/6393ee70-1545-11ee-a085-a368e58ec1a5/file-6y3zli1ljexxhaw.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh; vertical-align: baseline;" alt="JEE Main 2019 (Online) 9th January Morning Slot Physics - Waves Question 97 English Explanation">
<br><br>Resultant force on the ball of mass M when car is moving with a acceleration <i>a</i> is ,
<br><br>F<sub>net</sub> = $$\sqrt {{{\left( {Mg} \right)}^2} + {{\left( {Ma} \right)}^2}} $$
<br><br> = $$M\sqrt {{g^2} + {a^2}} $$
<br><br>$$ \therefore $$ T = M$$\sqrt {{g^2} + {a^2}} $$
<br><br>We know,
<br><br>Velocity, V = $$\sqrt {{T \over \mu }} $$
<br><br>When Car is at rest then,
<br><br> 60 = $$\sqrt {{{Mg} \over \mu }} $$ . . . . (1)
<br><br>and when is moving then
<br><br> 60.5 = $$\sqrt {{{M\sqrt {{g^2} + {a^2}} } \over \mu }} $$ . . . . (2)
<br><br>By dividing (2) by (1) we get,
<br><br>$${{60.5} \over {60}} = \sqrt {{{\sqrt {{g^2} + {a^2}} } \over g}} $$
<br><br>$$ \Rightarrow $$ $$\left( {1 + {{0.5} \over {60}}} \right)$$ = $${\left( {{{{g^2} + {a^2}} \over {{g^2}}}} \right)^{{1 \over 4}}}$$
<br><br>$$ \Rightarrow $$ $${{{g^2} + {a^2}} \over {{g^2}}}$$ = $${\left( {1 + {{0.5} \over {60}}} \right)^4}$$
<br><br>$$ \Rightarrow $$ $${{{g^2} + {a^2}} \over {{g^2}}}$$ = 1 + 4 $$ \times $$ $${{{0.5} \over {60}}}$$ [Using Binomial approximation]
<br><br>$$ \Rightarrow $$ $${{{g^2} + {a^2}} \over {{g^2}}}$$ = 1 + $${1 \over {30}}$$
<br><br>$$ \Rightarrow $$ 1 + $${{{a^2}} \over {{g^2}}}$$ = 1 + $${1 \over {30}}$$
<br><br>$$ \Rightarrow $$ $${a \over g}$$ = $${1 \over {\sqrt {30} }}$$
<br><br>$$ \Rightarrow $$ a = $${g \over {\sqrt {30} }}$$
<br><br>$$ \therefore $$ Closest answer, a = $${g \over 5}$$ | mcq | jee-main-2019-online-9th-january-morning-slot | 13,154 |
NvTjG9c3TSiLaGrvlNUf9 | physics | waves | basic-of-waves-and-progressive-waves | A string of length 1 m and mass 5 g is fixed at both ends. The tension in the string is 8.0 N. The string is set into vibration using an external vibrator of frequency 100 Hz. The separation between successive nodes on the string is close to -
| [{"identifier": "A", "content": "16.6 cm"}, {"identifier": "B", "content": "10.0 cm"}, {"identifier": "C", "content": "20.0 cm"}, {"identifier": "D", "content": "33.3 cm"}] | ["C"] | null | Velocity of wave on string
<br><br>$$V = \sqrt {{T \over \mu }} = \sqrt {{8 \over 5} \times 1000} = 40m/s$$
<br><br>Now, wavelength of wave
<br><br>$$\lambda = {v \over n} = {{40} \over {100}}m$$
<br><br>Separation b/w successive nodes,
<br><br>$${\lambda \over 2} = {{20} \over {100}}\,m$$ $$=$$ 20 cm | mcq | jee-main-2019-online-10th-january-morning-slot | 13,155 |
BYmFhNw4xdzmbhiLZpkm8 | physics | waves | basic-of-waves-and-progressive-waves | Equation of travelling wave on a stretched string of linear density 5 g/m is y = 0.03 sin(450 t – 9x) where distance and time are measured in SI units. The tension in the string is : | [{"identifier": "A", "content": "10 N"}, {"identifier": "B", "content": "7.5 N"}, {"identifier": "C", "content": "5 N"}, {"identifier": "D", "content": "12.5 N"}] | ["D"] | null | y = 0.03 sin(450 t $$-$$ 9x)
<br><br>v = $${\omega \over k} = {{450} \over 9}$$ = 50m/s
<br><br>v = $$\sqrt {{T \over \mu }} \Rightarrow {T \over \mu }$$ = 2500
<br><br>$$ \Rightarrow $$ T = 2500 $$ \times $$ 5 $$ \times $$ 10<sup>$$-$$3</sup>
<br><br>= 12.5 N | mcq | jee-main-2019-online-11th-january-morning-slot | 13,156 |
H61a7txwgKg17fypuWvj8 | physics | waves | basic-of-waves-and-progressive-waves | A travelling harmonic wave is represented by the equation y(x,t) = 10<sup>–3</sup>sin (50t + 2x), where, x and y are in mater and t is in seconds. Which of the following is a correct statement about the wave ? | [{"identifier": "A", "content": "The wave is propagating along the positive x-axis with speed 100 ms<sup>\u20131</sup>"}, {"identifier": "B", "content": "The wave is propagating along the positive x-axis with speed 25 ms<sup>\u20131</sup>\n"}, {"identifier": "C", "content": "The wave is propagating along the negative x-axis with speed 25 ms<sup>\u20131</sup>"}, {"identifier": "D", "content": "The wave is propagating along the negative x-axis with speed 100 ms<sup>\u20131</sup>"}] | ["C"] | null | y = a sin($$\omega $$t + kx)
<br><br>$$ \Rightarrow $$ wave is moving along $$-$$ve x-axis with speed
<br><br>v = $${\omega \over K}$$ $$ \Rightarrow $$ v = $${{50} \over 2}$$ = 25m/sec | mcq | jee-main-2019-online-12th-january-morning-slot | 13,157 |
SR6QkQKvEU37EApJ9Zw53 | physics | waves | basic-of-waves-and-progressive-waves | The pressure wave, P = 0.01 sin [1000t – 3x] Nm<sup>–2</sup>,
corresponds to the sound produced by a vibrating blade
on a day when atmospheric temperature is 0°C. On
some other day, when temperature is T, the speed of
sound produced by the same blade and at the same
frequency is found to be 336 ms<sup>–1</sup> . Approximate value
of T is | [{"identifier": "A", "content": "12\u00b0C"}, {"identifier": "B", "content": "15\u00b0C"}, {"identifier": "C", "content": "4\u00b0C"}, {"identifier": "D", "content": "11\u00b0C"}] | ["C"] | null | Speed of wave from wave equation<br><br>
$$v = - {{\left( {coeffecient{\rm{ }}of{\rm{ }}t} \right)} \over {\left( {coeffecient{\rm{ }}of{\rm{ }}x} \right){\rm{ }}}}$$<br><br>
$$v = - {{1000} \over {( - 3)}} = {{1000} \over 3}$$<br><br>
Since speed of wave $$ \propto \sqrt T $$<br><br>
So $$ = {{1000} \over {{3 \over {336}}}} = \sqrt {{{273} \over T}} $$<br><br>
$$ \Rightarrow $$ T = 277.41 K<br><br/>
T = 4.41°C
| mcq | jee-main-2019-online-9th-april-morning-slot | 13,158 |
xZ8LeI5xN7BtecvFcY3rsa0w2w9jx3puwbf | physics | waves | basic-of-waves-and-progressive-waves | A progressive wave travelling along the positive x-direction is represented by y(x,t) = Asin(kx – $$\omega $$t + $$\phi $$). Its
snapshot at t = 0 is given in the figure.
<img src="data:image/png;base64,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"/>
<br/>For this wave, the phase $$\phi $$ is : | [{"identifier": "A", "content": "$${\\pi \\over 2}$$"}, {"identifier": "B", "content": "$$\\pi $$"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "$$ - {\\pi \\over 2}$$"}] | ["B"] | null | y = A sin (kx – wt + $$\varphi $$)<br><br>
at x = 0, t = 0 and slope is negative<br><br>
$$ \Rightarrow \varphi = \pi $$ | mcq | jee-main-2019-online-12th-april-morning-slot | 13,159 |
EGhlHU9UoLuqxFE6E63rsa0w2w9jx7i5dk8 | physics | waves | basic-of-waves-and-progressive-waves | A small speaker delivers 2 W of audio output. At what distance from the speaker will one detect 120 dB
intensity sound ? [Given reference intensity of sound as 10<sup>–12</sup> W/m<sup>2</sup>
] | [{"identifier": "A", "content": "20 cm"}, {"identifier": "B", "content": "10 cm"}, {"identifier": "C", "content": "40 cm"}, {"identifier": "D", "content": "30 cm"}] | ["C"] | null | Sound level = 10$${\log _{10}}\left( {{I \over {{I_0}}}} \right)$$
<br><br>$$ \Rightarrow $$ 120 = 10$${\log _{10}}\left( {{I \over {{{10}^{ - 12}}}}} \right)$$
<br><br>$$ \Rightarrow $$ 12 = $${\log _{10}}\left( {{I \over {{{10}^{ - 12}}}}} \right)$$
<br><br>$$ \Rightarrow $$ 10<sup>12</sup> = $${{I \over {{{10}^{ - 12}}}}}$$
<br><br>$$ \Rightarrow $$ I = 1 W/m<sup>2</sup>
<br><br>Also we know,
<br><br>I = $${P \over {4\pi {r^2}}}$$
<br><br>$$ \Rightarrow $$ 1 = $${2 \over {4\pi {r^2}}}$$
<br><br>$$ \Rightarrow $$ r = 40 cm | mcq | jee-main-2019-online-12th-april-evening-slot | 13,160 |
B0Jecr48H1IJRiCs1Xjgy2xukev0j9v4 | physics | waves | basic-of-waves-and-progressive-waves | Two identical strings X and Z made of same
material have tension T<sub>X</sub> and T<sub>Z</sub> in them. If their
fundamental frequencies are 450 Hz and
300 Hz, respectively, then the ratio T<sub>X</sub>/T<sub>Z</sub> is
| [{"identifier": "A", "content": "2.25"}, {"identifier": "B", "content": "0.44"}, {"identifier": "C", "content": "1.25"}, {"identifier": "D", "content": "1.5"}] | ["A"] | null | f = $${1 \over {2l}}\sqrt {{T \over \mu }} $$
<br><br>For identical string $$l$$ and $$\mu $$ will be same
<br><br>f $$ \propto $$ $$\sqrt T $$
<br><br>$$ \therefore $$ $${{450} \over {300}} = \sqrt {{{{T_x}} \over {{T_y}}}} $$
<br><br>$$ \Rightarrow $$ $${{{{T_x}} \over {{T_y}}} = {9 \over 4}}$$ = 2.25 | mcq | jee-main-2020-online-2nd-september-morning-slot | 13,163 |
APAConvaEa3JPbTzze7k9k2k5hgyp97 | physics | waves | basic-of-waves-and-progressive-waves | A transverse wave travels on a taut steel wire
with a velocity of v when tension in it is
2.06 × 10<sup>4</sup> N. When the tension is changed to
T, the velocity changed to v/2. The value of T
is close to : | [{"identifier": "A", "content": "30.5 \u00d7 10<sup>4</sup> N"}, {"identifier": "B", "content": "2.50 \u00d7 10<sup>4</sup> N"}, {"identifier": "C", "content": "10.2 \u00d7 10<sup>2</sup> N"}, {"identifier": "D", "content": "5.15 \u00d7 10<sup>3</sup> N"}] | ["D"] | null | $$v = \sqrt {{T \over \mu }} $$
<br><br>$$ \therefore $$ $${{{v_1}} \over {{v_2}}} = \sqrt {{{{T_1}} \over {{T_2}}}} $$
<br><br>v<sub>1</sub> = v, v<sub>2</sub> = $${v \over 2}$$
<br><br>$$ \Rightarrow $$ $${v \over {{v \over 2}}} = $$ $$\sqrt {{{2.06 \times {{10}^4}} \over {{T_2}}}} $$
<br><br>$$ \Rightarrow $$ T<sub>2</sub> = $${{{2.06 \times {{10}^4}} \over 4}}$$ = 5.15 × 10<sup>3</sup> N | mcq | jee-main-2020-online-8th-january-evening-slot | 13,164 |
VzV2m7NC9fsU8uYnup7k9k2k5dkt9w7 | physics | waves | basic-of-waves-and-progressive-waves | Speed of a transverse wave on a straight wire (mass 6.0 g, length 60 cm and area of cross-section 1.0 mm<sup>2</sup>) is 90 ms<sup>-1</sup>. If the Young's modulus of wire is 16 $$ \times $$ 10<sup>11</sup> Nm<sup>-2</sup>, the extension of wire over its natural length is : | [{"identifier": "A", "content": "0.03 mm"}, {"identifier": "B", "content": "0.04 mm"}, {"identifier": "C", "content": "0.02 mm"}, {"identifier": "D", "content": "0.01 mm"}] | ["A"] | null | Velocity of the wave, v = $$\sqrt {{T \over \mu }} $$
<br><br>$$ \Rightarrow $$ T = v<sup>2</sup>$$\mu $$
<br><br>We know, Youngs modulus, <br><br>Y = $${{{F \over A}} \over {{{\Delta l} \over l}}}$$ = $${{{T \over A}} \over {{{\Delta l} \over l}}}$$
<br><br>[As here F = T]
<br><br>$$ \Rightarrow $$ $${Y{{\Delta l} \over l}}$$ = $${{T \over A}}$$ = $${{{{v^2}\mu } \over A}}$$
<br><br>$$ \Rightarrow $$ $$\Delta $$<i>l</i> = $${{{{v^2}\mu l} \over {AY}}}$$
<br><br>= $${{90 \times 90 \times {{60 \times {{10}^{ - 3}}} \over {60 \times {{10}^{ - 2}}}} \times 60 \times {{10}^{ - 2}}} \over {1 \times {{10}^{ - 6}} \times 16 \times {{10}^{11}}}}$$
<br><br>= 3 $$ \times $$ 10<sup>-5</sup> m
<br><br> = 0.03 mm | mcq | jee-main-2020-online-7th-january-morning-slot | 13,165 |
IedRVvf4ePGbKxJOmkjgy2xukeu3qoi4 | physics | waves | basic-of-waves-and-progressive-waves | For a transverse wave travelling along a straight line, the distance between two peaks (crests) is 5 m, while the distance between one crest and one trough is 1.5 m. The possible wavelengths (in m) of the are :
| [{"identifier": "A", "content": "1, 3, 5, ....."}, {"identifier": "B", "content": "$${1 \\over 1},{1 \\over 3},{1 \\over 5},$$ ....."}, {"identifier": "C", "content": "1, 2, 3, ....."}, {"identifier": "D", "content": "$${1 \\over 2},{1 \\over 4},{1 \\over 6},$$"}] | ["B"] | null | $$1.5 = \left( {2{n_1} + 1} \right){\lambda \over 2}$$ ......(1)
<br><br>5 = n<sub>2</sub>$$\lambda $$ .....(2)
<br><br>(1) $$ \div $$ (2)
<br><br>$${{1.5} \over 5} = {{\left( {2{n_1} + 1} \right)} \over {2{n_2}}}$$
<br><br>$$ \Rightarrow $$ 3n<sub>2</sub> = 10n<sub>1</sub> + 5
<br><br>n<sub>1</sub> = 1 ; n<sub>2</sub> = 5 $$ \Rightarrow $$ $$\lambda $$ = 1
<br><br>n<sub>1</sub> = 4 ; n<sub>2</sub> = 15 $$ \Rightarrow $$ $$\lambda $$ = 1/3
<br><br>n<sub>1</sub> = 7 ; n<sub>2</sub> = 25 $$ \Rightarrow $$ $$\lambda $$ = 1/5 | mcq | jee-main-2020-online-4th-september-morning-slot | 13,166 |
CSMNhNZ3by0JbBLv2p1klt3gom2 | physics | waves | basic-of-waves-and-progressive-waves | The percentage increase in the speed of transverse waves produced in a stretched string if the tension is increased by 4%, will be __________%. | [] | null | 2 | Speed of transverse wave is <br><br>$$V = \sqrt {{T \over \mu }} $$<br><br>$$\ln V = {1 \over 2}\ln T - {1 \over 2}\ln \mu $$<br><br>$${{\Delta V} \over V} = {1 \over 2}{{\Delta T} \over T}$$<br><br>$$ = {1 \over 2} \times 4$$<br><br>$$ \Rightarrow $$ $${{\Delta V} \over V} = 2\% $$ | integer | jee-main-2021-online-25th-february-evening-slot | 13,168 |
3d6HU9Wj9D6P3EXmzW1kltjplxc | physics | waves | basic-of-waves-and-progressive-waves | The mass per unit length of a uniform wire is 0.135 g/cm. A transverse wave of the form y = $$-$$ 0.21 sin (x + 30t) is produced in it, where x is in meter and t is in second. Then, the expected value of tension in the wire is x $$\times$$ 10<sup>$$-$$2</sup> N. Value of x is _________. (Round off to the nearest integer) | [] | null | 1215 | $$\mu = 0.135$$ gm/cm<br><br>$$\mu = 0.135 \times {{{{10}^{ - 3}}} \over {{{10}^{ - 2}}}}{{kg} \over m}$$<br><br>y = $$-$$0.21 sin (x + 30t)<br><br>$$v = {\omega \over K} = {{30} \over 1}$$ = 30 m/s<br><br>v = $$\sqrt {{T \over \mu }} $$<br><br>T = v<sup>2</sup> $$\times$$ $$\mu$$<br><br>T = (30)<sup>2</sup> $$\times$$ 0.135 $$\times$$ 10<sup>$$-$$1</sup><br><br>T = 900 $$\times$$ 0.135 $$\times$$ 10<sup>$$-$$1</sup><br><br>T = 12.15 N<br><br>T = 1215 $$\times$$ 10<sup>$$-$$2</sup> N<br><br>x = 1215 | integer | jee-main-2021-online-26th-february-morning-slot | 13,169 |
MP7OCgdzOAK4bfZ4Vj1kmk7usxc | physics | waves | basic-of-waves-and-progressive-waves | A sound wave of frequency 245 Hz travels with the speed of 300 ms<sup>$$-$$1</sup> along the positive x-axis. Each point of the wave moves to and from through a total distance of 6 cm. What will be the mathematical expression of this travelling wave? | [{"identifier": "A", "content": "Y(x, t) = 0.03 [ sin 5.1x $$-$$ (0.2 $$\\times$$ 10<sup>3</sup>)t ]"}, {"identifier": "B", "content": "Y(x, t) = 0.03 [ sin 5.1x $$-$$ (1.5 $$\\times$$ 10<sup>3</sup>)t ]"}, {"identifier": "C", "content": "Y(x, t) = 0.06 [ sin 5.1x $$-$$ (1.5 $$\\times$$ 10<sup>3</sup>)t ]"}, {"identifier": "D", "content": "Y(x, t) = 0.06 [ sin 0.8x $$-$$ (0.5 $$\\times$$ 10<sup>3</sup>)t ]"}] | ["B"] | null | $$Y = A\sin (kx - \omega t)$$<br><br>$$A = {6 \over 2}$$ = 3cm = 0.03 m<br><br>$$\omega = 2\pi f = 2\pi \times 245$$<br><br>$$\omega = 1.5 \times {10^3}$$<br><br>$$k = {\omega \over v} = {{1.5 \times {{10}^3}} \over {300}}$$<br><br>$$k = 5.1$$<br><br>$$y = 0.03\sin (5.1x - (1.5 \times {10^3})t)$$ | mcq | jee-main-2021-online-17th-march-evening-shift | 13,170 |
1krpqb0us | physics | waves | basic-of-waves-and-progressive-waves | The amplitude of wave disturbance propagating in the positive x-direction is given by $$y = {1 \over {{{(1 + x)}^2}}}$$ at time t = 0 and $$y = {1 \over {1 + {{(x - 2)}^2}}}$$ at t = 1 s, where x and y are in metres. The shape of wave does not change during the propagation. The velocity of the wave will be ___________ m/s. | [] | null | 2 | As per question, <br/><br/>at t = 0, y = $${1 \over {{{(1 + x)}^2}}}$$<br/><br/>and at t = 1 s, y = $${1 \over {1 + {{(x - 2)}^2}}}$$ .... (i)<br/><br/>As we know,<br/><br/>At t = t s, y = $${1 \over {1 + {{(x - vt)}^2}}}$$<br/><br/>So, at t = 1 s, y = $${1 \over {1 + {{(x - v)}^2}}}$$ .... (ii)<br/><br/>On comparing Eqs. (i) and (ii), we get<br/><br/>v = 2 ms<sup>$$-$$1</sup><br/><br/>Hence, the velocity of the wave will be 2 m/s. | integer | jee-main-2021-online-20th-july-morning-shift | 13,171 |
1l56vbh22 | physics | waves | basic-of-waves-and-progressive-waves | <p>If a wave gets refracted into a denser medium, then which of the following is true?</p> | [{"identifier": "A", "content": "wavelength, speed and frequency decreases."}, {"identifier": "B", "content": "wavelength increases, sped decreases and frequency remains constant."}, {"identifier": "C", "content": "wavelength and speed decreases but frequency remains constant."}, {"identifier": "D", "content": "wavelength, speed and frequency increases."}] | ["C"] | null | <p>Frequency is independent of medium. For denser medium, wavelength and speed both would decrease.</p> | mcq | jee-main-2022-online-27th-june-evening-shift | 13,174 |
1l5alcytx | physics | waves | basic-of-waves-and-progressive-waves | <p>The first overtone frequency of an open organ pipe is equal to the fundamental frequency of a closed organ pipe. If the length of the closed organ pipe is 20 cm. The length of the open organ pipe is _____________ cm.</p> | [] | null | 80 | <p>$$2 \times \left( {{V \over {2{L_0}}}} \right) = \left( {{V \over {4{L_c}}}} \right)$$</p>
<p>$$ \Rightarrow {L_0} = 4{L_c}$$</p>
<p>$$ = 4 \times 20$$</p>
<p>$$ = 80$$ cm</p> | integer | jee-main-2022-online-25th-june-morning-shift | 13,175 |
1l5w2qtor | physics | waves | basic-of-waves-and-progressive-waves | <p>Which of the following equations correctly represents a travelling wave having wavelength $$\lambda$$ = 4.0 cm, frequency v = 100 Hz and travelling in positive x-axis direction?</p> | [{"identifier": "A", "content": "$$y = A\\sin [(0.50\\,\\pi \\,c{m^{ - 1}})x - (100\\,\\pi \\,{s^{ - 1}})t]$$"}, {"identifier": "B", "content": "$$y = A\\sin \\,\\,2\\pi [(0.25\\,\\,c{m^{ - 1}})x - (50\\,{s^{ - 1}})t]$$"}, {"identifier": "C", "content": "$$y = A\\sin \\left[ {\\left( {{{2\\pi } \\over 4}\\,c{m^{ - 1}}} \\right)x - \\left( {{{2\\pi } \\over {100}}\\,{s^{ - 1}}} \\right)t} \\right]$$"}, {"identifier": "D", "content": "$$y = A\\sin \\,\\pi [(0.5\\,\\,c{m^{ - 1}})x - (200\\,\\,{s^{ - 1}})t]$$"}] | ["D"] | null | <p>We know, equation of wave travelling in positive x-direction is -</p>
<p>$$y = A\sin (kx - wt)$$</p>
<p>where $$k = {{2\pi } \over \lambda }$$</p>
<p>and $$w = 2\pi f$$</p>
<p>Here given $$\lambda$$ = 4 cm and frequency (f) = 100 Hz</p>
<p>$$\therefore$$ $$k = {{2\pi } \over 4} = 0.5\pi $$ cm<sup>$$-$$1</sup></p>
<p>and $$w = 2\pi \times 100 = 200\pi $$ s<sup>$$-$$1</sup></p>
<p>$$\therefore$$ Equation of travelling wave,</p>
<p>$$y = A\sin (0.5\pi x - 200\pi t)$$</p> | mcq | jee-main-2022-online-30th-june-morning-shift | 13,176 |
1l6ma0qrt | physics | waves | basic-of-waves-and-progressive-waves | <p>In the wave equation</p>
<p>$$
y=0.5 \sin \frac{2 \pi}{\lambda}(400 \mathrm{t}-x) \,\mathrm{m}
$$</p>
<p>the velocity of the wave will be:</p> | [{"identifier": "A", "content": "200 m/s"}, {"identifier": "B", "content": "200$$\\sqrt2$$ m/s"}, {"identifier": "C", "content": "400 m/s"}, {"identifier": "D", "content": "400$$\\sqrt2$$ m/s"}] | ["C"] | null | <p>$${v_{wave}} = \left| {{{coefficient\,of\,t} \over {coefficient\,of\,x}}} \right|$$</p>
<p>$$ = {{400} \over 1} = 400$$ m/s</p> | mcq | jee-main-2022-online-28th-july-morning-shift | 13,178 |
1ldofk8r1 | physics | waves | basic-of-waves-and-progressive-waves | <p>A steel wire with mass per unit length $$7.0 \times 10^{-3} \mathrm{~kg} \mathrm{~m}^{-1}$$ is under tension of $$70 \mathrm{~N}$$. The speed of transverse waves in the wire will be:</p> | [{"identifier": "A", "content": "$$10 \\mathrm{~m} / \\mathrm{s}$$"}, {"identifier": "B", "content": "$$50 \\mathrm{~m} / \\mathrm{s}$$"}, {"identifier": "C", "content": "$$100 \\mathrm{~m} / \\mathrm{s}$$"}, {"identifier": "D", "content": "$$200 \\pi\\mathrm{~m} / \\mathrm{s}$$"}] | ["C"] | null | Speed of transverse wave $=\sqrt{\frac{T}{\mu}}$ $=\sqrt{\frac{70}{7 \times 10^{-3}}}=100 \mathrm{~m} / \mathrm{s}$ | mcq | jee-main-2023-online-1st-february-morning-shift | 13,179 |
1ldujmgbz | physics | waves | basic-of-waves-and-progressive-waves | <p>The distance between two consecutive points with phase difference of 60$$^\circ$$ in a wave of frequency 500 Hz is 6.0 m. The velocity with which wave is travelling is __________ km/s</p> | [] | null | 18 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le5ky7xx/062183e2-42e3-4295-aca9-366c2da5a2a3/7ebd1e50-ad22-11ed-8bc1-d3bd0941e5b5/file-1le5ky7xy.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le5ky7xx/062183e2-42e3-4295-aca9-366c2da5a2a3/7ebd1e50-ad22-11ed-8bc1-d3bd0941e5b5/file-1le5ky7xy.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 25th January Morning Shift Physics - Waves Question 22 English Explanation"></p>
<p>$$
\begin{aligned}
& \Delta x=\frac{\lambda}{2 \pi} \times\left(\frac{\pi}{3}\right)=\left(\frac{\lambda}{6}\right) \\\\
& \Rightarrow \quad \frac{\lambda}{6}=6 \mathrm{~m} \\\\
& \quad \lambda=36 \mathrm{~m} \\\\
& U=f\lambda=500 \mathrm{~Hz} \times 36 \\\\
& =18000 \mathrm{~m} / \mathrm{s} \\\\
& =18 \mathrm{~km} / \mathrm{s}
\end{aligned}
$$</p> | integer | jee-main-2023-online-25th-january-morning-shift | 13,180 |
1ldye5kc3 | physics | waves | basic-of-waves-and-progressive-waves | <p>A travelling wave is described by the equation</p>
<p>$$y(x,t) = [0.05\sin (8x - 4t)]$$ m</p>
<p>The velocity of the wave is : [all the quantities are in SI unit]</p> | [{"identifier": "A", "content": "$$\\mathrm{4~ms^{-1}}$$"}, {"identifier": "B", "content": "$$\\mathrm{2~ms^{-1}}$$"}, {"identifier": "C", "content": "$$\\mathrm{8~ms^{-1}}$$"}, {"identifier": "D", "content": "$$\\mathrm{0.5~ms^{-1}}$$"}] | ["D"] | null | $\because y(x, t)=[0.05 \sin (8 x-4 t)] \mathrm{m}$
<br/><br/>
$$
\begin{aligned}
\text { Speed of wave } & =\left|\frac{\text { Coefficient of } t}{\text { Coefficient of } x}\right| \\\\
& =\frac{4}{8}=0.5 \mathrm{~ms}^{-1}
\end{aligned}
$$ | mcq | jee-main-2023-online-24th-january-morning-shift | 13,181 |
lgnyy5pt | physics | waves | basic-of-waves-and-progressive-waves | The fundamental frequency of vibration of a string stretched between two rigid support is $50 \mathrm{~Hz}$. The mass of the string is $18 \mathrm{~g}$ and its linear mass density is $20 \mathrm{~g} / \mathrm{m}$. The speed of the transverse waves so produced in the string is ___________ $\mathrm{ms}^{-1}$ | [] | null | 90 | To find the speed of the transverse waves produced in the string, we can use the formula for the fundamental frequency of a vibrating string:
<br/><br/>
$f = \frac{1}{2L} \cdot v$
<br/><br/>
where $f$ is the fundamental frequency, $L$ is the length of the string, and $v$ is the speed of the transverse waves.
<br/><br/>
First, we are given the mass of the string ($m = 18g$) and the linear mass density ($\mu = 20g/m$). We can find the length of the string by dividing the mass by the linear mass density:
<br/><br/>
$L = \frac{m}{\mu} = \frac{18g}{20g/m} = 0.9m$
<br/><br/>
Now we can plug in the values for the fundamental frequency ($f = 50Hz$) and the length of the string ($L = 0.9m$) into the formula:
<br/><br/>
$50Hz = \frac{1}{2(0.9m)} \cdot v$
<br/><br/>
To isolate $v$, we multiply both sides by $2(0.9m)$:
<br/><br/>
$v = 50Hz \cdot 2(0.9m) = 90 \mathrm{ms}^{-1}$
<br/><br/>
The speed of the transverse waves produced in the string is $90 ~\mathrm{ms}^{-1}$.<br/><br/>
<b>Alternate Method:</b><br/><br/>
To find the speed of the transverse waves produced in the string, we can use the formula for the fundamental frequency of a vibrating string:
<br/><br/>
$f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$
<br/><br/>
where $f_1$ is the fundamental frequency, $L$ is the length of the string, $T$ is the tension in the string, and $\mu$ is the linear mass density of the string.
<br/><br/>
We're given that the fundamental frequency $f_1 = 50 ,\text{Hz}$, the mass of the string $m = 18 ,\text{g}$, and the linear mass density $\mu = 20 ,\text{g/m}$. To find the speed of the transverse waves, we need to find the tension $T$ and the length $L$ of the string.
<br/><br/>
First, let's find the length $L$ of the string using the mass and linear mass density:
<br/><br/>
$L = \frac{m}{\mu} = \frac{18 ,\text{g}}{20 ,\text{g/m}} = 0.9 ~\text{m}$
<br/><br/>
Now, we can rearrange the formula for the fundamental frequency to solve for the tension $T$:
<br/><br/>
$T = \mu \left(\frac{2Lf_1}{1}\right)^2$
<br/><br/>
Substitute the known values:
<br/><br/>
$T = 20 ,\text{g/m} \cdot \left(\frac{2 \cdot 0.9 ~\text{m} \cdot 50 ~\text{Hz}}{1}\right)^2$
<br/><br/>
$T = 20 ,\text{g/m} \cdot (90 ~\text{m/s})^2$
<br/><br/>
$T = 20 ,\text{g/m} \cdot 8100 ~\text{m}^2/\text{s}^2$
<br/><br/>
$T = 162000 ,\text{g m}/\text{s}^2$
<br/><br/>
Now, we can find the speed of the transverse waves $v$ using the formula:
<br/><br/>
$v = \sqrt{\frac{T}{\mu}}$
<br/><br/>
Substitute the known values:
<br/><br/>
$v = \sqrt{\frac{162000}{20}}$
<br/><br/>
$v = \sqrt{8100} = 90~ \text{m/s}$
<br/><br/>
The speed of the transverse waves produced in the string is $90 ~\text{m/s}$. | integer | jee-main-2023-online-15th-april-morning-shift | 13,182 |
1lgp11j31 | physics | waves | basic-of-waves-and-progressive-waves | <p>In an experiment with sonometer when a mass of $$180 \mathrm{~g}$$ is attached to the string, it vibrates with fundamental frequency of $$30 \mathrm{~Hz}$$. When a mass $$\mathrm{m}$$ is attached, the string vibrates with fundamental frequency of $$50 \mathrm{~Hz}$$. The value of $$\mathrm{m}$$ is ___________ g.</p> | [] | null | 500 | We can use the fact that the ratio of frequencies is equal to the square root of the ratio of tensions:
<br/><br/>
$$\frac{f_2}{f_1}=\sqrt{\frac{T_2}{T_1}}$$
<br/><br/>
In the first case, the mass attached to the string is $$180 \mathrm{~g}$$ and the frequency is $$30 \mathrm{~Hz}$$, so we have:
<br/><br/>
$$\frac{f_2}{30~\mathrm{Hz}}=\sqrt{\frac{T_2}{T_1}}$$
<br/><br/>
In the second case, the frequency is $$50 \mathrm{~Hz}$$, so we have:
<br/><br/>
$$\frac{50~\mathrm{Hz}}{30~\mathrm{Hz}}=\sqrt{\frac{T_2}{T_1}}$$
<br/><br/>
Simplifying, we get:
<br/><br/>
$$\frac{5}{3}=\sqrt{\frac{T_2}{T_1}}$$
<br/><br/>
Squaring both sides, we get:
<br/><br/>
$$\frac{25}{9}=\frac{T_2}{T_1}$$
<br/><br/>
Since the tension in the string is proportional to the mass attached to it, we can write:
<br/><br/>
$$\frac{m}{180~\mathrm{g}}=\frac{T_2}{T_1}=\frac{25}{9}$$
<br/><br/>
Solving for $$m$$, we get:
<br/><br/>
$$m=\frac{25}{9}(180~\mathrm{g})=\boxed{500~\mathrm{g}}$$
<br/><br/>
Therefore, the mass attached to the string in the second case is $$500 \mathrm{~g}$$. | integer | jee-main-2023-online-13th-april-evening-shift | 13,183 |
1lgrjvfdb | physics | waves | basic-of-waves-and-progressive-waves | <p>For a certain organ pipe, the first three resonance frequencies are in the ratio of $$1:3:5$$ respectively. If the frequency of fifth harmonic is $$405 \mathrm{~Hz}$$ and the speed of sound in air is $$324 \mathrm{~ms}^{-1}$$ the length of the organ pipe is _________ $$\mathrm{m}$$.</p> | [] | null | 1 | <p>Given that the first three resonance frequencies are in the ratio of $$1:3:5$$, we can express them as follows:</p>
<p>$$f_1 = kf$$
$$f_3 = 3kf$$
$$f_5 = 5kf$$</p>
<p>Where $$k$$ is a constant and $$f_1, f_3$$, and $$f_5$$ are the first, third, and fifth resonance frequencies, respectively. We are given that the frequency of the fifth harmonic is $$405 \mathrm{~Hz}$$, so we can write:</p>
<p>$$f_5 = 5kf = 405 \mathrm{~Hz}$$</p>
<p>Now we can solve for the constant $$k$$:</p>
<p>$$k = \frac{405}{5} = 81 \mathrm{~Hz}$$</p>
<p>We also know that the speed of sound in air is $$v = 324 \mathrm{~ms}^{-1}$$. The relationship between the speed of sound, the frequency, and the wavelength of a standing wave in a closed pipe can be expressed as follows:</p>
<p>$$v = f\lambda$$</p>
<p>Where $$\lambda$$ is the wavelength of the wave. For the first harmonic in a closed pipe, the length of the pipe is equal to one-fourth of the wavelength:</p>
<p>$$L = \frac{1}{4}\lambda$$</p>
<p>We can now substitute the expression for the wavelength in terms of the length into the equation for the speed of sound:</p>
<p>$$v = f_1 \cdot 4L$$</p>
<p>Now, we can substitute the value of $$f_1 = kf = 81 \mathrm{~Hz}$$ and the speed of sound $$v = 324 \mathrm{~ms}^{-1}$$ into the equation:</p>
<p>$$324 = 81 \times 4L$$</p>
<p>Now we can solve for the length of the organ pipe $$L$$:</p>
<p>$$L = \frac{324}{81 \times 4} = \frac{324}{324} = 1 \mathrm{~m}$$</p>
<p>The length of the organ pipe is $$1 \mathrm{~m}$$.</p>
| integer | jee-main-2023-online-12th-april-morning-shift | 13,184 |
1lgvrlanb | physics | waves | basic-of-waves-and-progressive-waves | <p>For a periodic motion represented by the equation</p>
<p>$$y=\sin \omega \mathrm{t}+\cos \omega \mathrm{t}$$</p>
<p>the amplitude of the motion is</p> | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "$$\\sqrt2$$"}, {"identifier": "C", "content": "0.5"}, {"identifier": "D", "content": "2"}] | ["B"] | null | <p>We can write the given equation as:</p>
<p>$$y = \sqrt{(\sin\omega t)^2 + (\cos\omega t)^2} \cos\left(\omega t - \arctan\frac{\sin\omega t}{\cos\omega t}\right)$$</p>
<p>Using the identity $\sin^2\theta + \cos^2\theta = 1$, we get:</p>
<p>$$y = \sqrt{1 + \sin 2\omega t} \cos\left(\omega t - \frac{\pi}{4}\right)$$</p>
<p>The amplitude of the motion is the maximum value of $|y|$, which occurs when $\sin 2\omega t = 1$, i.e., at $t = \frac{\pi}{4\omega} + \frac{n\pi}{\omega}$, where $n$ is an integer. Substituting this value of $t$ in the above equation, we get:</p>
<p>$$|y_{\text{max}}| = \sqrt{1 + \sin \frac{\pi}{2}} = \sqrt{2}$$</p>
<p>Therefore, the amplitude of the motion is $\sqrt{2}$</p>
| mcq | jee-main-2023-online-10th-april-evening-shift | 13,186 |
1lgyfgqv6 | physics | waves | basic-of-waves-and-progressive-waves | <p>A transverse harmonic wave on a string is given by</p>
<p>$$y(x,t) = 5\sin (6t + 0.003x)$$</p>
<p>where x and y are in cm and t in sec. The wave velocity is _______________ ms$$^{-1}$$.</p> | [] | null | 20 | <p>The general equation for a transverse harmonic wave on a string is given by:</p>
<p>$$ y(x,t) = A \sin(kx - \omega t + \phi) $$</p>
<p>where $A$ is the amplitude of the wave, $k$ is the wave number, $\omega$ is the angular frequency, and $\phi$ is the phase constant. The wave velocity $v$ is related to the wave number and angular frequency by the formula:</p>
<p>$$ v = \frac{\omega}{k} $$</p>
<p>Comparing the given equation with the general equation, we can see that:</p>
<p>$$ A = 5 \, \text{cm} $$</p>
<p>$$ k = 0.003 \, \text{cm}^{-1} $$</p>
<p>$$ \omega = 6 \, \text{rad/s} $$</p>
<p>Therefore, the wave velocity is:</p>
<p>$$ v = \frac{\omega}{k} = \frac{6}{0.003} = 2000 \, \text{cm/s} = \boxed{20 \, \text{m/s}} $$</p>
| integer | jee-main-2023-online-10th-april-morning-shift | 13,187 |
jaoe38c1lse6e2vu | physics | waves | basic-of-waves-and-progressive-waves | <p>The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If length of the open pipe is $$60 \mathrm{~cm}$$, the length of the closed pipe will be:</p> | [{"identifier": "A", "content": "15 cm"}, {"identifier": "B", "content": "60 cm"}, {"identifier": "C", "content": "45 cm"}, {"identifier": "D", "content": "30 cm"}] | ["A"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsluo2az/0a46530b-3d75-49ab-a7dd-6ded73044d75/114787b0-cb40-11ee-ad47-a16d1086e690/file-6y3zli1lsluo2b0.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsluo2az/0a46530b-3d75-49ab-a7dd-6ded73044d75/114787b0-cb40-11ee-ad47-a16d1086e690/file-6y3zli1lsluo2b0.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 31st January Morning Shift Physics - Waves Question 7 English Explanation"></p>
<p>$$\begin{aligned}
& \mathrm{f}_1=\frac{\mathrm{v}}{4 \mathrm{~L}_1} \\
& \mathrm{f}_1=\mathrm{f}_2 \\
& \frac{\mathrm{v}}{4 \mathrm{~L}_1}=\frac{\mathrm{v}}{\mathrm{L}_2} \\
& \Rightarrow \mathrm{L}_2=4 \mathrm{~L}_1 \\
& 60=4 \times \mathrm{L}_1 \\
& \mathrm{~L}_1=15 \mathrm{~cm}
\end{aligned}$$</p> | mcq | jee-main-2024-online-31st-january-morning-shift | 13,188 |
1lsg71zjn | physics | waves | basic-of-waves-and-progressive-waves | <p>A point source is emitting sound waves of intensity $$16 \times 10^{-8} \mathrm{~Wm}^{-2}$$ at the origin. The difference in intensity (magnitude only) at two points located at a distances of $$2 m$$ and $$4 m$$ from the origin respectively will be _________ $$\times 10^{-8} \mathrm{~Wm}^{-2}$$.</p> | [] | null | 3 | <p>To solve this problem, we need to understand the relationship between the intensity of sound waves and the distance from the source. The intensity $I$ of sound waves from a point source decreases with the square of the distance $r$ from the source, according to the inverse square law, which can be expressed as:</p>
<p>$$ I \propto \frac{1}{r^2} $$</p>
<p>This means that if the distance is doubled, the intensity becomes one-fourth of its initial value because $ (2r)^2 = 4r^2 $.</p>
<p>The initial intensity given at the origin (source) is:</p>
<p>$$ I_0 = 16 \times 10^{-8} \mathrm{~Wm}^{-2} $$</p>
<p>Let's call $ I_1 $ the intensity at $ r = 2 \text{ m} $ and $ I_2 $ the intensity at $ r = 4 \text{ m} $. Using the inverse square law, we can write:</p>
<p>$$ I_1 = \frac{I_0}{(2)^2} = \frac{I_0}{4} $$</p>
<p>and</p>
<p>$$ I_2 = \frac{I_0}{(4)^2} = \frac{I_0}{16} $$</p>
<p>Now substitute the given value for $ I_0 $ to find $ I_1 $ and $ I_2 $:</p>
<p>$$ I_1 = \frac{16 \times 10^{-8}}{4} = 4 \times 10^{-8} \mathrm{~Wm}^{-2} $$</p>
<p>$$ I_2 = \frac{16 \times 10^{-8}}{16} = 1 \times 10^{-8} \mathrm{~Wm}^{-2} $$</p>
<p>Now to find the difference in intensity (magnitude only) between the two points, we subtract $ I_2 $ from $ I_1 $:</p>
<p>$$ \Delta I = | I_1 - I_2 | $$</p>
<p>$$ \Delta I = | 4 \times 10^{-8} - 1 \times 10^{-8} | $$</p>
<p>$$ \Delta I = 3 \times 10^{-8} \mathrm{~Wm}^{-2} $$</p>
<p>So the difference in intensity (magnitude only) at the two points is:</p>
<p>$$ 3 \times 10^{-8} \mathrm{~Wm}^{-2} $$</p></p> | integer | jee-main-2024-online-30th-january-evening-shift | 13,189 |
1lsgxgqv0 | physics | waves | basic-of-waves-and-progressive-waves | <p>In a closed organ pipe, the frequency of fundamental note is $$30 \mathrm{~Hz}$$. A certain amount of water is now poured in the organ pipe so that the fundamental frequency is increased to $$110 \mathrm{~Hz}$$. If the organ pipe has a cross-sectional area of $$2 \mathrm{~cm}^2$$, the amount of water poured in the organ tube is __________ g. (Take speed of sound in air is $$330 \mathrm{~m} / \mathrm{s}$$)</p> | [] | null | 400 | <p>$$\begin{aligned}
& \frac{V}{4 \ell_1}=30 \Rightarrow \ell_1=\frac{11}{4} m \\
& \frac{V}{4 \ell_2}=110 \Rightarrow \ell_2=\frac{3}{4} m \\
& \Delta \ell=2 m,
\end{aligned}$$</p>
<p>Change in volume $$=A \Delta \ell=400 \mathrm{~cm}^3$$</p>
<p>$$M=400 \mathrm{~g} ;\left(\because \rho=1 \mathrm{~g} / \mathrm{cm}^3\right)$$</p> | integer | jee-main-2024-online-30th-january-morning-shift | 13,190 |
lv3ve49a | physics | waves | basic-of-waves-and-progressive-waves | <p>A plane progressive wave is given by $$y=2 \cos 2 \pi(330 \mathrm{t}-x) \mathrm{m}$$. The frequency of the wave is :</p> | [{"identifier": "A", "content": "660 Hz"}, {"identifier": "B", "content": "340 Hz"}, {"identifier": "C", "content": "330 Hz"}, {"identifier": "D", "content": "165 Hz"}] | ["C"] | null | <p>To find the frequency of the plane progressive wave given by the equation $$y = 2 \cos 2 \pi(330 \mathrm{t} - x) \mathrm{m}$$, we start by analyzing the general form of a wave equation.</p>
<p>The general form of a wave equation is:</p>
$$y = A \cos (2 \pi ft - kx + \phi)$$
<p>where:</p>
<ul>
<li>$$A$$ is the amplitude of the wave.</li>
<li>$$f$$ is the frequency of the wave.</li>
<li>$$t$$ is the time variable.</li>
<li>$$k$$ is the wave number, defined as $$\frac{2 \pi}{\lambda}$$ where $$\lambda$$ is the wavelength.</li>
<li>$$x$$ is the spatial variable.</li>
<li>$$\phi$$ is the phase constant.</li>
</ul>
<p>By comparing the given wave equation with the general form, we have:</p>
$$y = 2 \cos 2 \pi (330 t - x)$$
<p>We observe that the term $$2 \pi(330t - x)$$ corresponds to $$2 \pi ft - kx$$ in the general form.</p>
<p>From this, it is clear that:</p>
<ul>
<li>$$2 \pi ft \rightarrow 2 \pi \cdot 330 t$$</li>
<li>Therefore, $$f = 330$$ Hz</li>
</ul>
<p>So, the frequency of the wave is 330 Hz.</p>
<p>Thus, the correct answer is:</p>
<p><strong>Option C: 330 Hz</strong></p> | mcq | jee-main-2024-online-8th-april-evening-shift | 13,191 |
lv9s25iu | physics | waves | basic-of-waves-and-progressive-waves | <p>A sonometer wire of resonating length $$90 \mathrm{~cm}$$ has a fundamental frequency of $$400 \mathrm{~Hz}$$ when kept under some tension. The resonating length of the wire with fundamental frequency of $$600 \mathrm{~Hz}$$ under same tension _______ $$\mathrm{cm}$$.</p> | [] | null | 60 | <p>The fundamental frequency of a string (in this case, a sonometer wire) when it is vibrating, is inversely proportional to its length, provided the tension in the string and the linear mass density (mass per unit length) remain constant. This relationship is given by the formula:</p>
<p>$$f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$$</p>
<p>where:</p>
<ul>
<li>$f$ is the frequency of the string,</li>
<li>$L$ is the length of the string,</li>
<li>$T$ is the tension in the string, and</li>
<li>$\mu$ is the linear mass density of the string.</li>
</ul>
<p>Given that the sonometer wire has a resonating length of 90 cm (0.9 m) with a fundamental frequency of $400 \mathrm{Hz}$, we can set up our first equation but note that we are comparing two states of the same string under the same tension and hence can eliminate the tension and density terms for comparison purposes:</p>
<p>$$400 = \frac{1}{2 \times 0.9} \sqrt{\frac{T}{\mu}}$$</p>
<p>For the second scenario where the fundamental frequency is $600 \mathrm{Hz}$, we are asked to find the new length $L_2$. We can set up the equation in a similar manner:</p>
<p>$$600 = \frac{1}{2L_2} \sqrt{\frac{T}{\mu}}$$</p>
<p>Since the tension $T$ and the linear density $\mu$ are constants, and they do not change between the two states, we can set up a proportion between the two states by dividing the second equation by the first, which yields:</p>
<p>$$\frac{600}{400} = \frac{\frac{1}{2L_2}}{\frac{1}{2 \times 0.9}}$$</p>
<p>Simplifying this equation gives:</p>
<p>$$\frac{600}{400} = \frac{0.9}{L_2}$$</p>
<p>or</p>
<p>$$\frac{3}{2} = \frac{0.9}{L_2}$$</p>
<p>Solving for $L_2$ gives:</p>
<p>$$L_2 = \frac{0.9 \times 2}{3}$$</p>
<p>Calculating the value:</p>
<p>$$L_2 = \frac{1.8}{3} = 0.6 \hspace{1mm} \text{meters}$$</p>
<p>Converting meters to centimeters (since 1 meter = 100 centimeters), we find:</p>
<p>$$L_2 = 0.6 \times 100 = 60 \hspace{1mm} \text{centimeters}$$</p>
<p>Therefore, the resonating length of the wire with a fundamental frequency of $600 \mathrm{Hz}$ under the same tension is <strong>60 cm</strong>.</p> | integer | jee-main-2024-online-5th-april-evening-shift | 13,192 |
lvb29gyo | physics | waves | basic-of-waves-and-progressive-waves | <p>Two open organ pipes of lengths $$60 \mathrm{~cm}$$ and $$90 \mathrm{~cm}$$ resonate at $$6^{\text {th }}$$ and $$5^{\text {th }}$$ harmonics respectively. The difference of frequencies for the given modes is _________ $$\mathrm{Hz}$$. (Velocity of sound in air $$=333 \mathrm{~m} / \mathrm{s}$$)</p> | [] | null | 740 | <p>To solve this problem, let's first understand how the harmonics of open organ pipes work. For an open organ pipe, the harmonics are given by the formula:
<p>$$f_n = n \frac{v}{2L}$$</p>
<p>where:</p>
<ul>
<li>$$f_n$$ is the frequency of the nth harmonic,</li><br>
<li>$$n$$ is the harmonic number (an integer),</li><br>
<li>$$v$$ is the speed of sound in air,</li><br>
<li>$$L$$ is the length of the organ pipe.</li>
</ul>
<p>Given that the speeds of sound in air $$v = 333 \, \text{m/s}$$, and the lengths of the two open organ pipes are $$60 \, \text{cm} = 0.60 \, \text{m}$$ and $$90 \, \text{cm} = 0.90 \, \text{m}$$, we can calculate the frequencies of the 6th harmonic for the 60 cm pipe and the 5th harmonic for the 90 cm pipe.</p>
<p>For the 60 cm pipe at the 6th harmonic ($$n = 6$$):</p>
<p>$$f_{6,60} = 6 \frac{333}{2 \times 0.60} = 6 \times \frac{333}{1.2} = 6 \times 277.5 = 1665 \, \text{Hz}$$</p>
<p>For the 90 cm pipe at the 5th harmonic ($$n = 5$$):</p>
<p>$$f_{5,90} = 5 \frac{333}{2 \times 0.90} = 5 \times \frac{333}{1.8} = 5 \times 185 = 925 \, \text{Hz}$$</p>
<p>The difference in frequencies between these two modes is:</p>
<p>$$\Delta f = f_{6,60} - f_{5,90} = 1665 \, \text{Hz} - 925 \, \text{Hz} = 740 \, \text{Hz}$$</p>
<p>Therefore, the difference of frequencies for the given modes is $$740 \, \text{Hz}$$.</p></p> | integer | jee-main-2024-online-6th-april-evening-shift | 13,193 |
4MofO2vKfhMQ6Pth | physics | waves | doppler-effect | An observer moves towards a stationary source of sound, with a velocity one-fifth of the velocity of sound. What is the percentage increase in the apparent frequency ? | [{"identifier": "A", "content": "$$0.5\\% $$ "}, {"identifier": "B", "content": "zero "}, {"identifier": "C", "content": "$$20\\% $$ "}, {"identifier": "D", "content": "$$5\\% $$ "}] | ["C"] | null | $$n' = n\left[ {{{v + {v_0}} \over v}} \right] = n\left[ {{{v + {v \over 5}} \over v}} \right] = n\left[ {{6 \over 5}} \right]$$
<br><br>$${{n'} \over n} = {6 \over 5};{{n' - n} \over n}$$
<br><br>$$ = {{6 - 5} \over 5} \times 100 = 20\% $$ | mcq | aieee-2005 | 13,194 |
SiKOIADsyZ9cP2gD | physics | waves | doppler-effect | A whistle producing sound waves of frequencies $$9500$$ $$Hz$$ and above is approaching a stationary person with speed $$v$$ $$m{s^{ - 1}}.$$ The velocity of sound in air is $$300\,m{s^{ - 1}}.$$ If the person can hear frequencies upto a maximum of $$10,000$$ $$HZ,$$ the maximum value of $$v$$ upto which he can hear whistle is | [{"identifier": "A", "content": "$$15\\sqrt 2 \\,\\,m{s^{ - 1}}$$ "}, {"identifier": "B", "content": "$${{15} \\over {\\sqrt 2 }}\\,m{s^{ - 1}}$$ "}, {"identifier": "C", "content": "$$15\\,\\,m{s^{ - 1}}$$ "}, {"identifier": "D", "content": "$$30\\,\\,m{s^{ - 1}}$$ "}] | ["C"] | null | $$v' = v\left[ {{v \over {v - {v_s}}}} \right] \Rightarrow 10000$$
<br><br>$$ = 9500\left[ {{{300} \over {300 - v}}} \right]$$
<br><br>$$ \Rightarrow 300 - v = 300 \times 0.95 \Rightarrow v$$
<br><br>$$ = 300 - 285 = 15\,m{s^{ - 1}}$$ | mcq | aieee-2006 | 13,195 |
MYrgUrSyS9emm9QI | physics | waves | doppler-effect | A motor cycle starts from rest and accelerates along a straight path at $$2m/{s^2}.$$ At the starting point of the motor cycle there is a stationary electric siren. How far has the motor cycle gone when the driver hears the frequency of the siren at $$94\% $$ of its value when the motor cycle was at rest? (Speed of sound $$ = 330\,m{s^{ - 1}}$$) | [{"identifier": "A", "content": "$$98$$ $$m$$ "}, {"identifier": "B", "content": "$$147$$ $$m$$ "}, {"identifier": "C", "content": "$$196\\,m$$ "}, {"identifier": "D", "content": "$$49$$ $$m$$ "}] | ["A"] | null | <img class="question-image" src="https://imagex.cdn.examgoal.net/1POnWfVR0Y8DB7eSi/pNjeKMnw9GzZpaWsxVEN8SzzsQbf6/2UepBhXoKr2RLCCACXLmHM/image.svg" loading="lazy" alt="AIEEE 2009 Physics - Waves Question 117 English Explanation">
<br><br>$$v_m^2 - {u^2} = 2as \Rightarrow v_m^2 = 2 \times 2 \times s$$
<br><br>$$\therefore$$ $${v_m} = 2\sqrt s $$
<br><br>According to Doppler's effect
<br><br>$$0.94v = v\left[ {{{330 - 2\sqrt s } \over {330}}} \right] \Rightarrow s = 98.01\,m$$ | mcq | aieee-2009 | 13,196 |
OExRi6Lu6uVj7AJu | physics | waves | doppler-effect | A train is moving on a straight track with speed $$20\,m{s^{ - 1}}.$$ It is blowing its whistle at the frequency of $$1000$$ $$Hz$$. The percentage change in the frequency heard by a person standing near the track as the train passes him is (speed of sound $$ = 320\,m{s^{ - 1}}$$) close to : | [{"identifier": "A", "content": "$$18\\% $$ "}, {"identifier": "B", "content": "$$24\\% $$"}, {"identifier": "C", "content": "$$6\\% $$"}, {"identifier": "D", "content": "$$12\\% $$"}] | ["D"] | null | $${f_1} = f\left[ {{v \over {v - {v_s}}}} \right] = f \times {{320} \over {300}}Hz$$
<br><br>$${f_2} = f\left[ {{v \over {v + {v_s}}}} \right] = f \times {{320} \over {340}}Hz$$
<br><br>$$\left( {{{{f_2}} \over {{f_1}}} - 1} \right) \times 100 = \left( {{{300} \over {340}} - 1} \right) \times 100 = 12\% $$ | mcq | jee-main-2015-offline | 13,197 |
60X1nzX2V8xmSQS3CmPfZ | physics | waves | doppler-effect | Two engines pass each other moving in opposite directions with uniform speed of 30 m/s. One of them is blowing a whistle of frequency 540 Hz. Calculate the
frequency heard by driver of second engine before they pass each other. Speed of sound is 330 m/sec : | [{"identifier": "A", "content": "450 Hz "}, {"identifier": "B", "content": "540 Hz"}, {"identifier": "C", "content": "648 Hz"}, {"identifier": "D", "content": "270 Hz"}] | ["C"] | null | Frequency heard by the driver of second engine,
<br><br>F' = $$\left( {{{v + {v_0}} \over {v - {v_S}}}} \right)$$ f
<br><br>given v<sub>0</sub> = v<sub>s</sub> = 30 m/s
<br><br>$$ \therefore $$ f' = $$\left( {{{330 + 30} \over {330 - 30}}} \right) \times 540$$
<br><br>= 648 Hz | mcq | jee-main-2016-online-9th-april-morning-slot | 13,198 |
bsRPs9UNc7doIecsOnarW | physics | waves | doppler-effect | A toy-car, blowing its horn, is moving with a steady speed of 5 m/s, away from a wall. An observer, towards whom the toy car is moving, is able to hear 5 beats per second. If the velocity of sound in air is 340 m/s, the frequency of the horn of the toy car is close to :
| [{"identifier": "A", "content": "680 Hz"}, {"identifier": "B", "content": "510 Hz"}, {"identifier": "C", "content": "340 Hz"}, {"identifier": "D", "content": "170 Hz"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266580/exam_images/b0i9i2scgvchdl3mnqwi.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2016 (Online) 10th April Morning Slot Physics - Waves Question 98 English Explanation">
<br><br>Let the frequency of the horn = f
<br><br>Apparent frequency heared by the observer directly,
<br><br>$${f_{dir}} = \left( {{V \over {V - {V_s}}}} \right)f = \left( {{{340} \over {340 - 5}}} \right)f = {{340} \over {335}}f$$
<br><br>Apparent frequency heared by the observer on reflection from the wall,
<br><br>$${f_{ind}} = \left( {{V \over {V + {V_s}}}} \right)f = \left( {{{340} \over {340 + 5}}} \right)f = {{340} \over {345}}f$$
<br><br>Also, given that,
<br><br>f<sub>ind</sub> $$-$$ f<sub>dir</sub> $$=$$ 5
<br><br>$$ \Rightarrow $$ $${{340f} \over {345}}$$ $$-$$ $${{340f} \over {335}}$$ $$=$$ 5
<br><br>$$ \Rightarrow $$ f $$=$$ $${5 \over {340}} \times {{335 \times 345} \over {10}}$$
<br><br>= 169.9 $$ \simeq $$ 170 Hz | mcq | jee-main-2016-online-10th-april-morning-slot | 13,199 |
2KcecQ7w09psggxN | physics | waves | doppler-effect | An observer is moving with half the speed of light towards a stationary microwave source emitting waves
at frequency 10 GHz. What is the frequency of the microwave measured by the observer? (speed of light =
3 ×10<sup>8</sup> ms<sup>–1</sup>) | [{"identifier": "A", "content": "15.3 GHz"}, {"identifier": "B", "content": "10.1 GHz"}, {"identifier": "C", "content": "12.1 GHz"}, {"identifier": "D", "content": "17.3 GHz"}] | ["D"] | null | This question is from Doppler's effect of light.
<br><br>When observer is moving towards the source then the frequency of wave measured by the observer will be
<br><br>f<sub>observed</sub> = f<sub>actual</sub>$$\sqrt {{{c + v} \over {c - v}}} $$
<br><br>where c = speed of light and v = speed of observer
<br><br>According to the question, v = $${c \over 2}$$
<br><br>$$\therefore$$ f<sub>observed</sub> = f<sub>actual</sub>$$\sqrt {{{c + {c \over 2}} \over {c - {c \over 2}}}} $$
<br><br> = 10$$ \times $$$$\sqrt {{{{{3c} \over 2}} \over {{c \over 2}}}} $$
<br><br> = 10$$\sqrt 3 $$ = 17.3 GHz | mcq | jee-main-2017-offline | 13,200 |
BVOU7UdVCNvscWtoYT6jT | physics | waves | doppler-effect | Two sitar strings, A and B, playing the note 'Dha' are slightly out of tune and produce beats of frequency 5 Hz. The tension of the string B s slightly increased and the beat frequency is found to decrease by 3Hz. If the frequency of A is 425 Hz, the original frequency of B is : | [{"identifier": "A", "content": "430 Hz"}, {"identifier": "B", "content": "420 Hz"}, {"identifier": "C", "content": "428 Hz"}, {"identifier": "D", "content": "422 Hz"}] | ["B"] | null | Frequency of B, f<sub>B</sub> = 425 $$ \pm $$ 5 = 420 or 430 Hz
<br><br>As tension of string B is increased
<br><br>So, frequency of B, f<sub>B</sub> should also increase [as f $$ \propto $$ $$\sqrt T $$]
<br><br>If initially f<sub>B</sub> = 430 Hz then when f<sub>B</sub> increases by increasing the tension then f<sub>B</sub> $$-$$ f<sub>A</sub> increases that means beat frequency increase.
<br><br>So, f<sub>B</sub> can't be 430 Hz
<br><br>When f<sub>B</sub> = 420 then when f<sub>B</sub> increases f<sub>A</sub> $$-$$ f<sub>B</sub> decreases means beat frequency decreases. So, correct f<sub>B</sub> = 420 Hz. | mcq | jee-main-2018-online-16th-april-morning-slot | 13,201 |
SpUIuJrk4YBMGww1p4Iko | physics | waves | doppler-effect | A musician using an open flute of length 50 cm producess second harmonic sound waves. A person runs towards the musician from another end of hall at a speed of 10 km/h. If the wave speed is 330 m/s, the frequency heard by the running person shall be close to : | [{"identifier": "A", "content": "666 Hz"}, {"identifier": "B", "content": "753 Hz"}, {"identifier": "C", "content": "500 Hz"}, {"identifier": "D", "content": "333 Hz"}] | ["A"] | null | Frequency of sound wave produce by flute
<br><br>= $${{2{V_S}} \over {2\ell }}$$
<br><br>= $${{2 \times 330} \over {2 \times 50 \times {{10}^{ - 2}}}}$$
<br><br>= 660 Hz
<br><br>Speed of the observer
<br><br>= 10km/hr
<br><br>= 10 $$ \times $$ $${5 \over {18}}$$ m/s
<br><br>= $${{25} \over 9}$$ m/s
<br><br>Frequency heard by the observer,
<br><br>f' = $$\left( {{{{V_S} + {V_o}} \over {{V_S}}}} \right)f$$
<br><br>= $$\left( {{{330 + {{25} \over 9}} \over {330}}} \right) \times 660$$
<br><br>= 666 Hz | mcq | jee-main-2019-online-9th-january-evening-slot | 13,202 |
ECXQtYbodUIg95yuaHdt1 | physics | waves | doppler-effect | A train moves towards a stationary observer with speed 34 m/s. The train sounds a whistle and its frequency registered by the observer is ƒ<sub>1</sub>. If the speed of the train is reduced to 17 m/s, the frequency registered is ƒ<sub>2</sub>. If speed of sound is 340 m/s, then the ratio ƒ<sub>1</sub>/ƒ<sub>2</sub> is - | [{"identifier": "A", "content": "19/18"}, {"identifier": "B", "content": "20/19"}, {"identifier": "C", "content": "21/20"}, {"identifier": "D", "content": "18/17"}] | ["A"] | null | f<sub>app</sub> = f<sub>0</sub> $$\left[ {{{{v_2} \pm {v_0}} \over {{v_2} \pm {v_s}}}} \right]$$
<br><br>f<sub>1</sub> = f<sub>0</sub> $$\left[ {{{340} \over {340 - 34}}} \right]$$
<br><br>f<sub>2</sub> = f<sub>0</sub> $$\left[ {{{340} \over {340 - 17}}} \right]$$
<br><br>$${{{f_1}} \over {{f_2}}} = {{340 - 17} \over {340 - 34}} = {{323} \over {306}} \Rightarrow {{{f_1}} \over {{f_2}}} = {{19} \over {18}}$$ | mcq | jee-main-2019-online-10th-january-morning-slot | 13,203 |
l25sS5PIYF23EuyYXSlOP | physics | waves | doppler-effect | Two cars A and B are moving away from each
other in opposite directions. Both the cars are
moving with a speed of 20 ms<sup>–1</sup> with respect
to the ground. If an observer in car A detects
a frequency 2000 Hz of the sound coming from
car B, what is the natural frequency of the sound
source in car B ?<br/>
(speed of sound in air = 340 ms<sup>–1</sup>) :- | [{"identifier": "A", "content": "2300 Hz"}, {"identifier": "B", "content": "2060 Hz"}, {"identifier": "C", "content": "2250 Hz"}, {"identifier": "D", "content": "2150 Hz"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265320/exam_images/kuqjpe6hob09ulco02v6.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th April Evening Slot Physics - Waves Question 87 English Explanation">
$$f = {{\left( {v \pm {u_0}} \right)} \over {\left( {v \pm {u_s}} \right)}}.{f_0} = {{\left( {v - 20} \right)} \over {\left( {v + 20} \right)}}.{f_0}$$<br><br>
$$ \Rightarrow 2000 = {{320} \over {360}}.{f_0}$$<br><br>
$$ \Rightarrow {{2000 \times 9} \over 8} = {f_0} = 2250\,Hz$$ | mcq | jee-main-2019-online-9th-april-evening-slot | 13,204 |
i5pnbyKL1RZWoPSsSe18hoxe66ijvztf4yl | physics | waves | doppler-effect | A stationary source emits sound waves of
frequency 500 Hz. Two observers moving
along a line passing through the source detect
sound to be of frequencies 480 Hz and 530 Hz.
Their respective speeds are, in ms–1,
(Given speed of sound = 300 m/s) | [{"identifier": "A", "content": "12, 18"}, {"identifier": "B", "content": "16, 14"}, {"identifier": "C", "content": "12, 16"}, {"identifier": "D", "content": "8, 18"}] | ["A"] | null | $$v = {{v + {v_o}} \over v}{v_o}$$<br><br>
$$ \Rightarrow {v_o} = \left( {{v \over {{v_o}}} - 1} \right)v$$<br><br>
$${v_o} = \left( {{{530} \over {500}} - 1} \right)300 = 18\,m/s$$<br><br>
$${v_o} = \left| {\left( {{{480} \over {500}} - 1} \right)300} \right| = 12\,m/s$$ | mcq | jee-main-2019-online-10th-april-morning-slot | 13,205 |
oFaY37SfwT0dHIv9g43rsa0w2w9jwzj67bh | physics | waves | doppler-effect | A source of sound S is moving with a velocity of 50 m/s towards a stationary observer. The observer
measures the frequency of the source as 1000 Hz. What will be the apparent frequency of the source when it
is moving away from the observer after crossing him? (Take velocity of sound in air is 350 m/s) | [{"identifier": "A", "content": "750 Hz"}, {"identifier": "B", "content": "857 Hz"}, {"identifier": "C", "content": "807 Hz"}, {"identifier": "D", "content": "1143 Hz"}] | ["A"] | null | $${f_a} = {V \over {V - {V_s}}}{f_o} = 1000\,Hz$$<br><br>
$$f_a^{'} = {V \over {V + {V_s}}}{f_o}$$<br><br>
$${{f_a^{'}} \over {{f_a}}} = {{V - {V_s}} \over {V + {V_s}}} = {{350 - 50} \over {350 + 50}} = {{300} \over {400}} = {3 \over 4}$$<br><br>
$$f_a^{'} = {3 \over 4} \times 1000 = 750\,Hz$$ | mcq | jee-main-2019-online-10th-april-evening-slot | 13,206 |
jFjezGAIX0ffZW5UK93rsa0w2w9jx3sa7hv | physics | waves | doppler-effect | A submarine (A) travelling at 18 km/hr is being chased along the line of its velocity by another submarine
(B) travelling at 27 km/hr. B sends a sonar signal of 500 Hz to detect A and receives a reflected sound of
frequency $$\upsilon $$. The value of $$\upsilon $$ is close to: (Speed of sound in water =1500 ms<sup>–1</sup>) | [{"identifier": "A", "content": "507 Hz"}, {"identifier": "B", "content": "502 Hz"}, {"identifier": "C", "content": "499 Hz"}, {"identifier": "D", "content": "504 Hz"}] | ["B"] | null | f<sub>1</sub> (frequency received by A)<br><br>
$$ = {v_0}\left[ {{{1500 - 5} \over {1500 - 7.5}}} \right]$$<br><br>
f<sub>2</sub> [frequency received by B]<br><br>
$$ = {v_0}{{1495} \over {1492.5}} \times {{1507.5} \over {1505}}$$<br><br>
$$ = 502\,Hz$$ | mcq | jee-main-2019-online-12th-april-morning-slot | 13,207 |
l48AhpYxhb8JRSAuT53rsa0w2w9jx7gevqh | physics | waves | doppler-effect | Two sources of sound S<sub>1</sub> and S<sub>2</sub> produce sound waves of same frequency 660 Hz. A listener is moving from
source S<sub>1</sub> towards S<sub>2</sub> with a constant speed u m/s and he hears 10 beats/s. The velocity of sound is 330 m/s.
Then, u equals : | [{"identifier": "A", "content": "10.0 m/s"}, {"identifier": "B", "content": "5.5 m/s"}, {"identifier": "C", "content": "15.0 m/s"}, {"identifier": "D", "content": "2.5 m/s"}] | ["D"] | null | As observer goes away from source S<sub>1</sub> so apparent frequency,
<br> $${f_1} = \left( {{{v - u} \over v}} \right)f$$
<br>here $$v$$ = speed of sound, $$u$$ = speed of observer
<br><br>As observer goes towards source S<sub>2</sub> so apparent frequency,
<br>$${f_2} = \left( {{{v + u} \over v}} \right)f$$
<br><br>Beat frequency = $${f_2} - {f_1}$$ = 10
<br>$$ \Rightarrow $$ $$\left( {{{v + u} \over v}} \right)f$$ - $$\left( {{{v - u} \over v}} \right)f$$ = 10
<br>$$ \Rightarrow $$ $$f\left( {{{v - u - v + u} \over v}} \right)$$ = 10
<br>$$ \Rightarrow $$ $$f\left( {{{2u} \over v}} \right)$$ = 10
<br>$$ \Rightarrow $$ $$u = {{10v} \over {2f}}$$ = $${{10 \times 330} \over {2 \times 660}}$$ = 2.5 m/s | mcq | jee-main-2019-online-12th-april-evening-slot | 13,208 |
4CgySbYagkP7F8xIJrjgy2xukfrs17g8 | physics | waves | doppler-effect | A sound source S is moving along a straight track with speed v, and is emitting, sound of frequency
v<sub>0</sub>
(see figure). An observer is standing at a finite distance, at the point O, from the track. The
time variation of frequency heard by the observer is best represented by: <sub></sub><br/>(t<sub>0</sub>
represents the
instant when the distance between the source and observer is minimum) | [{"identifier": "A", "content": "<img src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264493/exam_images/eacsmj6hqro4hgvcp4lk.webp\" style=\"max-width: 100%;height: auto;display: block;margin: 0 auto;\" loading=\"lazy\" alt=\"JEE Main 2020 (Online) 6th September Morning Slot Physics - Waves Question 64 English Option 1\">"}, {"identifier": "B", "content": "<img src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265700/exam_images/woau4cfxvyo07pkn2ghy.webp\" style=\"max-width: 100%;height: auto;display: block;margin: 0 auto;\" loading=\"lazy\" alt=\"JEE Main 2020 (Online) 6th September Morning Slot Physics - Waves Question 64 English Option 2\">"}, {"identifier": "C", "content": "<img src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266659/exam_images/zn99139xajbaeerdaqzg.webp\" style=\"max-width: 100%;height: auto;display: block;margin: 0 auto;\" loading=\"lazy\" alt=\"JEE Main 2020 (Online) 6th September Morning Slot Physics - Waves Question 64 English Option 3\">"}, {"identifier": "D", "content": "<img src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266489/exam_images/uaf0djru159hj8cq3hkv.webp\" style=\"max-width: 100%;height: auto;display: block;margin: 0 auto;\" loading=\"lazy\" alt=\"JEE Main 2020 (Online) 6th September Morning Slot Physics - Waves Question 64 English Option 4\">"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264867/exam_images/bq5qo3izegaburzb2gq3.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 6th September Morning Slot Physics - Waves Question 64 English Explanation">
<br><br>While approaching
<br><br>v = $${v_0}\left( {{c \over {c - v\cos \theta }}} \right)$$
<br><br>While receding
<br><br>v = $${v_0}\left( {{c \over {c + v\cos \theta }}} \right)$$
<br><br>When source is approaching towards observer then $$\theta $$ is inceasing so cos$$\theta $$ is decreasing. As cos$$\theta $$ is decreasing so c - vcos$$\theta $$ is increasing. Then v will decrease.
<br><br>When source is receding from observer then $$\theta $$ is decreasing so cos$$\theta $$ is increasing. As cos$$\theta $$ is increasing so c + vcos$$\theta $$ is increasing. Then v will also decrease.
<br><br>As v is changing depending on cos$$\theta $$, so curve should look like cos$$\theta $$ curve. | mcq | jee-main-2020-online-6th-september-morning-slot | 13,209 |
jxwkx4ONjuEghsZq9tjgy2xukfl34vez | physics | waves | doppler-effect | A driver in a car, approaching a vertical wall
notices that the frequency of his car horn, has
changed from 440 Hz to 480 Hz, when it gets
reflected from the wall. If the speed of sound in
air is 345 m/s, then the speed of the car is : | [{"identifier": "A", "content": "36 km/hr"}, {"identifier": "B", "content": "54 km/hr"}, {"identifier": "C", "content": "24 km/hr"}, {"identifier": "D", "content": "18 km/hr"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264373/exam_images/wwsww7opsrcdyoiqxmzd.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 5th September Evening Slot Physics - Waves Question 65 English Explanation">
<br><br>f<sub>1</sub> = frequency heard by wall
<br>= $$\left( {{{{v_s} - 0} \over {{v_s} - {v_c}}}} \right){f_0}$$ = $${{{{v_s}} \over {{v_s} - {v_c}}} \times 440}$$
<br><br>f<sub>2</sub> = frequency heard by driver after reflection from wall
<br><br>= $$\left( {{{{v_s} + {v_c}} \over {{v_s} - 0}}} \right){f_1}$$
<br><br>$$ \Rightarrow $$ 480 = $$\left( {{{{v_s} + {v_c}} \over {{v_s}}}} \right){f_1}$$
<br><br>$$ \Rightarrow $$ 480 = $$\left( {{{{v_s} + {v_c}} \over {{v_s}}}} \right)\left( {{{{v_s}} \over {{v_s} - {v_c}}}} \right) \times 440$$
<br><br>$$ \Rightarrow $$ 480 = $$\left( {{{{v_s} + {v_c}} \over {{v_s} - {v_c}}}} \right) \times 440$$
<br><br>$$ \Rightarrow $$ 480 = $$\left( {{{345 + {v_c}} \over {345 - {v_c}}}} \right) \times 440$$
<br><br>$$ \Rightarrow $$ 12 = $$\left( {{{345 + {v_c}} \over {345 - {v_c}}}} \right) \times 11$$
<br><br>$$ \Rightarrow $$ 11 × 345 + 11v<sub>c</sub> = 12 × 345 – 12v<sub>c</sub>
<br><br>$$ \Rightarrow $$ v<sub>c</sub> = $${{345} \over {23}}$$ m/s = $${{345} \over {23}} \times {{18} \over 5}$$ = 54 km/hr | mcq | jee-main-2020-online-5th-september-evening-slot | 13,210 |
SnKKcB2x4K79oGV17g7k9k2k5f7qel2 | physics | waves | doppler-effect | A stationary observer receives sound from two identical tuning forks, one of which approaches
and the other one recedes with the same speed (much less than the speed of sound). The
observer hears 2 beats/sec. The oscillation frequency of each tuning fork is v<sub>0</sub>
= 1400 Hz and the
velocity of sound in air is 350 m/s. The speed of each tuning fork is close to : | [{"identifier": "A", "content": "1 m/s"}, {"identifier": "B", "content": "$${1 \\over 8}$$ m/s"}, {"identifier": "C", "content": "$${1 \\over 4}$$ m/s"}, {"identifier": "D", "content": "$${1 \\over 2}$$ m/s"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263489/exam_images/uxtxlfdgjh4wbmcz7e4j.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 7th January Evening Slot Physics - Waves Question 76 English Explanation">
<br>f<sub>1</sub> = $$\left( {{c \over {c - v}}} \right){f_0}$$
<br><br>f<sub>1</sub> = $$\left( {{c \over {c + v}}} \right){f_0}$$
<br><br>beat frequency = f<sub>1</sub> – f<sub>2</sub>
<br><br>= $$c{f_0}\left( {{1 \over {c - v}} - {1 \over {c + v}}} \right)$$
<br><br>= $$c{f_0}\left( {{{2v} \over {{c^2} - {v^2}}}} \right)$$
<br><br>As c $$ \gg $$ v then $${{c^2} - {v^2}}$$ = $${{c^2}}$$
<br><br>= $$c{f_0}\left( {{{2v} \over {{c^2}}}} \right)$$
<br><br>= $${f_0}\left( {{{2v} \over c}} \right)$$
<br><br>$$ \therefore $$ $${f_0}\left( {{{2v} \over c}} \right)$$ = 2
<br><br>$$ \Rightarrow $$ v = $${{350} \over {1400}}$$ = $${1 \over 4}$$ m/s | mcq | jee-main-2020-online-7th-january-evening-slot | 13,211 |
hHz0hj0lEMSO0UBsvujgy2xukfahqmr0 | physics | waves | doppler-effect | The driver of a bus approaching a big wall notices that the frequency of his bus's horn changes
from 420 Hz to 490 Hz when he hears it after it gets reflected from the wall. Find the speed of the
bus if speed of the sound is 330 ms<sup>–1</sup>. | [{"identifier": "A", "content": "91 kmh<sup>\u20131</sup>"}, {"identifier": "B", "content": "81 kmh<sup>\u20131</sup>"}, {"identifier": "C", "content": "61 kmh<sup>\u20131</sup>"}, {"identifier": "D", "content": "71 kmh<sup>\u20131</sup>"}] | ["A"] | null | Frequency received by wall,<br><br>$${f_w} = \left( {{{330} \over {330 - v}}} \right){f_0}$$<br><br>Frequency after reflection, $$f' = \left( {{{330 + v} \over {330}}} \right){f_w}$$<br><br>$$ = \left( {{{330 + v} \over {330}}} \right) \times \left( {{{330} \over {330 - v}}} \right){f_0}$$<br><br>$$ \Rightarrow $$ $$490 = \left( {{{330 + v} \over {330 - v}}} \right)420$$<br><br>$$ \therefore $$ v = 25.2 m/s<br><br>= 91 km/h | mcq | jee-main-2020-online-4th-september-evening-slot | 13,212 |
gbmanywZXVz64mKueG1klrp3ah1 | physics | waves | doppler-effect | Two cars are approaching each other at an equal speed of 7.2 km/hr. When they see each other, both blow horns having frequency of 676 Hz. The beat frequency heard by each driver will be ___________ Hz. [Velocity of sound in air is 340 m/s.] | [] | null | 8 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265189/exam_images/rcpchyanvyz24kwmydls.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 24th February Evening Shift Physics - Waves Question 62 English Explanation"><br>Given, v<sub>A</sub> = v<sub>B</sub> = 7.2 kmh<sup>$$-$$1</sup><br><br>$$ = {{72} \over {10}} \times {5 \over {18}}$$ = 2 ms<sup>$$-$$1</sup><br><br>Frequency of source, f<sub>s</sub> = 676 Hz<br><br>Speed of sound in air, v = 340 ms<sup>$$-$$1</sup><br><br>Let f<sub>0</sub> be the frequency heard by each driver.<br><br>By using Doppler effect for A,<br><br>$$(v - {v_A}){f_s} = (v + {v_B}){f_0}$$<br><br>$$ \Rightarrow {f_0} = \left( {{{v + {v_A}} \over {v - {v_B}}}} \right){f_s} = \left( {{{340 + 2} \over {340 - 2}}} \right)676 = {{342} \over {338}} \times 676 = 684$$ Hz<br><br>Now, beat frequency $$ = {f_0} - {f_s} = 684 - 676 = 8$$ Hz | integer | jee-main-2021-online-24th-february-evening-slot | 13,213 |
1krpq7gh1 | physics | waves | doppler-effect | The frequency of a car horn encountered a change from 400 Hz to 500 Hz, when the car approaches a vertical wall. If the speed of sound is 330 m/s. Then the speed of car is ___________ km/h. | [] | null | 132 | $$\because$$ Since, frequency received by the wall,<br/><br/>$$f' = \left( {{{{v_s}} \over {{v_s} - v}}} \right){f_o}$$ .... (i)<br/><br/>where, v<sub>s</sub> = velocity of sound in air, v = velocity of car and f<sub>o</sub> = observed frequency of sound.<br/><br/>Reflected frequency received by man is<br/><br/>$$f'' = \left( {{{{v_s} + v} \over {{v_s}}}} \right)f'$$ ..... (ii)<br/><br/>From Eqs. (i) and (ii), we get<br/><br/>$$f'' = \left( {{{{v_s} + v} \over {{v_s}}}} \right)\left( {{{{v_s}} \over {{v_s} - v}}} \right){f_o} \Rightarrow f'' = \left( {{{{v_s} + v} \over {{v_s} - v}}} \right){f_o}$$<br/><br/>$$ \Rightarrow 500 = \left( {{{330 + v} \over {330 - v}}} \right) \times 400 \Rightarrow {{500} \over {400}} = {{(330 + v)} \over {(330 - v)}}$$<br/><br/>$$ \Rightarrow 5(330 - v) = 4(330 + v) \Rightarrow 1650 - 5v = 1320 + 4v$$<br/><br/>$$ \Rightarrow 9v = 330 \Rightarrow v = {{330} \over 9}$$ m/s<br/><br/>or $$v = {{330} \over 9} \times {{18} \over 5} = 132$$ km/h | integer | jee-main-2021-online-20th-july-morning-shift | 13,214 |
1ktag4ips | physics | waves | doppler-effect | A source and a detector move away from each other in absence of wind with a speed of 20 m/s with respect to the ground. If the detector detects a frequency of 1800 Hz of the sound coming from the source, then the original frequency of source considering speed of sound in air 340 m/s will be ............... Hz. | [] | null | 2025 | Image<br><br>$$f' = f\left( {{{C - {V_0}} \over {C + {V_s}}}} \right)$$<br><br>$$ \Rightarrow $$ $$1800 = f\left( {{{340 - 20} \over {340 + 20}}} \right)$$<br><br>f = 2025 Hz | integer | jee-main-2021-online-26th-august-morning-shift | 13,215 |
1kte79ots | physics | waves | doppler-effect | Two cars X and Y are approaching each other with velocities 36 km/h and 72 km/h respectively. The frequency of a whistle sound as emitted by a passenger in car X, heard by the passenger in car Y is 1320 Hz. If the velocity of sound in air is 340 m/s, the actual frequency of the whistle sound produced is .................. Hz. | [] | null | 1210 | Image<br><br>V<sub>x</sub> = 36 km/hr = 10 m/s<br><br>V<sub>y</sub> = 72 km/hr = 20 m/s<br><br>by doppler's effect<br><br>$$F' = {F_0}\left( {{{V \pm {V_0}} \over {V \pm {V_s}}}} \right)$$<br><br>$$1320 = {F_0}\left( {{{340 + 20} \over {340 - 10}}} \right) \Rightarrow {F_0} = 1210$$ Hz | integer | jee-main-2021-online-27th-august-morning-shift | 13,216 |
1l57qc1t6 | physics | waves | doppler-effect | <p>An observer moves towards a stationary source of sound with a velocity equal to one-fifth of the velocity of sound. The percentage change in the frequency will be :</p> | [{"identifier": "A", "content": "20%"}, {"identifier": "B", "content": "10%"}, {"identifier": "C", "content": "5%"}, {"identifier": "D", "content": "0%"}] | ["A"] | null | <p>$$f' = {f_0}\left[ {{{v - {v_0}} \over {v - {v_s}}}} \right]$$</p>
<p>$$ \Rightarrow f' = {f_0}\left[ {{{v + {v \over 5}} \over v}} \right]$$</p>
<p>$$ \Rightarrow f' = {{6{f_0}} \over 5}$$</p>
<p>$$\Rightarrow$$ % change = 20</p> | mcq | jee-main-2022-online-27th-june-morning-shift | 13,217 |
1l5w3dm0l | physics | waves | doppler-effect | <p>An employee of a factory moving away from his workplace by a car listens to the siren of the factory. He drives the car at the speed of 72 kmh<sup>$$-$$1</sup> in the direction of wind which is blowing at 72 kmh<sup>$$-$$1</sup> speed. Frequency of siren is 720 Hz. The employee hears an apparent frequency of ____________ Hz.</p>
<p>(Assume speed of sound to be 340 ms<sup>$$-$$1</sup>)</p> | [] | null | 680 | Here, the apparent frequency is given by
<br/><br/>
$$
\begin{aligned}
&f^{\prime}=f\left(\frac{V-V_{0}}{V+V_{s}}\right)=720\left(\frac{(340+20)-20}{(340+20)-0}\right) \\\\
&=\frac{720 \times 340}{360}=680 \mathrm{~Hz}
\end{aligned}
$$ | integer | jee-main-2022-online-30th-june-morning-shift | 13,218 |
1l6e01guz | physics | waves | doppler-effect | <p>An observer is riding on a bicycle and moving towards a hill at $$18 \,\mathrm{kmh}^{-1}$$. He hears a sound from a source at some distance behind him directly as well as after its reflection from the hill. If the original frequency of the sound as emitted by source is $$640 \mathrm{~Hz}$$ and velocity of the sound in air is $$320 \mathrm{~m} / \mathrm{s}$$, the beat frequency between the two sounds heard by observer will be _____________ $$\mathrm{Hz}$$.</p> | [] | null | 20 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l6tcr1h9/39768849-cae4-40df-acbc-f337b8ee6638/829a8ed0-1bd3-11ed-bd9f-71db3a4811b2/file-1l6tcr1ha.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l6tcr1h9/39768849-cae4-40df-acbc-f337b8ee6638/829a8ed0-1bd3-11ed-bd9f-71db3a4811b2/file-1l6tcr1ha.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 25th July Morning Shift Physics - Waves Question 34 English Explanation"></p>
<p>$${f_1} = {f_0}\left( {{{320 - 5} \over {320}}} \right) = 640\left( {{{315} \over {320}}} \right)$$</p>
<p>$$ = 630$$ Hz</p>
<p>$${f_3} = {f_0}$$ [No relative motion]</p>
<p>$${f_2} = {f_0}\left[ {{{320 + 5} \over {320}}} \right] = 640\left( {{{325} \over {320}}} \right)$$</p>
<p>$$ = 650$$</p>
<p>Beat frequency $$ = {f_2} - {f_1}$$</p>
<p>$$ = 650 - 630 = 20$$ Hz</p> | integer | jee-main-2022-online-25th-july-morning-shift | 13,219 |
1l6f5rrff | physics | waves | doppler-effect | <p>Two waves executing simple harmonic motions travelling in the same direction with same amplitude and frequency are superimposed. The resultant amplitude is equal to the $$\sqrt3$$ times of amplitude of individual motions. The phase difference between the two motions is ___________ (degree).</p> | [] | null | 60 | <p>$${A_{net}} = \sqrt {A_1^2 + A_2^2 + 2{A_1}{A_2}\cos \phi } $$</p>
<p>$$\sqrt 3 A = \sqrt {{A^2} + {A^2} + 2{A^2}\cos \phi } $$</p>
<p>$$3{A^2} = 2{A^2} + 2{A^2}\cos \phi $$</p>
<p>$$\cos \phi = {1 \over 2}$$</p>
<p>$$\phi = 60^\circ $$</p> | integer | jee-main-2022-online-25th-july-evening-shift | 13,220 |
1l6gnnztl | physics | waves | doppler-effect | <p>When a car is approaching the observer, the frequency of horn is $$100 \mathrm{~Hz}$$. After passing the observer, it is $$50 \mathrm{~Hz}$$. If the observer moves with the car, the frequency will be $$\frac{x}{3} \mathrm{~Hz}$$ where $$x=$$ ________________.</p> | [] | null | 200 | <p>$$100 = {v_0}{v \over {v - {v_c}}}$$</p>
<p>$$50 = {v_0}{v \over {v + {v_c}}}$$</p>
<p>$$2 = {{v + {v_c}} \over {v - {v_c}}}$$</p>
<p>$$2v - 2{v_c} = v + {v_c}$$</p>
<p>$${v_c} = {v \over 3}$$</p>
<p>$$100 = {v_0}{{v \times 3} \over {2v}} \Rightarrow {v_0} = {{200} \over 3} = {x \over 3}$$</p>
<p>$$ \Rightarrow x = 200$$</p> | integer | jee-main-2022-online-26th-july-morning-shift | 13,221 |
1l6mbki0f | physics | waves | doppler-effect | <p>The frequency of echo will be __________ Hz if the train blowing a whistle of frequency 320 Hz is moving with a velocity of 36 km/h towards a hill from which an echo is heard by the train driver. Velocity of sound in air is 330 m/s.</p> | [] | null | 340 | <p>$${v_s} = 36 \times {5 \over {18}} = 10$$ m/sec</p>
<p>$$f = {{v + {v_s}} \over {v - {v_s}}}{f_0}$$</p>
<p>$$ = {{340} \over {320}} \times 320$$</p>
<p>$$ = 340$$ Hz</p> | integer | jee-main-2022-online-28th-july-morning-shift | 13,222 |
1ldtzzfbb | physics | waves | doppler-effect | <p>A train blowing a whistle of frequency 320 Hz approaches an observer standing on the platform at a speed of 66 m/s. The frequency observed by the observer will be (given speed of sound = 330 ms$$^{-1}$$) __________ Hz.</p> | [] | null | 400 | $f=f_{0}\left(\frac{v}{v-v_{s}}\right)$
<br/><br/>
$$
\begin{aligned}
& f=320\left(\frac{330}{330-66}\right) \\\\
& =320 \times \frac{330}{264} \\\\
& =400 \mathrm{~Hz} .
\end{aligned}
$$ | integer | jee-main-2023-online-25th-january-evening-shift | 13,224 |
1lgswkwfl | physics | waves | doppler-effect | <p>A car P travelling at $$20 \mathrm{~ms}^{-1}$$ sounds its horn at a frequency of $$400 \mathrm{~Hz}$$. Another car $$\mathrm{Q}$$ is travelling behind the first car in the same direction with a velocity $$40 \mathrm{~ms}^{-1}$$.
The frequency heard by the passenger of the car $$\mathrm{Q}$$ is approximately [Take, velocity of sound $$=360 \mathrm{~ms}^{-1}$$ ]</p> | [{"identifier": "A", "content": "485 Hz"}, {"identifier": "B", "content": "514 Hz"}, {"identifier": "C", "content": "421 Hz"}, {"identifier": "D", "content": "471 Hz"}] | ["C"] | null | <p>Using the Doppler effect formula:</p>
<p>$$f = f_0\left(\frac{c + v_0}{c + v_s}\right)$$</p>
<p>where</p>
<ul>
<li>$$f$$ is the observed frequency (frequency heard by the passenger in car Q)</li>
<li>$$f_0$$ is the source frequency (400 Hz)</li>
<li>$$c$$ is the speed of sound (360 m/s)</li>
<li>$$v_0$$ is the velocity of the observer (passenger in car Q) relative to the medium (air) in the direction of the source (40 m/s)</li>
<li>$$v_s$$ is the velocity of the source (car P) relative to the medium (air) in the direction of the observer (20 m/s, since it's moving in the same direction as car Q)</li>
</ul>
<p>Plugging in the values:</p>
<p>$$f = 400\left(\frac{360 + 40}{360 + 20}\right)$$<br/><br/>
$$f = 400\left(\frac{400}{380}\right)$$<br/><br/>
$$f = 421$$</p>
<p>So, the observed frequency is 421 Hz.
| mcq | jee-main-2023-online-11th-april-evening-shift | 13,225 |
1lh262no6 | physics | waves | doppler-effect | <p>A person driving car at a constant speed of $$15 \mathrm{~m} / \mathrm{s}$$ is approaching a vertical wall. The person notices a change of $$40 \mathrm{~Hz}$$ in the frequency of his car's horn upon reflection from the wall. The frequency of horn is _______________ $$\mathrm{Hz}$$.</p>
<p>(Given: Speed of sound : $$330 \mathrm{~m} / \mathrm{s}$$ )</p> | [] | null | 420 | $$
\begin{aligned}
& \text { Frequency of reflected sound }=\left(\frac{v+v_{\mathrm{c}}}{v-v_c}\right) f_0 \\\\
& f=\left(\frac{330+15}{330-15}\right) \times f_0 \\\\
& =\frac{345}{315} f_0 \\\\
& \frac{345}{315} f_0-f_0=40 \\\\
& \frac{30}{315} f_0=40 \\\\
& f_0=\frac{4 \times 315}{3}=420 \mathrm{~Hz}
\end{aligned}
$$ | integer | jee-main-2023-online-6th-april-morning-shift | 13,227 |
9Ich0KInX8771D3Q | physics | waves | superposition-and-reflection-of-waves | A wave $$y=a$$ $$\sin \left( {\omega t - kx} \right)$$ on a string meets with another wave producing a node at $$x=0.$$ Then the equation of the unknown wave is | [{"identifier": "A", "content": "$$y = a\\,\\sin \\,\\left( {\\omega t + kx} \\right)$$ "}, {"identifier": "B", "content": "$$y = - a\\,\\sin \\,\\left( {\\omega t + kx} \\right)$$ "}, {"identifier": "C", "content": "$$y = a\\,\\sin \\,\\left( {\\omega t - kx} \\right)$$ "}, {"identifier": "D", "content": "$$y = - a\\,\\sin \\,\\left( {\\omega t - kx} \\right)$$ "}] | ["B"] | null | To form a node there should be superposition of this wave with the reflected wave. The reflected wave should travel in opposite direction with a phase change of $$\pi $$. The equation of the reflected wave will be
<br><br>$$y = a\sin \left( {\omega t + kx + \pi } \right)$$
<br><br>$$ \Rightarrow y = - a\sin \left( {\omega t + kx} \right)$$ | mcq | aieee-2002 | 13,228 |
pvjyp3ZHeVkb48LJ | physics | waves | superposition-and-reflection-of-waves | A tuning fork arrangement (pair) produces $$4$$ beats/sec with one fork of frequency $$288$$ $$cps.$$ A little wax is placed on the unknown fork and it then produces $$2$$ beats/sec. The frequency of the unknown fork is | [{"identifier": "A", "content": "$$286$$ $$cps$$ "}, {"identifier": "B", "content": "$$292$$ $$cps$$ "}, {"identifier": "C", "content": "$$294$$ $$cps$$ "}, {"identifier": "D", "content": "$$288$$ $$cps$$ "}] | ["B"] | null | A tuning fork produces $$4$$ beats/sec with another tuning fork of frequency $$288$$ cps. From this information we can conclude that the frequency of unknown fork is $$288+4$$ $$cps$$ or $$288-4$$ $$cps$$ i.e. $$292$$ $$cps$$ or $$284$$ $$cps.$$
<br><br>Here when a little wax is placed on the unknown fork, it decreases the frequency of unknown fork. Here also beats per second decreases to 2 from 4. So the difference between frequency decreases.
<br><br>This is possible only when before placing the wax, the frequency of unknown fork is greater than the frequency of the given tuning fork.
<br><br>So the frequency of the unknown tuninh fork is = 292 Hz | mcq | aieee-2002 | 13,229 |
pLQ7IRnOZQeVfyYK | physics | waves | superposition-and-reflection-of-waves | A tuning fork of known frequency $$256$$ $$Hz$$ makes $$5$$ beats per second with the vibrating string of a piano. The beat frequency decreases to $$2$$ beats per second when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was | [{"identifier": "A", "content": "$$256 + 2Hz$$ "}, {"identifier": "B", "content": "$$256 - 2Hz$$ "}, {"identifier": "C", "content": "$$256 - 5Hz$$ "}, {"identifier": "D", "content": "$$256 + 5Hz$$ "}] | ["C"] | null | A tuning fork of frequency $$256$$ $$Hz$$ makes $$5$$ beats/ second with the vibrating string of a piano. Therefore the frequency of the vibrating string of piano is $$\left( {256 \pm 5} \right)$$ $$Hz$$ ie either $$261$$$$Hz$$ or $$251$$ $$Hz.$$ When the tension in the piano string increases, its frequency will increases. Now since the beat frequency decreases, we can conclude that the frequency of piano string is $$251$$ $$Hz$$ | mcq | aieee-2003 | 13,230 |
QsfC42VBdLAKECVD | physics | waves | superposition-and-reflection-of-waves | A metal wire of linear mass density of $$9.8$$ $$g/m$$ is stretched with a tension of $$10$$ $$kg$$-$$wt$$ between two rigid supports $$1$$ metre 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 is | [{"identifier": "A", "content": "$$50$$ $$Hz$$ "}, {"identifier": "B", "content": "$$100$$ $$Hz$$ "}, {"identifier": "C", "content": "$$200$$ $$Hz$$ "}, {"identifier": "D", "content": "$$25$$ $$Hz$$ "}] | ["A"] | null | <b>KEY CONCEPT :</b> For a string vibrating between two rigid support, the fundamental frequency is given by
<br><br>$$n = {1 \over {2\ell }}\sqrt {{T \over \mu }} = {1 \over {2 \times }}\sqrt {{{10 \times 9.8} \over {9.8 \times {{10}^{ - 3}}}}} = 50Hz$$
<br><br>As the string is vibrating in resonance to a.c of frequency $$n,$$ therefore both the frequencies are same. | mcq | aieee-2003 | 13,231 |
R72cAxAJntbMHhze | physics | waves | superposition-and-reflection-of-waves | When two tuning forks (fork $$1$$ and fork $$2$$) are sounded simultaneously, $$4$$ beats per second are heated. Now, some tape is attached on the prong of the fork $$2.$$ When the tuning forks are sounded again, $$6$$ beats per second are heard. If the frequency of fork $$1$$ is $$200$$ $$Hz$$, then what was the original frequency of fork $$2$$ ? | [{"identifier": "A", "content": "$$202$$ $$Hz$$ "}, {"identifier": "B", "content": "$$200$$ $$Hz$$ "}, {"identifier": "C", "content": "$$204$$ $$Hz$$ "}, {"identifier": "D", "content": "$$196$$ $$Hz$$ "}] | ["D"] | null | No. of beats heard when fork $$2$$ is sounded with fork $$1$$ $$ = \Delta n = 4$$
<br><br>Now we know that if on loading (attaching tape) an unknown fork, the beat frequency increases (from $$4$$ to $$6$$ in this case) then the frequency of the unknown fork $$2$$ is given by,
<br><br>$$n = {n_0} - \Delta n = 200 - 4 = 196Hz$$ | mcq | aieee-2005 | 13,232 |
BOo8T7fZF1xCgKO3 | physics | waves | superposition-and-reflection-of-waves | A string is stretched between fixed points separated by $$75.0$$ $$cm.$$ It is observed to have resonant frequencies of $$420$$ $$Hz$$ and $$315$$ $$Hz$$. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is | [{"identifier": "A", "content": "$$105$$ $$Hz$$ "}, {"identifier": "B", "content": "$$1.05$$ $$Hz$$ "}, {"identifier": "C", "content": "$$1050$$ $$Hz$$ "}, {"identifier": "D", "content": "$$10.5$$ $$Hz$$"}] | ["A"] | null | Given $${{nv} \over {2\ell }} = 315$$ and $$\left( {n + 1} \right){v \over {2\ell }} = 420$$
<br><br>$$ \Rightarrow {{n + 1} \over n} = {{420} \over {315}} \Rightarrow n = 3$$
<br><br>Hence $$3 \times {v \over {2\ell }} = 315 \Rightarrow {v \over {2\ell }} = 105Hz$$
<br><br>Lowest resonant frequency is when $$n=1$$
<br><br>Therefore lowest resonant frequency $$ = 105\,Hz.$$ | mcq | aieee-2006 | 13,233 |
JrmC9oOmB9ar6zq3 | physics | waves | superposition-and-reflection-of-waves | 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. Then | [{"identifier": "A", "content": "$$18 > x$$ "}, {"identifier": "B", "content": "$$x > 54$$"}, {"identifier": "C", "content": "$$54 > x > 36$$ "}, {"identifier": "D", "content": "$$36 > x > 18$$ "}] | ["B"] | null | For first resonant length $$v = {v \over {4{\ell _1}}} = {v \over {4 \times 18}}$$ (in winter)
<br><br>For second resonant length
<br><br>$$v' = {{3v'} \over {4{\ell _2}}} = {{3v'} \over {4x}}$$ (in summer)
<br><br>$$\therefore$$ $${v \over {4 \times 18}} = {{3v'} \over {4 \times x}}$$
<br><br>$$\therefore$$ $$x = 3 \times 18 \times {{v'} \over v}$$
<br><br>$$\therefore$$ $$x = 54 \times {{v'} \over v}cm$$
<br><br>$$v' > v$$ because velocity of light is greater in summer as compared to winter
<br><br>$$\left( {v \propto \sqrt T } \right)$$
<br><br>$$\therefore$$ $$x > 54\,cm$$ | mcq | aieee-2008 | 13,234 |
cnyozyez95rIrvz7 | physics | waves | superposition-and-reflection-of-waves | A pipe of length $$85$$ $$cm$$ is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below $$1250$$ $$Hz$$. The velocity of sound in air is $$340$$ $$m/s$$. | [{"identifier": "A", "content": "$$12$$ "}, {"identifier": "B", "content": "$$8$$ "}, {"identifier": "C", "content": "$$6$$ "}, {"identifier": "D", "content": "$$4$$ "}] | ["C"] | null | Length of pipe $$=85$$ $$cm$$ $$=0.85m$$
<br><br>Pipe is closed from one end so it behaves as a closed organ pipe
<br><br>Frequency of oscillations of air column in closed organ pipe is given by,
<br><br>$$f = {{\left( {2n - 1} \right)\upsilon } \over {4L}}$$
<br><br>$$f = {{\left( {2n - 1} \right)\upsilon } \over {4L}} \le 1250$$
<br><br>$$ \Rightarrow {{\left( {2n - 1} \right) \times 340} \over {0.85 \times 4}} \le 1250$$
<br><br>$$ \Rightarrow 2n - 1 \le 12.5 \approx 6$$
<br><br>Possible value of n = 1, 2, 3, 4, 5, 6
<br><br>So, number of possible natural frequencies lie below 1250 Hz is 6. | mcq | jee-main-2014-offline | 13,236 |
rTKNtTaUgcoKrODk | physics | waves | superposition-and-reflection-of-waves | A pipe open at both ends has a fundamental frequency $$f$$ in air. The pipe is dipped vertically in water so that half of it is in water. The fundamental frequency of the air column is now : | [{"identifier": "A", "content": "$$2f$$ "}, {"identifier": "B", "content": "$$f$$ "}, {"identifier": "C", "content": "$${f \\over 2}$$ "}, {"identifier": "D", "content": "$${3f \\over 4}$$"}] | ["B"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l86iub7w/acfb8cc2-3aa0-40d9-b7e2-606784b6ac6e/bc26b8c0-36dd-11ed-a9f0-f56124f74cf4/file-1l86iub7x.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l86iub7w/acfb8cc2-3aa0-40d9-b7e2-606784b6ac6e/bc26b8c0-36dd-11ed-a9f0-f56124f74cf4/file-1l86iub7x.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2016 (Offline) Physics - Waves Question 109 English Explanation"><br>The fundamental frequency in case $$(a)$$ is $$f = {v \over {2\ell }}$$
<br><br>The fundamental frequency in case $$(b)$$ is
<br><br>$$f'{v \over {4\left( {\ell /2} \right)}} = {u \over {2\ell }} = f$$ | mcq | jee-main-2016-offline | 13,237 |
YJMd4iaaeH19BbSBnmZFC | physics | waves | superposition-and-reflection-of-waves | <img src="data:image/png;base64,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"/>
A wire of length 2L, is made by joining two
wires A and B of same length but different radii
r and 2r and made of the same material. It is
vibrating at a frequency such that the joint of
the two wires forms a node. If the number of
antinodes in wire A is p and that in B is q then
the ratio p : q is : | [{"identifier": "A", "content": "3 : 5"}, {"identifier": "B", "content": "4 : 9"}, {"identifier": "C", "content": "1 : 2"}, {"identifier": "D", "content": "1 : 4"}] | ["C"] | null | Let mass per unit length of wires are $$\mu $$<sub>1</sub> and $$\mu $$<sub>2</sub>
respectively<br><br>
$$ \because $$ Materials are same, so density $$\rho $$ will be same.<br><br>
$$ \therefore $$ $${\mu _1} = {{\rho \pi {r^2}L} \over L} = \mu $$ and $${\mu _2} = {{\rho 4\pi {r^2}L} \over L} = 4\mu $$<br><br>
Tension in both are same = T, let speed of wave in wires are V<sub>1</sub> and V<sub>2</sub><br><br>
$${V_1} = {{{V_1}} \over {2L}} = {V \over {2L}}\,\,\& \,{V_2} = {{{V_2}} \over {2L}} = {V \over {4L}}$$<br><br>
Frequency at which both resonate is L.C.M. of
both frequencies (i.e : $${V \over {2L}}$$ )<br><br>
Hence number of loops in wires are 1 and 2 respectively<br><br>
So, ratio of number of antinodes is 1 : 2. | mcq | jee-main-2019-online-8th-april-morning-slot | 13,238 |
4clTZBYdW9mQhKJbri3rsa0w2w9jx7h05fo | physics | waves | superposition-and-reflection-of-waves | A tuning fork of frequency 480 Hz is used in an experiment for measuring speed of sound (v) in air
by resonance tube method. Resonance is observed to occur at two successive lengths of the air column,
l<sub>1</sub> = 30 cm and l<sub>2</sub> = 70 cm. Then, v is equal to -
| [{"identifier": "A", "content": "338 ms<sup>\u20131</sup>"}, {"identifier": "B", "content": "384 ms<sup>\u20131</sup>"}, {"identifier": "C", "content": "379 ms<sup>\u20131</sup>"}, {"identifier": "D", "content": "332 ms<sup>\u20131</sup>"}] | ["B"] | null | We know,
<br/><br>$$\lambda $$ = 2($${l_2} - {l_1}$$)
<br/><br>given $${l_2}$$ = 70 cm and $${l_1}$$ = 30 cm
<br/><br>$$ \therefore $$ $$\lambda $$ = 2(70 - 30) = 80 cm
<br/><br>Also we know, v = $$\lambda $$$$f$$ = 0.8 $$ \times $$ 480 = 384 m/s | mcq | jee-main-2019-online-12th-april-evening-slot | 13,239 |
BSmHsmAOmza0pgAjen3rsa0w2w9jwziuwp7 | physics | waves | superposition-and-reflection-of-waves | The correct figure that shows, schematically, the wave pattern produced by superposition of two waves of
frequencies 9 Hz and 11 Hz, is : | [{"identifier": "A", "content": "<img src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264822/exam_images/x1ozctikexuxoqwqyhd4.webp\" style=\"max-width: 100%; height: auto;display: block;margin: 0 auto;\" loading=\"lazy\" alt=\"JEE Main 2019 (Online) 10th April Evening Slot Physics - Waves Question 84 English Option 1\">"}, {"identifier": "B", "content": "<img src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263807/exam_images/cbxb0am27eovnb7twu1e.webp\" style=\"max-width: 100%; height: auto;display: block;margin: 0 auto;\" loading=\"lazy\" alt=\"JEE Main 2019 (Online) 10th April Evening Slot Physics - Waves Question 84 English Option 2\">"}, {"identifier": "C", "content": "<img src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267584/exam_images/ixdqrortappwqeblt0p2.webp\" style=\"max-width: 100%; height: auto;display: block;margin: 0 auto;\" loading=\"lazy\" alt=\"JEE Main 2019 (Online) 10th April Evening Slot Physics - Waves Question 84 English Option 3\">"}, {"identifier": "D", "content": "<img src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266220/exam_images/yacwimdfple3ipqrqe3m.webp\" style=\"max-width: 100%; height: auto;display: block;margin: 0 auto;\" loading=\"lazy\" alt=\"JEE Main 2019 (Online) 10th April Evening Slot Physics - Waves Question 84 English Option 4\">"}] | ["C"] | null | Beat frequency = |f<sub>1</sub> – f<sub>2</sub>| = 11 – 9 = 2 Hz | mcq | jee-main-2019-online-10th-april-evening-slot | 13,240 |
dDpBFrPerkbjbOX4uj3AN | physics | waves | superposition-and-reflection-of-waves | A string 2.0 m long and fixed at its ends is
driven by a 240 Hz vibrator. The string vibrates
in its third harmonic mode. The speed of the
wave and its fundamental frequency is :- | [{"identifier": "A", "content": "180m/s, 80 Hz"}, {"identifier": "B", "content": "180m/s, 120 Hz"}, {"identifier": "C", "content": "320m/s, 120 Hz"}, {"identifier": "D", "content": "320m/s, 80 Hz"}] | ["D"] | null | We have:<br><br/>
$$f = {{nv} \over {2l}}$$<br><br/>
$$240 = {{3 \times v} \over {2 \times 2}}$$<br><br/>
$$ \Rightarrow $$ v = 320 m/s<br><br/>
Fundamental frequency = $${v \over {2l}}$$ = 80 Hz. | mcq | jee-main-2019-online-9th-april-evening-slot | 13,241 |
8db4RgUWh6H8WZ7RsBLsv | physics | waves | superposition-and-reflection-of-waves | A resonance tube is old and has jagged end. It is still used in the laboratory to determine velocity of sound in air. A tuning fork of frequency 512 Hz produces first resonance when the tube is filled with water to a mark 11 cm below a reference mark, near the open end of the tube. The experiment is repeated with another fork of frequency 256 Hz which produces first resonance when water reaches a mark 27 cm below the reference mark. The velocity of sound in air, obtained in the experiment, is close to : | [{"identifier": "A", "content": "335 ms<sup>\u20131</sup>"}, {"identifier": "B", "content": "328 ms<sup>\u20131</sup>"}, {"identifier": "C", "content": "341 ms<sup>\u20131</sup>"}, {"identifier": "D", "content": "322 ms<sup>\u20131</sup>"}] | ["B"] | null | <p>In first resonance, length of air column $$ = {\lambda \over 4}$$.</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l3318jt5/c00c244f-33d6-4e91-865a-3851e6d314eb/74f29390-d1f5-11ec-b83f-ebfea682138a/file-1l3318jt6.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l3318jt5/c00c244f-33d6-4e91-865a-3851e6d314eb/74f29390-d1f5-11ec-b83f-ebfea682138a/file-1l3318jt6.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2019 (Online) 12th January Evening Slot Physics - Waves Question 91 English Explanation"></p>
<p>So, $${l_1} + e = {\lambda \over 4}$$ or $$11 \times 4 + 4e = \lambda $$</p>
<p>So, speed of sound is</p>
<p>$$ \Rightarrow v = {f_1}\lambda = 512(44 + 4e)$$ ...... (i)</p>
<p>And in second case,</p>
<p>$$l{'_1} + e = {{\lambda '} \over 4}$$ or $$27 \times 4 + 4e = \lambda '$$</p>
<p>$$ \Rightarrow v = {f_2}\lambda ' = 256(108 + 4e)$$ ..... (ii)</p>
<p>Dividing both Eqs. (i) and (ii), we get</p>
<p>$$1 = {{512(44 + 4e)} \over {256(108 + 4e)}} \Rightarrow e = 5$$ cm</p>
<p>Substituting value of e in Eq. (i), we get</p>
<p>Speed of sound $$v = 512(44 + 4e)$$</p>
<p>$$ = 512(44 + 4 \times 5)$$</p>
<p>$$ = 512 \times 64$$ cm s<sup>$$-$$1</sup> = 327.68 ms<sup>$$-$$1</sup> $$\approx$$ 328 ms<sup>$$-$$1</sup></p> | mcq | jee-main-2019-online-12th-january-evening-slot | 13,243 |
v0ZzHvZpHuNn1LTcyH7k9k2k5imbtht | physics | waves | superposition-and-reflection-of-waves | Three harmonic waves having equal frequency
$$\nu $$ and same intensity $${I_0}$$, have phase angles 0, $${\pi \over 4}$$ and $$ - {\pi \over 4}$$ respectively. When they are
superimposed the intensity of the resultant wave
is close to : | [{"identifier": "A", "content": "5.8 I<sub>0</sub>"}, {"identifier": "B", "content": "3 I<sub>0</sub>"}, {"identifier": "C", "content": "0.2 I<sub>0</sub>"}, {"identifier": "D", "content": "I<sub>0</sub>"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267159/exam_images/snh5yrmauvkx1yba5vej.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 9th January Morning Slot Physics - Waves Question 73 English Explanation">
<br><br>I<sub>0</sub> = CA<sup>2</sup>
<br><br>A<sub>R</sub> = A + A$$\sqrt 2 $$ = A(1 + $$\sqrt 2 $$)
<br><br>I<sub>R</sub> = C$$A_R^2$$
<br><br>$$ \therefore $$ I<sub>R</sub> = CA<sup>2</sup>$${\left( {\sqrt 2 + 1} \right)^2}$$ = 5.8 I<sub>0</sub> | mcq | jee-main-2020-online-9th-january-morning-slot | 13,245 |
IZxmnnFa29NVFrr8Kw7k9k2k5kukxcz | physics | waves | superposition-and-reflection-of-waves | A wire of length L and mass per unit length
6.0 × 10<sup>–3</sup> kgm<sup>–1</sup> is put under tension of
540 N. Two consecutive frequencies that it
resonates at are : 420 Hz and 490 Hz. Then L
in meters is : | [{"identifier": "A", "content": "5.1 m"}, {"identifier": "B", "content": "2.1 m"}, {"identifier": "C", "content": "1.1 m"}, {"identifier": "D", "content": "8.1 m"}] | ["B"] | null | Fundamental frequency = 70 Hz.
<br><br>70 = $${1 \over {2l}}\sqrt {{T \over \mu }} $$
<br><br>$$ \Rightarrow $$ $$l$$ = 2.14 m | mcq | jee-main-2020-online-9th-january-evening-slot | 13,246 |
s0OKiThrw7Ybc7Wqljjgy2xukfg7w1tf | physics | waves | superposition-and-reflection-of-waves | In a resonance tube experiment when the tube
is filled with water up to a height of 17.0 cm
from bottom, it resonates with a given tuning
fork. When the water level is raised the next
resonance with the same tuning fork occurs at
a height of 24.5 cm. If the velocity of sound in
air is 330 m/s, the tuning fork frequency is : | [{"identifier": "A", "content": "2200 Hz"}, {"identifier": "B", "content": "3300 Hz"}, {"identifier": "C", "content": "1100 Hz"}, {"identifier": "D", "content": "550 Hz"}] | ["A"] | null | $${l_1} = l - 17$$<br><br>$${l_2} = l - 24.5$$<br><br>We know, $$v = 2f({l_1} - {l_2})$$<br><br>$$ \Rightarrow $$ $$330 = 2 \times f \times [(f \times [(l - 17) - (l - 24.5)] \times 10^{ - 2}$$<br><br>$$ \Rightarrow $$ $$165 = f \times 7.5 \times {10^{ - 2}}$$<br><br>$$ \Rightarrow $$ $$f = {{165 \times 1000} \over {7.5}}$$<br><br>$$ \Rightarrow $$ $$f = 2200\,Hz$$ | mcq | jee-main-2020-online-5th-september-morning-slot | 13,247 |
UcGyT2yc2Dc0vbsolj1klryf526 | physics | waves | superposition-and-reflection-of-waves | A student is performing the experiment of resonance column. The diameter of the column tube is 6 cm. The frequency of the tuning fork is 504 Hz. Speed of the sound at the given temperature is 336 m/s. The zero of the metre scale coincides with the top end of the resonance column tube. The reading of the water level in the column when the first resonance occurs is : | [{"identifier": "A", "content": "13 cm"}, {"identifier": "B", "content": "18.4 cm"}, {"identifier": "C", "content": "16.6 cm"}, {"identifier": "D", "content": "14.8 cm"}] | ["D"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267357/exam_images/rx0jxhuhynqqlhhy78ju.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264544/exam_images/hdjqmpmajoueugr8suum.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263498/exam_images/ysed3uhotmb20ozfzn99.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 25th February Morning Shift Physics - Waves Question 63 English Explanation"></picture>
<br>$$ \therefore $$ $$l + 1.8 = {\lambda \over 4}$$<br><br>Also $$\lambda = {v \over f} = {{336} \over {504}}$$<br><br>$$ \Rightarrow l + 1.8 = {{336} \over {4 \times 504}}$$<br><br>$$ \Rightarrow l = 14.86$$ cm | mcq | jee-main-2021-online-25th-february-morning-slot | 13,248 |
0zXuZhUDKLmKtGvzuf1klunfw2z | physics | waves | superposition-and-reflection-of-waves | A tuning fork A of unknown frequency produces 5 beats/s with a fork of known frequency 340 Hz. When fork A is filed, the beat frequency decreases to 2 beats/s. What is the frequency of fork A? | [{"identifier": "A", "content": "335 Hz"}, {"identifier": "B", "content": "345 Hz"}, {"identifier": "C", "content": "338 Hz"}, {"identifier": "D", "content": "342 Hz"}] | ["A"] | null | Initially beat frequency = 5Hz<br><br>so, $$\rho$$<sub>A</sub> = 340 $$ \pm $$ 5 = 345 Hz, or 335 Hz<br><br>after filing frequency increases slightly so, new value of frequency of A > $$\rho$$<sub>A</sub><br><br>Now, beat frequency = 2Hz<br><br>$$ \Rightarrow $$ new $$\rho$$<sub>A</sub> = 340 $$ \pm $$ 2 = 342 Hz, or 338 Hz<br><br>hence, original frequency of A is $$\rho$$<sub>A</sub> = 335 Hz | mcq | jee-main-2021-online-26th-february-evening-slot | 13,249 |
4zNX5MOlnum4HLonYF1kmiq1enu | physics | waves | superposition-and-reflection-of-waves | A closed organ pipe of length L and an open organ pipe contain gases of densities $$\rho$$<sub>1</sub> and $$\rho$$<sub>2</sub> respectively. The compressibility of gases are equal in both the pipes. Both the pipes are vibrating in their first overtone with same frequency. The length of the open pipe is $${x \over 3}L\sqrt {{{{\rho _1}} \over {{\rho _2}}}} $$ where x is ___________. (Round off to the Nearest Integer) | [] | null | 4 | <p>First overtone of open pipe $$ = {{{v_2}} \over {{L_2}}}$$</p>
<p>First overtone of closed pipe at one end $$ = {{3v} \over {4L}}$$</p>
<p>As per question,</p>
<p>$${{3V} \over {4L}} = {{{V_2}} \over L}$$</p>
<p>$$ \Rightarrow \sqrt {{B \over {{\rho _1}}}} \,.\,{3 \over {4L}} = \sqrt {{B \over {{\rho _2}}}} \,.\,{1 \over {{L_2}}}$$ ($$\because$$ $$V = \sqrt {{B \over \rho }} $$)</p>
<p>$$ \Rightarrow {L_2} = {{4L} \over 3}\sqrt {{{{\rho _1}} \over {{\rho _2}}}} $$ ..... (i)</p>
<p>According to question, the length of the open pipe is</p>
<p>$${x \over 3}L\sqrt {{{{\rho _1}} \over {{\rho _2}}}} $$ ..... (ii)</p>
<p>Comparing Eqs. (i) and (ii), we get</p>
<p>$$x = 4$$</p> | integer | jee-main-2021-online-16th-march-evening-shift | 13,250 |
1ktagv595 | physics | waves | superposition-and-reflection-of-waves | Two travelling waves produces a standing wave represented by equation,<br/><br/>y = 1.0 mm cos(1.57 cm<sup>$$-$$1</sup>) x sin(78.5 s<sup>$$-$$1</sup>)t. <br/><br/>The node closest to the origin in the region x > 0 will be at x = .............. cm. | [] | null | 1 | For node<br><br>cos(1.57 cm<sup>$$-$$1</sup>)x = 0<br><br>(1.57 cm<sup>$$-$$1</sup>)x = $${\pi \over 2}$$<br><br>x = $${\pi \over {2(1.57)}}$$ cm = 1 cm | integer | jee-main-2021-online-26th-august-morning-shift | 13,251 |
1ktbuqgfe | physics | waves | superposition-and-reflection-of-waves | Two waves are simultaneously passing through a string and their equations are : <br/><br/>y<sub>1</sub> = A<sub>1</sub> sin k(x $$-$$ vt), y<sub>2</sub> = A<sub>2</sub> sin k(x $$-$$ vt + x<sub>0</sub>). Given amplitudes A<sub>1</sub> = 12 mm and A<sub>2</sub> = 5 mm, x<sub>0</sub> = 3.5 cm and wave number k = 6.28 cm<sup>$$-$$1</sup>. The amplitude of resulting wave will be ................ mm. | [] | null | 7 | y<sub>1</sub> = A<sub>1</sub> sin k(x $$-$$ vt)<br><br>y<sub>1</sub> = 12 sin 6.28 (x $$-$$ vt)<br><br>y<sub>2</sub> = 5 sin 6.28 (x $$-$$ vt + 3.5)<br><br>$$\Delta \phi = {{2\pi } \over \lambda }(\Delta x)$$<br><br>$$ = K(\Delta x)$$<br><br>$$ = 6.28 \times 3.5 = {7 \over 2} \times 2\pi = 7\pi $$<br><br>$${A_{net}} = \sqrt {A_1^2 + A_2^2 + 2{A_1}{A_2}\cos \phi } $$<br><br>$${A_{net}} = \sqrt {{{(12)}^2} + {{(5)}^2} + 2(12)(5)\cos (7\pi )} $$<br><br>$$ = \sqrt {144 + 25 - 120} $$ | integer | jee-main-2021-online-26th-august-evening-shift | 13,252 |
1ktfom98x | physics | waves | superposition-and-reflection-of-waves | A tuning fork is vibrating at 250 Hz. The length of the shortest closed organ pipe that will resonate with the tuning fork will be __________ cm. (Take speed of sound in air as 340 ms<sup>$$-$$1</sup>) | [] | null | 34 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266296/exam_images/cpnbbosehfgnwf0sd9d7.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th August Evening Shift Physics - Waves Question 48 English Explanation"><br>$${\lambda \over 4}$$ = l $$\Rightarrow$$ $$\lambda$$ = 4l <br><br>f = $${V \over \lambda } = {V \over {4l}}$$<br><br>$$\Rightarrow$$ 250 = $${{340} \over {4l}}$$<br><br>$$\Rightarrow$$ l = $${{34} \over {4 \times 25}}$$ = 0.34 m<br><br>l = 34 cm | integer | jee-main-2021-online-27th-august-evening-shift | 13,253 |
1kth60nhw | physics | waves | superposition-and-reflection-of-waves | A wire having a linear mass density 9.0 $$\times$$ 10<sup>$$-$$4</sup> kg/m is stretched between two rigid supports with a tension of 900 N. The wire resonates at a frequency of 500 Hz. The next higher frequency at which the same wire resonates is 550 Hz. The length of the wire is ____________ m. | [] | null | 10 | $$\mu = 9.0 \times {10^{ - 4}}{{kg} \over m}$$<br><br>T = 900 N<br><br>$$V = \sqrt {{T \over \mu }} = \sqrt {{{900} \over {9 \times {{10}^{ - 4}}}}} = 1000$$ m/s<br><br>f<sub>1</sub> = 500 Hz<br><br>f = 550<br><br>$${{nV} \over {2l}} = 500$$ .... (i)<br><br>$${{(n + 1)V} \over {2l}} = 500$$ .... (ii)<br><br>(ii) (i) $${V \over {2l}} = 50$$<br><br>$$l = {{1000} \over {2 \times 50}} = 10$$ | integer | jee-main-2021-online-31st-august-morning-shift | 13,254 |
1l54vnq26 | physics | waves | superposition-and-reflection-of-waves | <p>In an experiment to determine the velocity of sound in air at room temperature using a resonance tube, the first resonance is observed when the air column has a length of 20.0 cm for a tuning fork of frequency 400 Hz is used. The velocity of the sound at room temperature is 336 ms<sup>$$-$$1</sup>. The third resonance is observed when the air column has a length of _____________ cm.</p> | [] | null | 104 | <p>$$400 = {v \over {4({L_1} + e)}}$$ ..... (i)</p>
<p>$$400 = {{5v} \over {4({L_2} + e)}}$$ ..... (ii)</p>
<p>$$ \Rightarrow {L_1} + e = {\lambda \over 4} = 21$$ cm</p>
<p>$${L_2} + e = {{5\lambda } \over 4} = 105$$ cm</p>
<p>$$\Rightarrow$$ e = 1 cm & L<sub>2</sub> = 104 cm</p> | integer | jee-main-2022-online-29th-june-evening-shift | 13,255 |
1l55m8uyl | physics | waves | superposition-and-reflection-of-waves | <p>A tunning fork of frequency 340 Hz resonates in the fundamental mode with an air column of length 125 cm in a cylindrical tube closed at one end. When water is slowly poured in it, the minimum height of water required for observing resonance once again is ___________ cm.</p>
<p>(Velocity of sound in air is 340 ms<sup>$$-$$1</sup>)</p> | [] | null | 50 | <p>Given $$340 = {n \over {4 \times 125}}v$$</p>
<p>$$ \Rightarrow n = 5$$</p>
<p>So $$\lambda = 100$$ cm</p>
<p>So minimum height is $${\lambda \over 2} = 50$$ cm</p> | integer | jee-main-2022-online-28th-june-evening-shift | 13,256 |
1l56a1cwz | physics | waves | superposition-and-reflection-of-waves | <p>The velocity of sound in a gas, in which two wavelengths 4.08 m and 4.16 m produce 40 beats in 12s, will be :</p> | [{"identifier": "A", "content": "282.8 ms<sup>$$-$$1</sup>"}, {"identifier": "B", "content": "175.5 ms<sup>$$-$$1</sup>"}, {"identifier": "C", "content": "353.6 ms<sup>$$-$$1</sup>"}, {"identifier": "D", "content": "707.2 ms<sup>$$-$$1</sup>"}] | ["D"] | null | <p>$${v \over {4.08}} - {v \over {4.16}} = {{40} \over {12}}$$</p>
<p>$$v = {{40} \over {12}} \times {{4.08 \times 4.16} \over {0.08}}$$</p>
<p>$$ = 707.2$$ m/s</p> | mcq | jee-main-2022-online-28th-june-morning-shift | 13,257 |
1l58ikuis | physics | waves | superposition-and-reflection-of-waves | <p>A set of 20 tuning forks is arranged in a series of increasing frequencies. If each fork gives 4 beats with respect to the preceding fork and the frequency of the last fork is twice the frequency of the first, then the frequency of last fork is _________ Hz.</p> | [] | null | 152 | <p>Given $${v_{20}} = 2{v_1}$$</p>
<p>Also $${v_{20}} = 4 \times 19 + {v_1}$$</p>
<p>So $${v_{20}} = 152\,Hz$$</p> | integer | jee-main-2022-online-26th-june-evening-shift | 13,258 |
1l5c3wc9l | physics | waves | superposition-and-reflection-of-waves | <p>The equations of two waves are given by :</p>
<p>y<sub>1</sub> = 5 sin 2$$\pi$$(x - vt) cm</p>
<p>y<sub>2</sub> = 3 sin 2$$\pi$$(x $$-$$ vt + 1.5) cm</p>
<p>These waves are simultaneously passing through a string. The amplitude of the resulting wave is :</p> | [{"identifier": "A", "content": "2 cm"}, {"identifier": "B", "content": "4 cm"}, {"identifier": "C", "content": "5.8 cm"}, {"identifier": "D", "content": "8 cm"}] | ["A"] | null | <p>$${y_1} = 5\sin (2\pi x - 2\pi vt)$$</p>
<p>$${y_2} = 3\sin (2\pi x - 2\pi vt + 3\pi )$$</p>
<p>$$\Rightarrow$$ Phase difference = 3$$\pi$$</p>
<p>$$ \Rightarrow {A_{net}} = \sqrt {A_1^2 + A_2^2 + 2{A_1}{A_2}\cos (3\pi )} $$</p>
<p>$$ \Rightarrow {A_{net}} = 2$$ cm</p> | mcq | jee-main-2022-online-24th-june-morning-shift | 13,260 |
1l6koful5 | physics | waves | superposition-and-reflection-of-waves | <p>A wire of length 30 cm, stretched between rigid supports, has it's n<sup>th</sup> and (n + 1)<sup>th</sup> harmonics at 400 Hz and 450 Hz, respectively. If tension in the string is 2700 N, it's linear mass density is ____________ kg/m.</p> | [] | null | 3 | <p>$${v \over {2l}} = 50$$ Hz</p>
<p>$$ \Rightarrow T = {\left[ {100 \times \left( {{{30} \over {100}}} \right)} \right]^2} \times \mu $$</p>
<p>$$ \Rightarrow \mu = {{2700} \over {900}} = 3$$</p> | integer | jee-main-2022-online-27th-july-evening-shift | 13,261 |
1ldspztjv | physics | waves | superposition-and-reflection-of-waves | <p>Two simple harmonic waves having equal amplitudes of 8 cm and equal frequency of 10 Hz are moving along the same direction. The resultant amplitude is also 8 cm. The phase difference between the individual waves is _________ degree.</p> | [] | null | 120 | $A_{R}=\sqrt{A_{1}^{2}+A_{2}^{2}+2 A_{1} A_{2} \cos \phi}$
<br/><br/>
$$
\begin{aligned}
& 8=\sqrt{8^{2}+8^{2}+2 \times 8 \times 8 \cos \phi} \\\\
& \Rightarrow \cos \phi=-\frac{1}{2} \\\\
& \Rightarrow \phi=120^{\circ}
\end{aligned}
$$
| integer | jee-main-2023-online-29th-january-morning-shift | 13,263 |
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