172749
If frequencies are $(v-1),(v+1)$, then find beats
1 2
2 1
3 3
4 4
Explanation:
A The formula for the beat frequency is, $\mathrm{f}_{\mathrm{b}}=\mathrm{f}_{2}-\mathrm{f}_{1}$ $\mathrm{f}_{\mathrm{b}}=(\mathrm{v}+1)-(\mathrm{v}-1)$ $\mathrm{f}_{\mathrm{b}}=\mathrm{v}+1-\mathrm{v}+1$ $\mathrm{f}_{\mathrm{b}}=2$ The number of beat produced will be 2 .
JIPMER-2019
WAVES
172750
Assertion: $V_{\text {rms }}$ and $V_{\text {mean }}$ of gases molecules is nearly of the order of velocity of sound. Reason: The sound travels in air because of vibrational molecular motion.
1 If both assertion and reason are true and reason is the correct explanation of assertion.
2 If both assertion and reason are true but reason is not the correct explanation of assertion.
3 If assertion is true but reason is false.
4 If both assertion and reason are false.
Explanation:
B Since we know that, Speed of sound in gaseous medium, $v=\sqrt{\frac{\mathrm{RT}}{\mathrm{M}}}$ Thus, we conclude that $\mathrm{v}_{\mathrm{rms}} \& \mathrm{v}_{\text {mean }}$ of gaseous molecules is nearly of same order of velocity of sound. Hence, assertion is true. Also, the sound travels in air because their molecules vibrate about their mean position. Hence, reason is also correct but not the right explanation for assertion.
AIIMS-26.05.2019(M) Shift-1
WAVES
172751
If speed of sound in air is $330 \mathrm{~m} / \mathrm{s}$ then find the number of tones present in an open organ pipe of length $1 \mathrm{~m}$ whose frequency in $\leq \mathbf{1 0 0 0}$.
1 2
2 4
3 8
4 6
Explanation:
D Given, Speed of sound $(\mathrm{v})=330 \mathrm{~m} / \mathrm{s}$ $l=1 \mathrm{~m} \text {. }$ We know that fundamental frequency of an open organ pipe is given by, $f =\frac{v}{2 l}$ $=\frac{330}{2 \times 1}=165 \mathrm{~Hz}$ Therefore, no. of tones present $(n)=\frac{1000}{165}$ $=6.06$ $\mathrm{n}=6 \text { tones }$
AIIMS-25.05.2019(M) Shift-1
WAVES
172752
Two sources $A$ and $B$ are producing notes of frequency $680 \mathrm{~Hz}$. A listener moves from $A$ to B with a constant velocity $\mu$. If the speed of sound in air is $340 \mathrm{~ms}^{-1}$, the value of $u$ so that he hears 10 beats per second is
1 $2.0 \mathrm{~ms}^{-1}$
2 $2.5 \mathrm{~ms}^{-1}$
3 $3.0 \mathrm{~ms}^{-1}$
4 $3.5 \mathrm{~ms}^{-1}$
Explanation:
B We have that, notes of frequency produced by the sources $A$ and $B$ is $680 \mathrm{~Hz}$, i.e., $f_{A}$ and $f_{B}=680 \mathrm{~Hz}$ Velocity of listener moves from $A$ to $B$ is constant $=v$, speed of sound $\left(\mathrm{v}_{\mathrm{s}}\right)=340 \mathrm{~m} / \mathrm{s}$, and beats per second $(\mathrm{n})=10$ Now, beats per second from point $\mathrm{A}$ to $\mathrm{B}$ is given as $n=f_{A}\left(\frac{v_{s}+u}{v_{s}}\right)-f_{B}\left(\frac{v_{s}-u}{v_{s}}\right)$ $10=680\left(\frac{340+u}{340}\right)-680\left(\frac{340-u}{340}\right)$ $10=680\left[\left(\frac{340+u}{340}\right)-\left(\frac{340-u}{340}\right)\right]$ $1=68\left[\frac{340+u-340+u}{340}\right]$ $1=68\left(\frac{2 u}{340}\right)$ $2 u=\frac{340}{68}$ $u=\frac{340}{68 \times 2}$ $u=2.5 \mathrm{~m} / \mathrm{s}$
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WAVES
172749
If frequencies are $(v-1),(v+1)$, then find beats
1 2
2 1
3 3
4 4
Explanation:
A The formula for the beat frequency is, $\mathrm{f}_{\mathrm{b}}=\mathrm{f}_{2}-\mathrm{f}_{1}$ $\mathrm{f}_{\mathrm{b}}=(\mathrm{v}+1)-(\mathrm{v}-1)$ $\mathrm{f}_{\mathrm{b}}=\mathrm{v}+1-\mathrm{v}+1$ $\mathrm{f}_{\mathrm{b}}=2$ The number of beat produced will be 2 .
JIPMER-2019
WAVES
172750
Assertion: $V_{\text {rms }}$ and $V_{\text {mean }}$ of gases molecules is nearly of the order of velocity of sound. Reason: The sound travels in air because of vibrational molecular motion.
1 If both assertion and reason are true and reason is the correct explanation of assertion.
2 If both assertion and reason are true but reason is not the correct explanation of assertion.
3 If assertion is true but reason is false.
4 If both assertion and reason are false.
Explanation:
B Since we know that, Speed of sound in gaseous medium, $v=\sqrt{\frac{\mathrm{RT}}{\mathrm{M}}}$ Thus, we conclude that $\mathrm{v}_{\mathrm{rms}} \& \mathrm{v}_{\text {mean }}$ of gaseous molecules is nearly of same order of velocity of sound. Hence, assertion is true. Also, the sound travels in air because their molecules vibrate about their mean position. Hence, reason is also correct but not the right explanation for assertion.
AIIMS-26.05.2019(M) Shift-1
WAVES
172751
If speed of sound in air is $330 \mathrm{~m} / \mathrm{s}$ then find the number of tones present in an open organ pipe of length $1 \mathrm{~m}$ whose frequency in $\leq \mathbf{1 0 0 0}$.
1 2
2 4
3 8
4 6
Explanation:
D Given, Speed of sound $(\mathrm{v})=330 \mathrm{~m} / \mathrm{s}$ $l=1 \mathrm{~m} \text {. }$ We know that fundamental frequency of an open organ pipe is given by, $f =\frac{v}{2 l}$ $=\frac{330}{2 \times 1}=165 \mathrm{~Hz}$ Therefore, no. of tones present $(n)=\frac{1000}{165}$ $=6.06$ $\mathrm{n}=6 \text { tones }$
AIIMS-25.05.2019(M) Shift-1
WAVES
172752
Two sources $A$ and $B$ are producing notes of frequency $680 \mathrm{~Hz}$. A listener moves from $A$ to B with a constant velocity $\mu$. If the speed of sound in air is $340 \mathrm{~ms}^{-1}$, the value of $u$ so that he hears 10 beats per second is
1 $2.0 \mathrm{~ms}^{-1}$
2 $2.5 \mathrm{~ms}^{-1}$
3 $3.0 \mathrm{~ms}^{-1}$
4 $3.5 \mathrm{~ms}^{-1}$
Explanation:
B We have that, notes of frequency produced by the sources $A$ and $B$ is $680 \mathrm{~Hz}$, i.e., $f_{A}$ and $f_{B}=680 \mathrm{~Hz}$ Velocity of listener moves from $A$ to $B$ is constant $=v$, speed of sound $\left(\mathrm{v}_{\mathrm{s}}\right)=340 \mathrm{~m} / \mathrm{s}$, and beats per second $(\mathrm{n})=10$ Now, beats per second from point $\mathrm{A}$ to $\mathrm{B}$ is given as $n=f_{A}\left(\frac{v_{s}+u}{v_{s}}\right)-f_{B}\left(\frac{v_{s}-u}{v_{s}}\right)$ $10=680\left(\frac{340+u}{340}\right)-680\left(\frac{340-u}{340}\right)$ $10=680\left[\left(\frac{340+u}{340}\right)-\left(\frac{340-u}{340}\right)\right]$ $1=68\left[\frac{340+u-340+u}{340}\right]$ $1=68\left(\frac{2 u}{340}\right)$ $2 u=\frac{340}{68}$ $u=\frac{340}{68 \times 2}$ $u=2.5 \mathrm{~m} / \mathrm{s}$
172749
If frequencies are $(v-1),(v+1)$, then find beats
1 2
2 1
3 3
4 4
Explanation:
A The formula for the beat frequency is, $\mathrm{f}_{\mathrm{b}}=\mathrm{f}_{2}-\mathrm{f}_{1}$ $\mathrm{f}_{\mathrm{b}}=(\mathrm{v}+1)-(\mathrm{v}-1)$ $\mathrm{f}_{\mathrm{b}}=\mathrm{v}+1-\mathrm{v}+1$ $\mathrm{f}_{\mathrm{b}}=2$ The number of beat produced will be 2 .
JIPMER-2019
WAVES
172750
Assertion: $V_{\text {rms }}$ and $V_{\text {mean }}$ of gases molecules is nearly of the order of velocity of sound. Reason: The sound travels in air because of vibrational molecular motion.
1 If both assertion and reason are true and reason is the correct explanation of assertion.
2 If both assertion and reason are true but reason is not the correct explanation of assertion.
3 If assertion is true but reason is false.
4 If both assertion and reason are false.
Explanation:
B Since we know that, Speed of sound in gaseous medium, $v=\sqrt{\frac{\mathrm{RT}}{\mathrm{M}}}$ Thus, we conclude that $\mathrm{v}_{\mathrm{rms}} \& \mathrm{v}_{\text {mean }}$ of gaseous molecules is nearly of same order of velocity of sound. Hence, assertion is true. Also, the sound travels in air because their molecules vibrate about their mean position. Hence, reason is also correct but not the right explanation for assertion.
AIIMS-26.05.2019(M) Shift-1
WAVES
172751
If speed of sound in air is $330 \mathrm{~m} / \mathrm{s}$ then find the number of tones present in an open organ pipe of length $1 \mathrm{~m}$ whose frequency in $\leq \mathbf{1 0 0 0}$.
1 2
2 4
3 8
4 6
Explanation:
D Given, Speed of sound $(\mathrm{v})=330 \mathrm{~m} / \mathrm{s}$ $l=1 \mathrm{~m} \text {. }$ We know that fundamental frequency of an open organ pipe is given by, $f =\frac{v}{2 l}$ $=\frac{330}{2 \times 1}=165 \mathrm{~Hz}$ Therefore, no. of tones present $(n)=\frac{1000}{165}$ $=6.06$ $\mathrm{n}=6 \text { tones }$
AIIMS-25.05.2019(M) Shift-1
WAVES
172752
Two sources $A$ and $B$ are producing notes of frequency $680 \mathrm{~Hz}$. A listener moves from $A$ to B with a constant velocity $\mu$. If the speed of sound in air is $340 \mathrm{~ms}^{-1}$, the value of $u$ so that he hears 10 beats per second is
1 $2.0 \mathrm{~ms}^{-1}$
2 $2.5 \mathrm{~ms}^{-1}$
3 $3.0 \mathrm{~ms}^{-1}$
4 $3.5 \mathrm{~ms}^{-1}$
Explanation:
B We have that, notes of frequency produced by the sources $A$ and $B$ is $680 \mathrm{~Hz}$, i.e., $f_{A}$ and $f_{B}=680 \mathrm{~Hz}$ Velocity of listener moves from $A$ to $B$ is constant $=v$, speed of sound $\left(\mathrm{v}_{\mathrm{s}}\right)=340 \mathrm{~m} / \mathrm{s}$, and beats per second $(\mathrm{n})=10$ Now, beats per second from point $\mathrm{A}$ to $\mathrm{B}$ is given as $n=f_{A}\left(\frac{v_{s}+u}{v_{s}}\right)-f_{B}\left(\frac{v_{s}-u}{v_{s}}\right)$ $10=680\left(\frac{340+u}{340}\right)-680\left(\frac{340-u}{340}\right)$ $10=680\left[\left(\frac{340+u}{340}\right)-\left(\frac{340-u}{340}\right)\right]$ $1=68\left[\frac{340+u-340+u}{340}\right]$ $1=68\left(\frac{2 u}{340}\right)$ $2 u=\frac{340}{68}$ $u=\frac{340}{68 \times 2}$ $u=2.5 \mathrm{~m} / \mathrm{s}$
172749
If frequencies are $(v-1),(v+1)$, then find beats
1 2
2 1
3 3
4 4
Explanation:
A The formula for the beat frequency is, $\mathrm{f}_{\mathrm{b}}=\mathrm{f}_{2}-\mathrm{f}_{1}$ $\mathrm{f}_{\mathrm{b}}=(\mathrm{v}+1)-(\mathrm{v}-1)$ $\mathrm{f}_{\mathrm{b}}=\mathrm{v}+1-\mathrm{v}+1$ $\mathrm{f}_{\mathrm{b}}=2$ The number of beat produced will be 2 .
JIPMER-2019
WAVES
172750
Assertion: $V_{\text {rms }}$ and $V_{\text {mean }}$ of gases molecules is nearly of the order of velocity of sound. Reason: The sound travels in air because of vibrational molecular motion.
1 If both assertion and reason are true and reason is the correct explanation of assertion.
2 If both assertion and reason are true but reason is not the correct explanation of assertion.
3 If assertion is true but reason is false.
4 If both assertion and reason are false.
Explanation:
B Since we know that, Speed of sound in gaseous medium, $v=\sqrt{\frac{\mathrm{RT}}{\mathrm{M}}}$ Thus, we conclude that $\mathrm{v}_{\mathrm{rms}} \& \mathrm{v}_{\text {mean }}$ of gaseous molecules is nearly of same order of velocity of sound. Hence, assertion is true. Also, the sound travels in air because their molecules vibrate about their mean position. Hence, reason is also correct but not the right explanation for assertion.
AIIMS-26.05.2019(M) Shift-1
WAVES
172751
If speed of sound in air is $330 \mathrm{~m} / \mathrm{s}$ then find the number of tones present in an open organ pipe of length $1 \mathrm{~m}$ whose frequency in $\leq \mathbf{1 0 0 0}$.
1 2
2 4
3 8
4 6
Explanation:
D Given, Speed of sound $(\mathrm{v})=330 \mathrm{~m} / \mathrm{s}$ $l=1 \mathrm{~m} \text {. }$ We know that fundamental frequency of an open organ pipe is given by, $f =\frac{v}{2 l}$ $=\frac{330}{2 \times 1}=165 \mathrm{~Hz}$ Therefore, no. of tones present $(n)=\frac{1000}{165}$ $=6.06$ $\mathrm{n}=6 \text { tones }$
AIIMS-25.05.2019(M) Shift-1
WAVES
172752
Two sources $A$ and $B$ are producing notes of frequency $680 \mathrm{~Hz}$. A listener moves from $A$ to B with a constant velocity $\mu$. If the speed of sound in air is $340 \mathrm{~ms}^{-1}$, the value of $u$ so that he hears 10 beats per second is
1 $2.0 \mathrm{~ms}^{-1}$
2 $2.5 \mathrm{~ms}^{-1}$
3 $3.0 \mathrm{~ms}^{-1}$
4 $3.5 \mathrm{~ms}^{-1}$
Explanation:
B We have that, notes of frequency produced by the sources $A$ and $B$ is $680 \mathrm{~Hz}$, i.e., $f_{A}$ and $f_{B}=680 \mathrm{~Hz}$ Velocity of listener moves from $A$ to $B$ is constant $=v$, speed of sound $\left(\mathrm{v}_{\mathrm{s}}\right)=340 \mathrm{~m} / \mathrm{s}$, and beats per second $(\mathrm{n})=10$ Now, beats per second from point $\mathrm{A}$ to $\mathrm{B}$ is given as $n=f_{A}\left(\frac{v_{s}+u}{v_{s}}\right)-f_{B}\left(\frac{v_{s}-u}{v_{s}}\right)$ $10=680\left(\frac{340+u}{340}\right)-680\left(\frac{340-u}{340}\right)$ $10=680\left[\left(\frac{340+u}{340}\right)-\left(\frac{340-u}{340}\right)\right]$ $1=68\left[\frac{340+u-340+u}{340}\right]$ $1=68\left(\frac{2 u}{340}\right)$ $2 u=\frac{340}{68}$ $u=\frac{340}{68 \times 2}$ $u=2.5 \mathrm{~m} / \mathrm{s}$