NEET Test Series from KOTA - 10 Papers In MS WORD
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WAVES
172734
On producing the waves of frequency $1000 \mathrm{~Hz}$ in a Kundt's tube, the total distance between 6 successive nodes is $85 \mathrm{~cm}$. Then, the speed of sound in the gas filled in the tube is
1 $330 \mathrm{~ms}^{-1}$
2 $340 \mathrm{~ms}^{-1}$
3 $350 \mathrm{~ms}^{-1}$
4 $300 \mathrm{~ms}^{-1}$
Explanation:
B Given, Frequency of wave $(\mathrm{f})=1000 \mathrm{~Hz}$ Distance between six successive node $=\frac{5 \lambda}{2}=85$ $\lambda=\frac{2 \times 85}{5}=34 \mathrm{~cm}=0.34 \mathrm{~m}$ Then speed of sound in gas $=\mathrm{f} \lambda=1000 \times 0.34$ $=340 \mathrm{~m} / \mathrm{sec} \text {. }$
AP EAMCET (17.09.2020) Shift-I
WAVES
172737
Two identical wires are vibrating in unison. If the tension in one of the wire is increased by $2 \%$, five beats are produced per second by the two vibrating wires. The initial frequency of each wire is $(\sqrt{1.02}=\mathbf{1 . 0 1})$
172738
The frequency of two tuning forks $A$ and $B$ are $1.5 \%$ more and $2.5 \%$ less than that of the tuning fork $C$. When $A$ and $B$ are sounded together, 12 beats are produced in 1 second. The frequency of tuning fork $\mathbf{C}$ is
1 $300 \mathrm{~Hz}$
2 $240 \mathrm{~Hz}$
3 $360 \mathrm{~Hz}$
4 $200 \mathrm{~Hz}$
Explanation:
A Let, $f_{A}, f_{B}$ and $f_{C}$ are the frequencies of forks $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ respectively. Then, $\mathrm{f}_{\mathrm{A}}=\mathrm{f}_{\mathrm{C}} \times 1.015, \quad \mathrm{f}_{\mathrm{B}}=\mathrm{f}_{\mathrm{C}} \times 0.975$ According to question- $f_{A}-f_{B}=12$ $(1.015-0.975) f_{C}=12$ $0.040 f_{C}=12$ $\therefore \quad \mathrm{f}_{\mathrm{C}}=\frac{12}{0.040}=300 \mathrm{~Hz}$
MHT-CET 2020
WAVES
172740
A source of sound gives 5 beats per second, when sounded with another source of frequency $100 \mathrm{sec}^{-1}$. The second harmonic of the source, together with a source of frequency $205 \mathrm{sec}^{-1}$ gives 5 beats per second. What is frequency of the source?
1 $95 \mathrm{sec}^{-1}$
2 $105 \mathrm{sec}^{-1}$
3 $100 \mathrm{sec}^{-1}$
4 $205 \mathrm{sec}^{-1}$
Explanation:
B According to the question, Frequency of source $=100 \pm 5 \mathrm{~Hz}=105$ or $95 \mathrm{~Hz}$ Frequency of source in $2^{\text {nd }}$ harmonic $=2 \times$ frequency of source, Since, $=190 \text { or } 210 \mathrm{~Hz}$ $205 \mathrm{~Hz}$ frequency produces 5 beats/sec in second harmonic. Then, Actual frequency $=205 \pm 5 \mathrm{~Hz}$ $=200 \text { or } 210 \mathrm{~Hz}$ Hence, frequency of source $=105 \mathrm{~Hz}$.
172734
On producing the waves of frequency $1000 \mathrm{~Hz}$ in a Kundt's tube, the total distance between 6 successive nodes is $85 \mathrm{~cm}$. Then, the speed of sound in the gas filled in the tube is
1 $330 \mathrm{~ms}^{-1}$
2 $340 \mathrm{~ms}^{-1}$
3 $350 \mathrm{~ms}^{-1}$
4 $300 \mathrm{~ms}^{-1}$
Explanation:
B Given, Frequency of wave $(\mathrm{f})=1000 \mathrm{~Hz}$ Distance between six successive node $=\frac{5 \lambda}{2}=85$ $\lambda=\frac{2 \times 85}{5}=34 \mathrm{~cm}=0.34 \mathrm{~m}$ Then speed of sound in gas $=\mathrm{f} \lambda=1000 \times 0.34$ $=340 \mathrm{~m} / \mathrm{sec} \text {. }$
AP EAMCET (17.09.2020) Shift-I
WAVES
172737
Two identical wires are vibrating in unison. If the tension in one of the wire is increased by $2 \%$, five beats are produced per second by the two vibrating wires. The initial frequency of each wire is $(\sqrt{1.02}=\mathbf{1 . 0 1})$
172738
The frequency of two tuning forks $A$ and $B$ are $1.5 \%$ more and $2.5 \%$ less than that of the tuning fork $C$. When $A$ and $B$ are sounded together, 12 beats are produced in 1 second. The frequency of tuning fork $\mathbf{C}$ is
1 $300 \mathrm{~Hz}$
2 $240 \mathrm{~Hz}$
3 $360 \mathrm{~Hz}$
4 $200 \mathrm{~Hz}$
Explanation:
A Let, $f_{A}, f_{B}$ and $f_{C}$ are the frequencies of forks $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ respectively. Then, $\mathrm{f}_{\mathrm{A}}=\mathrm{f}_{\mathrm{C}} \times 1.015, \quad \mathrm{f}_{\mathrm{B}}=\mathrm{f}_{\mathrm{C}} \times 0.975$ According to question- $f_{A}-f_{B}=12$ $(1.015-0.975) f_{C}=12$ $0.040 f_{C}=12$ $\therefore \quad \mathrm{f}_{\mathrm{C}}=\frac{12}{0.040}=300 \mathrm{~Hz}$
MHT-CET 2020
WAVES
172740
A source of sound gives 5 beats per second, when sounded with another source of frequency $100 \mathrm{sec}^{-1}$. The second harmonic of the source, together with a source of frequency $205 \mathrm{sec}^{-1}$ gives 5 beats per second. What is frequency of the source?
1 $95 \mathrm{sec}^{-1}$
2 $105 \mathrm{sec}^{-1}$
3 $100 \mathrm{sec}^{-1}$
4 $205 \mathrm{sec}^{-1}$
Explanation:
B According to the question, Frequency of source $=100 \pm 5 \mathrm{~Hz}=105$ or $95 \mathrm{~Hz}$ Frequency of source in $2^{\text {nd }}$ harmonic $=2 \times$ frequency of source, Since, $=190 \text { or } 210 \mathrm{~Hz}$ $205 \mathrm{~Hz}$ frequency produces 5 beats/sec in second harmonic. Then, Actual frequency $=205 \pm 5 \mathrm{~Hz}$ $=200 \text { or } 210 \mathrm{~Hz}$ Hence, frequency of source $=105 \mathrm{~Hz}$.
172734
On producing the waves of frequency $1000 \mathrm{~Hz}$ in a Kundt's tube, the total distance between 6 successive nodes is $85 \mathrm{~cm}$. Then, the speed of sound in the gas filled in the tube is
1 $330 \mathrm{~ms}^{-1}$
2 $340 \mathrm{~ms}^{-1}$
3 $350 \mathrm{~ms}^{-1}$
4 $300 \mathrm{~ms}^{-1}$
Explanation:
B Given, Frequency of wave $(\mathrm{f})=1000 \mathrm{~Hz}$ Distance between six successive node $=\frac{5 \lambda}{2}=85$ $\lambda=\frac{2 \times 85}{5}=34 \mathrm{~cm}=0.34 \mathrm{~m}$ Then speed of sound in gas $=\mathrm{f} \lambda=1000 \times 0.34$ $=340 \mathrm{~m} / \mathrm{sec} \text {. }$
AP EAMCET (17.09.2020) Shift-I
WAVES
172737
Two identical wires are vibrating in unison. If the tension in one of the wire is increased by $2 \%$, five beats are produced per second by the two vibrating wires. The initial frequency of each wire is $(\sqrt{1.02}=\mathbf{1 . 0 1})$
172738
The frequency of two tuning forks $A$ and $B$ are $1.5 \%$ more and $2.5 \%$ less than that of the tuning fork $C$. When $A$ and $B$ are sounded together, 12 beats are produced in 1 second. The frequency of tuning fork $\mathbf{C}$ is
1 $300 \mathrm{~Hz}$
2 $240 \mathrm{~Hz}$
3 $360 \mathrm{~Hz}$
4 $200 \mathrm{~Hz}$
Explanation:
A Let, $f_{A}, f_{B}$ and $f_{C}$ are the frequencies of forks $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ respectively. Then, $\mathrm{f}_{\mathrm{A}}=\mathrm{f}_{\mathrm{C}} \times 1.015, \quad \mathrm{f}_{\mathrm{B}}=\mathrm{f}_{\mathrm{C}} \times 0.975$ According to question- $f_{A}-f_{B}=12$ $(1.015-0.975) f_{C}=12$ $0.040 f_{C}=12$ $\therefore \quad \mathrm{f}_{\mathrm{C}}=\frac{12}{0.040}=300 \mathrm{~Hz}$
MHT-CET 2020
WAVES
172740
A source of sound gives 5 beats per second, when sounded with another source of frequency $100 \mathrm{sec}^{-1}$. The second harmonic of the source, together with a source of frequency $205 \mathrm{sec}^{-1}$ gives 5 beats per second. What is frequency of the source?
1 $95 \mathrm{sec}^{-1}$
2 $105 \mathrm{sec}^{-1}$
3 $100 \mathrm{sec}^{-1}$
4 $205 \mathrm{sec}^{-1}$
Explanation:
B According to the question, Frequency of source $=100 \pm 5 \mathrm{~Hz}=105$ or $95 \mathrm{~Hz}$ Frequency of source in $2^{\text {nd }}$ harmonic $=2 \times$ frequency of source, Since, $=190 \text { or } 210 \mathrm{~Hz}$ $205 \mathrm{~Hz}$ frequency produces 5 beats/sec in second harmonic. Then, Actual frequency $=205 \pm 5 \mathrm{~Hz}$ $=200 \text { or } 210 \mathrm{~Hz}$ Hence, frequency of source $=105 \mathrm{~Hz}$.
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
WAVES
172734
On producing the waves of frequency $1000 \mathrm{~Hz}$ in a Kundt's tube, the total distance between 6 successive nodes is $85 \mathrm{~cm}$. Then, the speed of sound in the gas filled in the tube is
1 $330 \mathrm{~ms}^{-1}$
2 $340 \mathrm{~ms}^{-1}$
3 $350 \mathrm{~ms}^{-1}$
4 $300 \mathrm{~ms}^{-1}$
Explanation:
B Given, Frequency of wave $(\mathrm{f})=1000 \mathrm{~Hz}$ Distance between six successive node $=\frac{5 \lambda}{2}=85$ $\lambda=\frac{2 \times 85}{5}=34 \mathrm{~cm}=0.34 \mathrm{~m}$ Then speed of sound in gas $=\mathrm{f} \lambda=1000 \times 0.34$ $=340 \mathrm{~m} / \mathrm{sec} \text {. }$
AP EAMCET (17.09.2020) Shift-I
WAVES
172737
Two identical wires are vibrating in unison. If the tension in one of the wire is increased by $2 \%$, five beats are produced per second by the two vibrating wires. The initial frequency of each wire is $(\sqrt{1.02}=\mathbf{1 . 0 1})$
172738
The frequency of two tuning forks $A$ and $B$ are $1.5 \%$ more and $2.5 \%$ less than that of the tuning fork $C$. When $A$ and $B$ are sounded together, 12 beats are produced in 1 second. The frequency of tuning fork $\mathbf{C}$ is
1 $300 \mathrm{~Hz}$
2 $240 \mathrm{~Hz}$
3 $360 \mathrm{~Hz}$
4 $200 \mathrm{~Hz}$
Explanation:
A Let, $f_{A}, f_{B}$ and $f_{C}$ are the frequencies of forks $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ respectively. Then, $\mathrm{f}_{\mathrm{A}}=\mathrm{f}_{\mathrm{C}} \times 1.015, \quad \mathrm{f}_{\mathrm{B}}=\mathrm{f}_{\mathrm{C}} \times 0.975$ According to question- $f_{A}-f_{B}=12$ $(1.015-0.975) f_{C}=12$ $0.040 f_{C}=12$ $\therefore \quad \mathrm{f}_{\mathrm{C}}=\frac{12}{0.040}=300 \mathrm{~Hz}$
MHT-CET 2020
WAVES
172740
A source of sound gives 5 beats per second, when sounded with another source of frequency $100 \mathrm{sec}^{-1}$. The second harmonic of the source, together with a source of frequency $205 \mathrm{sec}^{-1}$ gives 5 beats per second. What is frequency of the source?
1 $95 \mathrm{sec}^{-1}$
2 $105 \mathrm{sec}^{-1}$
3 $100 \mathrm{sec}^{-1}$
4 $205 \mathrm{sec}^{-1}$
Explanation:
B According to the question, Frequency of source $=100 \pm 5 \mathrm{~Hz}=105$ or $95 \mathrm{~Hz}$ Frequency of source in $2^{\text {nd }}$ harmonic $=2 \times$ frequency of source, Since, $=190 \text { or } 210 \mathrm{~Hz}$ $205 \mathrm{~Hz}$ frequency produces 5 beats/sec in second harmonic. Then, Actual frequency $=205 \pm 5 \mathrm{~Hz}$ $=200 \text { or } 210 \mathrm{~Hz}$ Hence, frequency of source $=105 \mathrm{~Hz}$.